This section covers:
 Factoring Methods
 Completing the Square (Square Root Method)
 Completing the Square to get Vertex Form
 Obtaining Quadratic Equations from a Graph or Points
 Quadratics Review
 More Practice
Note that factoring the sum and difference of cubes, and more advanced polynomial factoring and exponential factoring can be found in the Advanced Factoring section.
Again, there are several ways to solve quadratics, or find the solution (also known as the \(x\)intercepts, roots, zeros, or values) to a quadratic equation. (Remember that we solve quadratic equations most easily by getting everything to one side of the equal sign, which sets the quadratic to 0).
In the Quadratics section, we learned two ways to solve quadratics:
 Quadratic Formula: For \(0=a{{x}^{2}}+bx+c\), solutions are \(\displaystyle \frac{{b\pm \sqrt{{{{b}^{2}}4ac}}}}{{2a}}\).
 Graphing Quadratics: Looking for zeros by pinpointing where the quadratic graph (\(y=a{{x}^{2}}+bx+c\)) crosses the \(x\)axis.
In this section, we’ll talk about two other ways:
 Factoring and setting factors to 0 (there are many ways to factor, but not everything can be factored)
 Completing the square and taking the square root of each side (a way where we don’t have to set the quadratic to 0!)
Factoring Methods
Just the way we learned how to multiply binomials (via FOILING), we need to learn how to do the opposite, or factor (or “unfoil”) the resulting trinomials. After we’ve done enough multiplying binomials and factoring trinomials, this will become second nature.
We’ll also learn other basic polynomial factoring methods, like taking out the Greatest Common Factors (GCF) of polynomials, and factoring the difference of two squares and factoring perfect square trinomials.
Think of factoring as just “pulling apart” things that are multiplied together. It’s the same principle of factoring 35 and getting 5 and 7. When we factor in algebra, we do the same thing, but with variables.
NOTE: Remember that when we factor, we want to set each factor with a variable in it to 0, and solve for the variable to get the roots. This is because any factor that becomes 0 makes the whole expression 0. (This is the zero product property: if \(ab=0\), than \(a=0\) and/or \(b=0\)).
Also remember that when we factor to solve quadratics or any polynomials, we can never just divide by factors (with variables) on both sides to get rid of them. If we do this, we may be missing solutions!
Note that not every quadratic can be factored; if it can’t, we’ll have to use one of the other methods.
But let’s learn the factoring techniques so we can solve the quadratics that can be factored; it’s a pretty simple way to solve them. And later, when we learn how to solve more generic polynomials in the Graphing and Solving Polynomial Functions and Advanced Factoring sections, we’ll know how to factor!
Taking out the Greatest Common Factor (GCF)
Let’s first start with polynomials that need some simple factoring. We always need to pull out the GCF (largest coefficients/variables that go into all the terms) first. (We learned about the GCF with regular numbers in the Multiplying and Dividing Section here.)
Remember from the Exponents and Radicals in Algebra section that we add exponents when we multiply with the same base, and subtract exponents when we divide with the same base.
Here are some examples of factoring polynomials by taking the GCF out:
Polynomial  GCF  What’s Left  Factored Polynomial 
\(2{{x}^{3}}+4x\) 
\(2x\) 
\({{x}^{2}}+2\) 
\(2x\left( {{{x}^{2}}+2} \right)\) 
\(9{{x}^{4}}18{{x}^{2}}+12x+36\) 
\(3\) 
\(3{{x}^{4}}6{{x}^{2}}+4x+12\) 
\(3\left( {3{{x}^{4}}6{{x}^{2}}+4x+12} \right)\) 
\(4{{x}^{4}}{{y}^{3}}+64{{x}^{2}}y12{{x}^{3}}{{y}^{8}}\) 
\(4{{x}^{2}}y\) 
\({{x}^{2}}{{y}^{2}}+163x{{y}^{7}}\) 
\(4{{x}^{2}}y\left( {{{x}^{2}}{{y}^{2}}+163x{{y}^{7}}} \right)\) 
See how it’s reversing the distributive property? It’s “undistributing”. And it’s always a good idea to multiply back to check your answers!
Factoring Trinomials (Quadratics)
When we factor quadratics, we try to “unfoil” to get two binomials. Again, remember that when factoring trinomials, we always need to take out any GCF’s first!
And remember that this method does not always work; certain quadratics are “unfactorable” (prime, or irreducible), and must be solved with another method. Also remember that, with a trinomial, if the discriminant \({{b}^{2}}4ac\), where \(a\), \(b\), and \(c\) are from \(a{{x}^{2}}+bx+c\), isn’t a perfect square (like 25 or 49), you can’t factor it.
Here are some methods that we use:
Special Products of Binomials – Easily Factored
There are two special cases where you can use a shortcut to factor the trinomials: Difference of Two Squares (actually not a trinomial, but a binomial!), and Perfect Square Trinomials. We saw these in the Introduction to Polynomials section in the table here.
Here are the two cases and how to factor them:
Quadratic Factoring 
Notes 
\(\displaystyle \color{#800000}{{{{x}^{2}}{{y}^{2}}}}=\left( {xy} \right)\left( {x+y} \right)\,\)
Examples:
\(\displaystyle \begin{array}{c}\,\color{#800000}{{81{{x}^{2}}1}}=\left( {9a1} \right)\left( {9a+1} \right)\\\color{#800000}{{36{{x}^{2}}{{y}^{6}}9{{x}^{2}}{{y}^{2}}}}=\left( {6x{{y}^{3}}3xy} \right)\left( {6x{{y}^{3}}+3xy} \right)\\\,\color{#800000}{{4{{{\left( {x4} \right)}}^{2}}16}}=4\left[ {{{{\left( {x4} \right)}}^{2}}4} \right]\\=4\left[ {\left( {x4} \right)2} \right]\left[ {\left( {x4} \right)+2} \right]\\\,=4\left( {x6} \right)\left( {x2} \right)\end{array}\) 
Difference of Two Squares
This is a special case when you have the difference of two perfect square terms.
When you factor this, you don’t have to worry about the middle terms, since we learned earlier that they just cancel themselves out when multiplying.
You end up with two binomials: the square root of the first term minus the square root of the second term, multiplied by the square root of the first term plus the square root of the second term. Remember that when we take the square root of a term with an exponent, we divide the exponent by 2.
Multiply them back to see how they work! 
\(\displaystyle \begin{array}{l}\color{#800000}{{{{x}^{2}}+2xy+{{y}^{2}}}}=\left( {x+y} \right)\left( {x+y} \right)={{\left( {x+y} \right)}^{2}}\\\color{#800000}{{{{x}^{2}}2xy+{{y}^{2}}}}=\left( {xy} \right)\left( {xy} \right)={{\left( {xy} \right)}^{2}}\end{array}\)
Examples:
\(\displaystyle \begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\color{#800000}{{81{{x}^{2}}18x+1}}={{\left( {9x1} \right)}^{2}}\\\color{#800000}{{36{{x}^{2}}{{y}^{6}}+36{{x}^{2}}{{y}^{4}}+9{{x}^{2}}{{y}^{2}}}}={{\left( {6x{{y}^{3}}+3xy} \right)}^{2}}\end{array}\) See how the middle term is the twice the product of the square root of the first term (\(6x{{y}^{3}}\)) and the square root of the last term (\(3xy\)). 
Perfect Square Trinomial
This is a special case where the first and last terms of the trinomial are perfect squares and the middle term is twice the product of the square roots of those respective squares. Watch the signs: a minus before the second term in the trinomial means a minus in the binomial that is squared.
You end up with a perfect square binomial: the square root of the first term of the trinomial added to or subtracted from the square root of the last term of the trinomial.
Multiply them back to see how they work! 
Guess and Check
“Guess and Check” is just what it sounds; we have certain rules, but we try combinations to see what will work.
NOTE: Always take a quick look to see if the trinomial is a perfect square trinomial, but you try the guess and check. In these cases, the middle term will be twice the product of the respective square roots of the first and last terms, as we saw above. For example, \(\displaystyle {{x}^{2}}12xy+36{{y}^{2}}=\left( {x6y} \right)\left( {x6y} \right)={{\left( {x6y} \right)}^{2}}\).
Let’s start with an example; let’s “UNFOIL” or factor one of the first examples that we worked with in Quadratics. Here’s how we FOIL’ed, and how we “UNFOIL“:
FOIL  UnFOILing – coefficient of \(\boldsymbol{{{x}^{2}}}\) is 1 
Note that the middle term of the trinomial (the \(3x\)) is the sum of the last terms in the binomials (the “1” and “2”), and the last term (the 2) is the product of the last terms in the binomials (the “1” and “2”).
To “unfoil”, we’ll need to find 2 numbers that when multiplied together equals “2” and added together equals “3”.This works since we have nothing (no coefficient) before the (actually the coefficient is 1).
We can see that \(2\times 1=2\) and \(2+1=3\).
So 2 and 1 work! We can foil back to see that we got the right factors! 
If we were to solve this trinomial, we would set the factors equal to zero:
\(\displaystyle \begin{array}{c}x+2=0;\,\,\,\,\,\,\,x=2\\x+1=0;\,\,\,\,\,\,\,x=1\\\text{ }\left\{ {\,x:x=2,\,1\,} \right\}\end{array}\)
Note the signs when you “unfoil”; multiply these back to make sure they work:
Quadratic  Factoring  Example  Notes 
\(a{{x}^{2}}+bx+c\) 
\(\displaystyle \left( {x~+~\_\_} \right)\left( {x~+~\_\_} \right)\) 
\(\displaystyle \begin{array}{c}{{x}^{2}}+3x+2=\\\left( {x~\,+~\,\underline{2}} \right)\left( {x~\,+\,~\underline{1}} \right)\end{array}\)  When you have a “\(+\)” before the \(\displaystyle c\): If there’s a “\(+\)” before the \(\displaystyle bx\), use 2 “\(+\)”s. 
\(a{{x}^{2}}bx+c\)  \(\displaystyle \left( {x~~\_\_} \right)\left( {x~~\_\_} \right)\)  \(\displaystyle \begin{array}{c}{{x}^{2}}3x+2=\\\left( {x~\,~\,\underline{2}} \right)\left( {x~\,\,~\underline{1}} \right)\end{array}\)  When you have a “\(+\)” before the \(c\): If there’s a “\(\)” before the \(\displaystyle bx\), use 2 “\(\)”s. 
\(a{{x}^{2}}+bxc\)  \(\displaystyle \left( {x~+~\_\_} \right)\left( {x~~\_\_} \right)\)  \(\displaystyle \begin{array}{c}{{x}^{2}}+3x10=\\\left( {x~\,+~\,\underline{5}} \right)\left( {x~\,\,~\underline{2}} \right)\end{array}\)  When you have a “\(\)” before the \(c\): If there’s a “\(+\)” before the \(\displaystyle bx\), the “\(+\)” is before the bigger number (5); the other sign is a “\(\)”.
Note that \(\displaystyle +5x+2x=3x\). 
\(a{{x}^{2}}bxc\)  \(\displaystyle \left( {x~~\_\_} \right)\left( {x~+~\_\_} \right)\)  \(\displaystyle \begin{array}{c}{{x}^{2}}3x10=\\\left( {x~\,~\,\underline{5}} \right)\left( {x~\,+\,~\underline{2}} \right)\end{array}\)  When you have a “\(\)” before the \(c\): If there’s a “\(\)” before the \(\displaystyle bx\), the “\(\)” is before the bigger number (5); the other sign is a “\(+\)”.
Note that \(\displaystyle 5x+2x=3x\). 
Sometimes it’s easier and more visual if you use an “X” when you begin to factor these; we’ll first show this when there are no numbers (coefficients) before the \({{x}^{2}}\) term. We can put the coefficient of the constant term (“c”) on the top, middle term (“b”) on the bottom, and then use the two sides to find 2 numbers multiplied together to equal the bottom, but added together to equal the top.
(This is sometimes called the “A and M” or “AM” method, since we are looking for numbers that both add and multiply to certain numbers.)
Let’s try this for the last problem above: \({{x}^{2}}3x10=\)
Let’s try another one that’s a little bit more complicated.
Note again that to make the factoring easier, we always want to factor the GCF out of the polynomial before we “unfoil”.
Example:
Factor \(8{{x}^{2}}22x+14\). First we take out the GCF (2) to get \(2(4{{x}^{2}}11x+7)\).
We still need to “unfoil” \(4{{x}^{2}}11x+7\). This is a little more complicated, since we have the \(4{{x}^{2}}\). Now we need two factors of 4 and two factors of 7, so that multiplying the inside terms and multiplying the outside terms and adding them together will give us –11.
We can see that this trinomial is factorable, since \({{b}^{2}}4ac\) (the discriminant) is \({{11}^{2}}4\left( 4 \right)\left( 7 \right)=9\), which is a perfect square.
When we’re “guessing and checking”, we’re actually “foiling back” to see if we got the right answer! You may want to use your factor trees (that we learned in the Multiplying and Dividing section) to write down all the factors of the first and last coefficients.
Here are all the combinations we could have. Note that the factors of 4 are 2 and 2, or 4 and 1. The factors of 7 are just 7 and 1.
Quadratic 
Factors to Try 
Guess and Check 
\(4{{x}^{2}}11x+7=\) 
\(\displaystyle \left( {x~~\_\_} \right)\left( {x~~\_\_} \right)\) 
Outside + Inside Products: want to equal \(\boldsymbol{11x}\) 
\(\displaystyle \left( {1x~\,\,~\,\,\underline{7}} \right)\left( {4x~\,\,\,\,~\underline{1}} \right)\) 
\(1x+28x=29x\) NO! 

\(\displaystyle \left( {2x~\,\,~\,\,\underline{7}} \right)\left( {2x~\,\,\,\,~\underline{1}} \right)\) 
\(2x+14x=16x\) NO! 

\(\displaystyle \left( {2x~\,\,~\,\,\underline{1}} \right)\left( {2x~\,\,\,\,~\underline{7}} \right)\) 
\(14x+2x=16x\) NO! 

\(\displaystyle \left( {4x~\,\,~\,\,\underline{7}} \right)\left( {1x~\,\,\,\,~\underline{1}} \right)\) 
\(\displaystyle \,\,\,4x+7x=11x\) YES!!!! 
The complete factoring is: \(2(4x7)(x1)\). If we were to solve this trinomial, we would get \(\displaystyle \left\{ {x:\text{ }x=\frac{7}{4},\text{ }1} \right\}\) by setting each factor to 0 and solving for \(x\). (Ignore the factor of 2, since 2 can never be 0. If we had an \(x\) on the outside, an additional factor would be 0).
Note that if we had \(2(4{{x}^{2}}11xy+7{{y}^{2}})\), we’d also have to split the \(y\)’s at the end, so we would factor to \(2(4x7y)(xy)\). Multiply it all to together to show that it works! (This is one we wouldn’t be asked to solve).
Grouping Method, or the “ac” Method
The Grouping Method for factoring trinomials has gotten more popular since the FOILing hasn’t been taught as much recently. (The reason it’s not taught as much is because it can be only used with binomial multiplication. I still like FOILing since I believe learning to factor is easier if we learn FOILing first).
We will turn the trinomial into a quadratic with four terms, to be able to do the grouping. Then we have to find a pattern of binomials so we can use the distributive property to put them together (like a puzzle!).
Let’s look at the same problem as above (with the 2 already factored out, but let’s keep the \(y\)’s at the end):
\(4{{x}^{2}}11xy+7{{y}^{2}}\)
Let’s use our “X” again to help us solve by the grouping method.
We can put the middle term’s coefficient “b” (–11) on the bottom part of the “X”, but this time we multiply the first and last coefficients (the “a” and “c” of \(a{{x}^{2}}+bx+c\) and put it on the top part of an “X”: \(4\times 7=28\). Remember that the sign of a term comes before it, and pay attention to signs.
Then we do the “guessing and checking” as follows:
So since we found that –4 and –7 “work”, we can rewrite the trinomial and separate the middle term, so we can do some grouping and factor with the grouping method:
Quadratic Factoring 
Notes 
\(\begin{array}{c}\,\color{#800000}{{\,4{{x}^{2}}11xy+7{{y}^{2}}}}\\=4{{x}^{2}}4xy7xy+7{{y}^{2}}\end{array}\) 
Separate the middle terms with the coefficients that we found: two numbers multiplied together to get “ac”, which is \(–28\), but added together to get “b”, which is \(–11\). It doesn’t matter which of the separated middle term you put first. 
\(\displaystyle \begin{array}{c}\left( {4{{x}^{2}}4xy} \right)\left( {7xy+7{{y}^{2}}} \right)\\=\,4x\left( {xy} \right)7y\left( {xy} \right)\end{array}\) 
We need to separate the quadratic into 2 groups of 2 terms, so we can take out the GCF in each set of terms. It doesn’t really matter which 2 you put together. Watch signs when taking out factors; remember that two “\(\)”s multiplied together is a “\(+\)”.
Note: In problems that work like these, you’ll see that what’s left after the GCF is taken out are the same! 
\(\displaystyle \begin{array}{c}4x\left( {xy} \right)7y\left( {xy} \right)\\=\,\left( {4x7y} \right)\left( {xy} \right)\end{array}\) 
Now, remember that \(acbc=\left( {ab} \right)\left( c \right)\) (Distributive Property). Think of the \((x–y)\) as the \(c\); you only include it once. 
\(\displaystyle \begin{array}{l}\,\color{#800000}{{4{{x}^{2}}11xy+7{{y}^{2}}}}\\=\,\left( {4x7y} \right)\left( {xy} \right)\end{array}\) 
We did it! Make sure to FOIL or distribute back to make sure we did it correctly. 
Grouping Method with Four Terms
Here are some examples of the grouping method when we start out with 4 terms. Notice that the first one is a 4term quadratic and the second is a cubic polynomial that includes factoring with the difference of squares.
Remember the difference of squares factoring is \(\displaystyle {{x}^{2}}{{y}^{2}}=\left( {xy} \right)\left( {x+y} \right)\).
Quadratic Factoring  Notes 
\(\displaystyle \begin{array}{l}\color{#800000}{{15{{x}^{2}}3xz+5yxyz}}\\\,\,\,\,=3x\left( {5xz} \right)+y\left( {5xz} \right)\\\,\,\,\,=\left( {3x+y} \right)\left( {5xz} \right)\end{array}\)  Since we have 4 terms, we don’t have to separate the middle terms; they are already separated for us!
Divide into 2 groups of 2 terms (look for a pattern – see the 15/–3 and 5/–1 coefficients!), and take out the GCFs.
Use the (inverse of) Distributive Property to finish the factoring. 
\(\begin{array}{l}\color{#800000}{{3{{x}^{3}}27x2{{x}^{2}}+18}}\\\,\,\,\,=3x\left( {{{x}^{2}}9} \right)2\left( {{{x}^{2}}9} \right)\\\,\,\,\,=\left( {3x2} \right)\left( {{{x}^{2}}9} \right)\\\,\,\,\,=\left( {3x2} \right)\left( {x+3} \right)\left( {x3} \right)\end{array}\)  We can perform this type of factoring (if it works), even if we have a polynomial with a degree higher than 2.
Note that we had to use the difference of squares to factor further after using the grouping method. 
Note that having four terms to factor doesn’t always work with the grouping method above; sometimes we have to look for differences of squares (sometimes combined with the grouping method):
Quadratic Factoring  Notes 
\(\begin{array}{l}\color{#800000}{{9{{x}^{2}}24xy+16{{y}^{2}}25{{z}^{2}}}}\\\,\,\,\,={{\left( {3x4y} \right)}^{2}}25{{z}^{2}}\\\,\,\,\,=\left( {\left( {3x4y} \right)5z} \right)\left( {\left( {3x4y} \right)+5z} \right)\\\,\,\,\,=\left( {3x4y5z} \right)\left( {3x4y+5z} \right)\end{array}\)  In this problem, we can’t really see a pattern, but we do notice that we have a lot of squares in the terms. This means we’ll probably use difference of squares to factor.
Note that the first three terms is a perfect square, and so is the last term. We can use difference of squares to factor.
Use the (inverse of) Distributive Property to finish the factoring. 
\(\displaystyle \begin{array}{l}\color{#800000}{{{{t}^{2}}{{s}^{2}}rt+rs}}\\\,\,\,\,\,=\left( {t+s} \right)\left( {ts} \right)r\left( {ts} \right)\\\,\,\,\,\,=\left( {t+sr} \right)\left( {ts} \right)\end{array}\)  This is another example where we can’t find a pattern to separate 2 groups of terms, but we notice that we have a difference of squares in the first 2 terms, and we can take out an \(r\) from the last 2 terms. Watch signs!
Then it just turns out that we can factor using the (inverse of) Distributive Property! 
The Box Method
One more method that is getting popular is called the “Box” method. This is a modified “ac” method, but we use greatest common factors (GCF) to help us factor.
This is sort of the opposite of what we did with the “multiplication box” earlier.
We put the first term in the upper left hand corner, the last term in the lower right hand corner, and we divide up the middle term in the remaining two corners so they add up to the middle term, and are factors of the first term times the last term (thus, the “ac” method). You can put the middle terms (upper right and lower left corners) in any order, but make sure the signs are correct so they add up to the middle term.
If you have set up the box correctly, the diagonals should multiply to the same product.
Then we get the GCFs across the columns and down the rows, using the same sign of the closest box (boxes either on the left or the top).
Let’s try this for \(10{{x}^{2}}11x6\):
Then read across and down to get the factors: \((5x+2)(2x–3)\). Foil it back, and we see that we got it correct! If we were to solve this trinomial, we would get \(\displaystyle \left\{ {\,x:x=\frac{2}{5},\,\,\,\frac{3}{2}\,} \right\}\) by setting each factor to 0 and solving for \(x\).
Note: When we do the box method, we have to make sure the \({{x}^{2}}\) has a positive coefficient; otherwise, we have to take the negative out across all three terms, do the factoring, and then put the negative in the front of the factored answer.
The “Star” Method
Another method for factoring trinomials is the “Star” Method, which is kind of a combination of the GuessandCheck “X” method and the “ac” method. Let’s use \(10{{x}^{2}}11x6\), setting up the “star” as follows:
The “ac” Method without Grouping
There’s another new method to factor trinomials out there; it’s similar to the “star” method, but without drawing the “star”. Again, let’s use \(10{{x}^{2}}11x6\):
Quadratic Factoring  Notes 
\(\displaystyle \begin{array}{c}\color{#800000}{{10{{x}^{2}}11x6}}\\\\\,\,\color{#800000}{{{{x}^{2}}11x60}}\\\,\left( {x15} \right)\left( {x+4} \right)\end{array}\)  Rewrite the trinomial without the 10 before the \({{x}^{2}}\), but put at the end. This is the coefficient of the first term (10) multiplied by the coefficient of the last term (– 6).
Then factor like you normally would: find two numbers that add together to get 11 and multiply to get 60. 
\(\displaystyle \begin{align}\left( {x\frac{{15}}{{10}}} \right)&\left( {x+\frac{4}{{10}}} \right)\\\,\,\,\left( {x\frac{3}{2}} \right)&\left( {x+\frac{2}{5}} \right)\end{align}\)  We’re not done yet; since we removed the 10 from the first term, we have to put it back!
The way we do it is to put the 10 (the “\(a\)”, or what is in front of the ) under the 15 and 4, and we need to reduce the fractions if we can. 
\(\begin{align}\left( {x\frac{3}{2}} \right)&\left( {x+\frac{2}{5}} \right)\\\left( {2x3} \right)&\left( {5x+2} \right)\end{align}\)  Now, you can either solve at that point by setting each factor to 0, or to get in factored form, move the denominators over back to the front of the \(x\)’s. Weird, but it works! 
Completing the Square (Square Root Method)
Completing the square is what is says: we take a quadratic in standard form \((y=a{{x}^{2}}+bx+c)\) and manipulate it to have a binomial square in it, like \(y=a{{\left( {x+b} \right)}^{2}}+c\). This way we can solve it by isolating the binomial square (getting it on one side) and taking the square root of each side. This is commonly called the square root method.
We can also complete the square to find the vertex more easily, since the vertex form is \(y=a{{\left( {xh} \right)}^{2}}+k\).
What we want to do for the square root method is to make a square out of the side with the variable, and move the numbers (constants) to the other side, so we can take the square root of both sides. Then we don’t have to use the quadratic equation, or “unfoil” to solve.
Let’s first think about what happens when we square a binomial by looking at:
\(\displaystyle \begin{array}{c}{{\left( {x+3} \right)}^{2}}=\left( {x+3} \right)\left( {x+3} \right)={{x}^{2}}+3x+3x+9\\={{x}^{2}}+6x+9\end{array}\)
(We first saw these perfect square trinomials in Introduction To Polynomials.)
Since we add the product of the middle terms twice, we have twice the product of the first (\(x\)) and second (3) terms in the middle (to get \(6x\)). See also how we have the square of the second term (3) at the end (9).
To complete the square of a trinomial that isn’t a perfect square, we need to halve the second term and take the square of it – and then add that number so the square can be complete. Then we have to make sure to add the same thing to the other side.
Then we take the square root of each side, remember that we need to include the plus and minus of the right hand side, since by definition, the square root is just the positive. Another way to think of it is the absolute value of the left side equals the right side, so we have to include the plus and minus of the right side.
(Note: you may want to review Solving Exponential and Radical Equations to review how to solve square root equations.)
Let’s work with \({{x}^{2}}+8x+15=0\) first, since it just starts with an \({{x}^{2}}\) (coefficient of \({{x}^{2}}\) is 1):
Completing the Square Problem 
Notes 
\(\displaystyle \begin{array}{c}\color{#800000}{{{{x}^{2}}+8x+15=0}}\\\\{{x}^{2}}+8x=15\end{array}\)  Since the coefficient of the \({{x}^{2}}\) is 1, we can proceed to move the constant to right side, so we only have \(x\)’s on the left side.
We are ready to complete the square! 
\(\require{cancel} \begin{align}\left( {{{x}^{2}}+8x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=15+\,\,\underline{{\,\,\,\,\,\,}}\\\left( {{{x}^{2}}+8x+\color{#2E8B57}{{\underline{{{{{\left( {\frac{8}{2}} \right)}}^{2}}}}}}\,} \right)&=15+\color{#2E8B57}{{\underline{{{{{\left( {\frac{8}{2}} \right)}}^{2}}}}}}\\{{\left( {x+\underline{4}} \right)}^{2}}&=15+\underline{{{{4}^{2}}}}=15+16=1\end{align}\)  Now we need to divide coefficient of the “\(x\)” term (8) by two (to get 4) to start building the square. We square this number (to get 16) and add it to both sides.
(As an example of why we can do this: \(2=2\), so \(\displaystyle 2+\underline{4}=2+\underline{4}\).)
Remember that the number inside the square (4) is the same number as the middle term (8) of the original divided by 2. 
\(\begin{array}{l}\,\,\,\,\,{{\left( {x+4} \right)}^{2}}=1\\\sqrt{{{{{\left( {x+4} \right)}}^{2}}}}=\pm \sqrt{1}\end{array}\)  Now we just want to take the square root of each side.
Remember when we take the square root of the right side, we have to include the plus and the minus, since, by definition, the square root of something is just the positive number. 
\(\begin{align}\sqrt{{{{{\left( {x+4} \right)}}^{2}}}}&=+\,\sqrt{1}\\x+4&=1\\x&=3\end{align}\) \(\begin{align}\sqrt{{{{{\left( {x+4} \right)}}^{2}}}}&=\,\sqrt{1}\\x+4&=1\\x&=5\end{align}\) 
At this point, I like to divide up the equation into two equations: one with the plus, and one with the minus.
When we solve for \(x\), we get –3 and 5. 
Note that it would have been much easier to factor this quadratic, but, like the quadratic equation, we can use the completing the square method for any quadratic. (We may not always get a real number for the answer(s); we’ll talk about imaginary numbers later).
Here’s one that’s a little trickier, because of the radical in the coefficient of \(x\):
Completing the Square Problem  Notes 
\({{x}^{2}}4x\sqrt{5}+12=0\)
\({{x}^{2}}4\sqrt{5}x=12\) 
We need to put the 4 and \(\sqrt{5}\) together and then \(4\sqrt{5}\) is the coefficient of \(x\).
We move the constant to the right side, and we are ready to complete the square! 
\(\begin{align}\left( {{{x}^{2}}4\sqrt{5}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=12+\,\,\underline{{\,\,\,\,\,\,}}\\\left( {{{x}^{2}}4\sqrt{5}x+\color{#03A89E}{{\underline{{{{{\left( {2\sqrt{5}} \right)}}^{2}}}}}}\,} \right)&=12+\color{#03A89E}{{\underline{{{{{\left( {2\sqrt{5}} \right)}}^{2}}}}}}\\{{\left( {x\underline{{2\sqrt{5}}}} \right)}^{2}}&=12+\underline{{{{{\left( {2\sqrt{5}} \right)}}^{2}}}}=12+{{\left( 2 \right)}^{2}}{{\left( {\sqrt{5}} \right)}^{2}}\\{{\left( {x2\sqrt{5}} \right)}^{2}}&=12+\left( 4 \right)\left( 5 \right)\\{{\left( {x2\sqrt{5}} \right)}^{2}}&=8\end{align}\)  When we divide \(4\sqrt{5}\) by 2, we get \(2\sqrt{5}\) (think of the \(\sqrt{5}\) sort of like a variable).
When we square \(2\sqrt{5}\), we have to “push through” the exponent to get \({{\left( 2 \right)}^{2}}{{\left( {\sqrt{5}} \right)}^{2}}\), which is 20. 
\(\displaystyle \begin{align}{{\left( {x2\sqrt{5}} \right)}^{2}}&=8\\\sqrt{{{{{\left( {x2\sqrt{5}} \right)}}^{2}}}}&=\pm \sqrt{8}\\x2\sqrt{5}&=\pm \sqrt{8}=\pm 2\sqrt{2}\\x&=\pm 2\sqrt{2}+2\sqrt{5}\end{align}\)  Now we just want to take the square root of each side. The left side is just \(x2\sqrt{5}\), and it’s good to simplify \(\sqrt{8}\) to \(2\sqrt{2}\).
We then add \(2\sqrt{5}\) to both sides to get \(x\) by itself. The answers are sort of complicated since we can’t combine the \(2\sqrt{2}\) and \(2\sqrt{5}\) terms, but the answers are “exact”.

Here’s one where the coefficient of the \({{x}^{2}}\) isn’t 1. (Remember again that if we can take out any factors across the whole trinomial, do it first and complete the square with the trinomial only.) This gets a little more complicated, but it’s not too bad:
Completing the Square Problem  Notes 
\(\displaystyle 2{{x}^{2}}5x12=0\)
\(\displaystyle \begin{align}2{{x}^{2}}5x&=12\\2\left( {{{x}^{2}}\frac{5}{2}x} \right)&=12\\{{x}^{2}}\frac{5}{2}x&=6\end{align}\) 
We first want to move the constant to right side, so we only have \(x\)’s on the left side.
To complete the square, we must divide through by whatever’s in front (the coefficient) of the \({{x}^{2}}\). Notice that we have to divide the second term by 2, also, to get \(\displaystyle \frac{5}{2}\). At this point, you can divide both sides by the 2. 
\(\displaystyle \begin{align}{{x}^{2}}\frac{5}{2}x+\underline{{\,\,\,\,\,\,}}&=6+\underline{{\,\,\,\,\,\,}}\\{{x}^{2}}\frac{5}{2}x+\color{#03A89E}{{\underline{{{{{\left( {\frac{5}{4}} \right)}}^{2}}}}}}&=6+\underline{{\color{#03A89E}{{{{{\left( {\frac{5}{4}} \right)}}^{2}}}}}}\\{{\left( {x\underline{{\frac{5}{4}}}} \right)}^{2}}&=6+\underline{{{{{\left( {\frac{5}{4}} \right)}}^{2}}}}=6+\underline{{\frac{{25}}{{16}}}}=\frac{{96}}{{16}}+\underline{{\frac{{25}}{{16}}}}=\frac{{121}}{{16}}\end{align}\)  Now complete the square the same way we did above. Halve the middle term to get \(\displaystyle \frac{5}{4}\) (this number will go in the binomial that will be squared), and square it to complete the square. Thus, \(\displaystyle {{\left( {\frac{5}{4}} \right)}^{2}}\) is added to both sides of the equation. Simplify the righthand side. 
\(\begin{align}{{\left( {x\frac{5}{4}} \right)}^{2}}&=\frac{{121}}{{16}}\\\sqrt{{{{{\left( {x\frac{5}{4}} \right)}}^{2}}}}&=\pm \sqrt{{\frac{{121}}{{16}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,x\frac{5}{4}&=\pm \frac{{11}}{4}\end{align}\)  Again, we just want to take the square root of each side.
Remember when we take the square root of the right side, we have to include the plus and the minus, since, by definition, the square root of something is just the positive number. 
\(\begin{align}x\frac{5}{4}&=\frac{{11}}{4}\\x&=\frac{{16}}{4}=4\end{align}\) \(\begin{align}x\frac{5}{4}&=\frac{{11}}{4}\\x&=\frac{6}{4}=\frac{3}{2}\end{align}\)  Divide up the equation into two equations: one with the plus and one with the minus.
When we solve for \(x\), we get 4 and \(\displaystyle \frac{3}{2}\). 
Note: There is another method for Completing the Square from Standard Form to solve. For standard equation \(\displaystyle a{{x}^{2}}+bxc=0\), first subtract \(c\) from both sides to get in the form \(\displaystyle a{{x}^{2}}+bx=c\). If \(a\) is a perfect square, add \(\displaystyle \frac{{{{b}^{2}}}}{{4a}}\) to both sides and then factor and simplify. If \(a\) is not a perfect square, first multiply both sides by anything that makes \(a\) a perfect square, add \(\displaystyle \frac{{{{b}^{2}}}}{{4a}}\) to both sides and then factor and simplify. Try this for the example above: \(a=2\), \(b=5\), and \(c=12\):
\(\displaystyle \begin{array}{c}2{{x}^{2}}5x12=0;\,\,\,\,\,\,\,\,2{{x}^{2}}5x=12\,\,\\\,(\text{multipy both sides by }2\text{ to make a perfect square)}\\4{{x}^{2}}10x=24\\(\text{add }\frac{{{{b}^{2}}}}{{4a}}=\frac{{{{{\left( {10} \right)}}^{2}}}}{{4\left( 4 \right)}}\text{=}\frac{{100}}{{16}}\text{ to both sides)}\\4{{x}^{2}}10x+\frac{{100}}{{16}}=24+\frac{{100}}{{16}}\\{{\left( {2x\frac{{10}}{4}} \right)}^{2}}=\frac{{121}}{4}\\2x\frac{{10}}{4}=\pm \frac{{11}}{2};\,\,\,\,\,\,\,x=4,\,\frac{3}{2}\,\,\end{array}\)
I’m not a fan of these “memorization methods” since I’d rather understand what I’m doing! But if it’s easier for you, go for it!
Note: Even though complex solutions are easier to obtain using the Quadratic Formula, here are examples on how we can get imaginary solutions by Completing the Square:
Completing the Square with Complex (Imaginary) Solutions  
\(\displaystyle {{x}^{2}}2x+2=0\)
\(\displaystyle \begin{align}{{x}^{2}}2x&=2\\\,\,\,\,\,\,{{x}^{2}}2x+\,\,\underline{{\,\,\,\,\,\,}}&=2+\,\,\underline{{\,\,\,\,\,\,}}\\{{x}^{2}}2x+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}&=2+\underline{{\color{#03A89E}{{{{{\left( {\frac{2}{2}} \right)}}^{2}}}}}}\\\,\,\,{{x}^{2}}2x+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}&=2+\underline{{\color{#03A89E}{{{{{\left( 1 \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x1} \right)}^{2}}&=1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x1} \right)}}^{2}}}}&=\sqrt{{1}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x1&=\pm \sqrt{{1}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\pm \,i+1=1\pm \,\,i\end{align}\) 
\(\displaystyle 3{{x}^{2}}2x+2=0\)
\(\require{cancel} \displaystyle \begin{align}3\left( {{{x}^{2}}\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=2+\,\,\underline{{\,\,\,\,\,\,}}\\\frac{{\cancel{3}}}{{\cancel{3}}}\left( {{{x}^{2}}\frac{2}{3}x+\,\,\underline{{\,\,\,\,\,\,}}\,} \right)&=\frac{{2}}{3}+\,\,\underline{{\,\,\,\,\,\,}}\\\,\,\,\,{{x}^{2}}\frac{2}{3}x+\underline{{\color{#03A89E}{{{{{\left( {\frac{{\cancel{{{}^{1}2}}}}{{\cancel{{{}^{3}6}}}}} \right)}}^{2}}}}}}&=\frac{{2}}{3}+\underline{{\color{#03A89E}{{{{{\left( {\frac{1}{3}} \right)}}^{2}}}}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{\left( {x\frac{1}{3}} \right)}^{2}}&=\frac{{{}^{{6}}\cancel{{2}}}}{{\cancel{{{}^{9}3}}}}+\frac{1}{9}=\frac{{5}}{9}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sqrt{{{{{\left( {x\frac{1}{3}} \right)}}^{2}}}}&=\pm \sqrt{{\frac{5}{9}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\frac{1}{3}&=\pm \frac{{\sqrt{5}}}{3}\,\,i\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x&=\,\frac{1}{3}\,\,\pm \frac{{\sqrt{5}}}{3}\,\,i\end{align}\) 
Completing the Square to get Vertex Form
We can use the same technique to put \(y=2{{x}^{2}}5x12\) (standard form, or \(y=a{{x}^{2}}+bx+c\)) into vertex form (\(y=a{{(xh)}^{2}}+k\)). In this case, we want to leave everything on one side, so we have to undo what we’ve added or subtracted by completing the square:
Completing the Square Problem  Notes 
\(\displaystyle y=2{{x}^{2}}5x12\,\)
\(\displaystyle \begin{align}y&=2\left( {{{x}^{2}}\frac{5}{2}x+\underline{{\,\,\,\,\,\,\,\,}}\,} \right)12\,\,\underline{{\,\,\,\,\,\,\,\,}}\\y&=2{{\left( {x\frac{5}{4}} \right)}^{2}}12\underline{{\,\,\,\,\,\,\,\,}}\end{align}\) 
We must first divide through by the coefficient of the \({{x}^{2}}\), which is 2. We keep the –12 on the outside (don’t divide the –12 by 2).
(Some methods like you to move the –12 to the other side, and then move it back after completing the square, but I like to just keep it there.)
To complete the square, divide \(\displaystyle \frac{5}{2}\) by 2 to get \(\displaystyle \frac{5}{4}\), and then we’ll square it to get \(\displaystyle \frac{5}{4}\). We have to be careful though, since whatever we add, we must subtract. 
\(\require {cancel} \displaystyle \begin{align}y&=\color{#03A89E}{2}\left( {{{x}^{2}}\frac{5}{2}x+\color{#03A89E}{{\underline{{\frac{{25}}{{16}}}}}}\,} \right)12\underline{{\color{#03A89E}{{(2)\frac{{25}}{{16}}}}}}\\y&=2{{\left( {x\frac{5}{4}} \right)}^{2}}12\underline{{\,({}^{1}\cancel{2})\frac{{25}}{{\cancel{{{{{16}}_{8}}}}}}}}\\y&=2{{\left( {x\frac{5}{4}} \right)}^{2}}12\underline{{\frac{{25}}{8}}}\end{align}\)  Note that since we added an extra 2 times \(\displaystyle \frac{{25}}{{16}}\) at the beginning, we have to subtract 2 times \(\displaystyle \frac{{25}}{{16}}\) at the end.
The reason we’re subtracting it is because all of this is on one side, so what we add, we must subtract. I like to simplify the fraction before combining it.
We then need to add the –12 to the fraction. 
\(\displaystyle \begin{align}y&=2{{\left( {x\frac{5}{4}} \right)}^{2}}\frac{{96}}{8}\underline{{\frac{{25}}{8}}}\\y&=2{{\left( {x\frac{5}{4}} \right)}^{2}}\frac{{121}}{8}\end{align}\)  We find a common denominator for the last two numbers (8), put them together, and we have the quadratic in vertex form!
Then, since \((h,k)\) is the vertex in the equation \(y=a{{(xh)}^{2}}+k\), we know that \(\displaystyle \left( {\frac{5}{4},\frac{{121}}{8}} \right)\) is the vertex.

Note: There is another way to convert from Standard Form to Vertex Form. For standard equation \(\displaystyle y=a{{x}^{2}}+bx+c\), we can use a General Vertex Form equation \(\displaystyle y=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}\frac{{{{b}^{2}}4ac}}{{4a}}\) to convert to vertex form. Let’s try this for the example above: \(a=2\), \(b=5\), and \(c=12\): \(\displaystyle y=a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}\frac{{{{b}^{2}}4a}}{{4a}}=2{{\left( {x+\frac{{5}}{{2\left( 2 \right)}}} \right)}^{2}}\frac{{{{{\left( {5} \right)}}^{2}}4\left( 2 \right)\left( {12} \right)}}{{4\left( 2 \right)}}=2{{\left( {x\frac{5}{4}} \right)}^{2}}\frac{{121}}{8}\,\). Again, I’m not a fan of these “memorization methods” since I’d rather understand what I’m doing!
Obtaining Quadratic Equations from a Graph or Points
Sometimes you will be asked to look at a quadratic graph (or given the vertex and a point) and write the equation (in all three forms) for that graph. The easiest way to do this is to find the vertex from the graph (usually it’s obvious!), put the equation in vertex form, and then compute the “\(a\)” (coefficient of \({{x}^{2}}\)) from another point on the graph, such as a root or \(y\)intercept, if given.
If you’re given the \(x\)intercepts (roots), you can also put it in factored form, and use another point (not one of the roots) to find the “\(a\)” part of the equation. Remember that the “\(a\)” in all three forms (standard, factored, and vertex) will be the same.
Here are some examples:
Quadratics Graphs Given 
Finding Equations 
Find an equation for this graph:

We know from the graph that the vertex is \((1,8)\), the roots are \((3,0)\) and \((1,0)\), and the \(y\)intercept is \((0,6)\). We know that “\(a\)” is positive, since the parabola faces upwards. We also know that the “\(c\)” (constant) is –6, because of the \(y\)intercept.
So far, the vertex form is \(y=a{{\left( {x+1} \right)}^{2}}8\) and the factored form is \(y=a\left( {x+3} \right)\left( {x1} \right)\).
Now we just have to get the “\(a\)”. We can use either equation above, but if we use the vertex form, we can’t plug in the vertex, and if we use the factored form, we can’t plug in a root. Let’s use the factored form, and plug in the vertex for \((x,y)\): \(\begin{align}y&=a\left( {x+3} \right)\left( {x1} \right)\\8&=a\left( {1+3} \right)\left( {11} \right)\\8&=a\left( {4} \right)\\a&=2\end{align}\)
The factored form is \(y=2\left( {x+3} \right)\left( {x1} \right)\), the vertex form is \(y=2{{\left( {x+1} \right)}^{2}}8\), and the standard form is \(y=2{{x}^{2}}+4x6\). You can check all three forms by putting them all in a graphing calculator (like in \(\displaystyle {{Y}_{1}},\,\,\,{{Y}_{2}},\,\,\,{{Y}_{3}}\)) and making sure you get the same graph. 
Find an equation for this graph:

We know from the graph that the vertex is \((–1,7.875)\), and the roots are \((–2.5,0)\) and \((.5,0)\). We know that “\(a\)” is negative, since the parabola faces downwards.
So far, the vertex form is \(y=a{{\left( {x+1} \right)}^{2}}+7.875\) and the factored form is \(y=a\left( {x+2.5} \right)\left( {x.5} \right)\).
For this one, let’s start with the vertex form, and plug in the \((.5,0)\) root: \(\begin{align}y&=a{{\left( {x+1} \right)}^{2}}+7.875\\0&=a{{\left( {.5+1} \right)}^{2}}+7.875\\7.875&=2.25a\\a&=3.5\end{align}\)
The factored form is \(y=3.5\left( {x+2.5} \right)\left( {x.5} \right)\), the vertex form is \(y=3.5{{\left( {x+1} \right)}^{2}}+7.875\), and the standard form is \(y=3.5{{x}^{2}}7x+4.375\). 
Here’s another type of problem you might see where you have to write a Quadratic Function given a Parabola’s Axis of Symmetry and two NonVertex Points. Note that we have to use a System of Equations:
Quadratics Problem and Graph 
FInding Equation 
Write the equation of the quadratic in vertex form that has \(x=3\) as the line of symmetry (LOS), and contains the (nonvertex) points \(\left( {1,7} \right)\) and \(\left( {0,16} \right)\).
Let’s first attempt to draw a graph (from the LOS and the points, it looks like it will open downward):

We know that since the line of symmetry (LOS) is \(x=2\), the vertex is on that line, so it has a –2 for the \(x\). Thus, the equation of the quadratic will be in the form:
\(y=a{{\left( {x+2} \right)}^{2}}+k\)
Let’s use the two points to create a system of equations: \(\begin{align}y&=a{{\left( {x+2} \right)}^{2}}+k\\7&=a{{\left( {1+2} \right)}^{2}}+k\\7&=a+k\end{align}\) \(\begin{align}y&=a{{\left( {x+2} \right)}^{2}}+k\\16&=a{{\left( {0+2} \right)}^{2}}+k\\16&=4a+k\end{align}\)
The system is \(\left\{ \begin{array}{l}a+k=7\\4a+k=16\end{array} \right.\). Solve the first equation for \(k\) $(latex k=7a)$, and then substitute in the second: \(4a+\left( {7a} \right)=16;\,\,\,\,\,a=3\). Then \(k=7a=7\left( {3} \right)=4\).
The quadratic is: \(y=3{{\left( {x+2} \right)}^{2}}4\,\,\,\,\,\,\,\,\,\,\,\surd \). 
Here’s one more where we are given three points and need to find the equation of the parabola:
Problem:
Find the quadratic (standard form) that passes through the points \(\left( {1,12} \right)\), \(\left( {1,4} \right)\) and \(\left( {3,36} \right)\).
Solution:
We know the quadratic is in the form \(y=a{{x}^{2}}+bx+c\), so we need to find these coefficients. Let’s just set up a system of equations with the \(\left( {x,y} \right)\) points:
\(\begin{align}12&=a{{\left( {1} \right)}^{2}}+b\left( {1} \right)+c\\4&=a{{\left( 1 \right)}^{2}}+b\left( 1 \right)+c\\36&=a{{\left( 3 \right)}^{2}}+b\left( 3 \right)+c\end{align}\) or \(\begin{align}ab+c&=12\\a+b+c&=4\\9a+3b+c&=36\end{align}\)
Solve using elimination; we could also solve using substitution or matrices.
\(\displaystyle \begin{array}{l}\,\,ab+c=12\\\,\,\underline{{a+b+c=4}}\\2a\,\,\,\,+\,\,2c=16\end{array}\) \(\displaystyle \begin{array}{l}3\left( {ab+c=12} \right)\\\\3a3b+3c=36\\\underline{{9a+3b+\,\,c=36}}\\12a\,\,\,\,\,\,\,\,\,+4c=72\end{array}\) \(\displaystyle \begin{array}{l}2\left( {2a+2c=16} \right)\\\\\,\,4a4c=32\\\,\,\,\,\underline{{12a+4c=72}}\\\,\,\,\,\,\,8a\,\,\,\,\,\,\,\,\,\,\,\,=40\\\,\,\,\,\,\,\,\,\,\,\,\,\,a=5\end{array}\) \(\begin{array}{c}a=5\\12\left( 5 \right)+4c=72\\4c=12\\c=3\end{array}\) \(\begin{array}{c}ab+c=12\\5b+3=12\\b=4\\b=4\end{array}\)
Since \(a=5,\,\,b=4\), and \(c=3\), the quadratic is \(y=5{{x}^{2}}4x+3\).
Quadratics Review
Let’s review the different forms of Quadratics and also the methods for finding roots of Quadratics.
Note that the “\(\boldsymbol{a}\)” (coefficient of the \(\boldsymbol{{{{x}^{2}}}}\)) is the same for all three forms!
Quadratics 
Vertex Form 
Standard Form 
Factored Form (Intercept Form) 
Equation 
\(\begin{array}{l}y=a{{(xh)}^{2}}+k\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{or}\\yk=a{{(xh)}^{2}}\end{array}\)
Multiply out to get from Vertex Form to Standard Form. 
\(y=a{{x}^{2}}+bx+c\)
Complete the square to get from Standard Form to Vertex Form. 
\(y=a(x{{r}_{1}})(x{{r}_{2}})\)
Multiply out to get from Factored Form to Standard Form. 
Vertex 
\(\left( {h,\,k} \right)\) 
\(\displaystyle \left( {\frac{b}{{2a}},f\left( {\frac{b}{{2a}}} \right)} \right)\) (\(\displaystyle \frac{b}{{2a}}\) is the \(x\), plug \(\displaystyle \frac{b}{{2a}}\) into the \(x\) to get the \(y\))
(Can also use Graphing Calculator.) 
Convert to Standard Form and use
\(\displaystyle \left( {\frac{b}{{2a}},f\left( {\frac{b}{{2a}}} \right)} \right)\)
(Can also use Graphing Calculator.) 
Roots or \(\boldsymbol{x}\)–intercepts (when \(y=0\)) 
Convert to Standard Form or Factored Form to get roots.  See Method of Finding Roots table below. 
\(({{r}_{1}},0),\,\,\,\,({{r}_{2}},0)\) 
\(\boldsymbol{y}\)–intercepts (when \(x=0\)) 
Evaluate for \(x=0\), or convert to Standard Form to get \(y\)intercept.  \(\left( {0,\,c} \right)\)  Evaluate for \(x=0\), or convert to Standard Form to get \(y\)intercept.

Note: Sometimes we’ll have a Quadratic in “almost” vertex form and play around with it to get it in vertex form. For example, if we have \(\displaystyle y=8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3\) and want to change it to vertex form, we can take out the \(\displaystyle \frac{1}{2}\) coefficient of \(x\) and do some algebra: \(\displaystyle y=8{{\left( {\frac{1}{2}x+2} \right)}^{2}}+3=8{{\left( {\left( {\frac{1}{2}} \right)\left( {x+4} \right)} \right)}^{2}}+3=8{{\left( {\frac{1}{2}} \right)}^{2}}{{\left( {x+4} \right)}^{2}}+3=2{{\left( {x+4} \right)}^{2}}+3\), so the vertex is \((–4, 3)\).
Method of Finding Roots 
Set Up 
Roots 
\(a\left( {x{{r}_{1}}} \right)\left( {x{{r}_{2}}} \right)=0\) 
\(({{r}_{1}},0),\,\,\,\,({{r}_{2}},0)\) 

\(\displaystyle \begin{array}{c}a{{x}^{2}}+bx+c=0\\\text{Example:}\\a\left( {{{x}^{2}}+\frac{b}{a}x} \right)=c\\a\left[ {{{x}^{2}}+\frac{b}{a}x+{{{\left( {\frac{b}{{2a}}} \right)}}^{2}}} \right]=c+a{{\left( {\frac{b}{{2a}}} \right)}^{2}}\\a{{\left( {x+\frac{b}{{2a}}} \right)}^{2}}=c+a{{\left( {\frac{b}{{2a}}} \right)}^{2}}\\\text{Solve for }x\text{!}\end{array}\)  Complete the square and solve for \(x\). Example:
\(\displaystyle \begin{array}{c}\text{4}{{\text{x}}^{2}}+8x16=0\\\text{4}\left( {{{\text{x}}^{2}}+2x} \right)=16\\\text{4}\left( {{{\text{x}}^{2}}+2x+1} \right)=16+4\\4{{\left( {x+1} \right)}^{2}}=20;\,\,\,\,\,\,{{\left( {x+1} \right)}^{2}}=5\,\\x=\pm \sqrt{5}1\end{array}\) 

\(a{{x}^{2}}+bx+c=0\) 
\(\displaystyle \left( {\frac{{b\,\,\sqrt{{{{b}^{2}}4ac}}}}{{2a}},\,\,0} \right),\,\,\,\,\,\,\left( {\frac{{b\,+\,\sqrt{{{{b}^{2}}4ac}}}}{{2a}},\,\,0} \right)\) 

Put equation in “Y=“, use “2^{nd} TRACE” (CALC), and then either scroll down or push 2 for ZERO.  Use “Left Bound?” (move cursor to left of that zero and hit ENTER), “Right Bound?” (move cursor to right of that zero and hit ENTER), “Guess?” (hit ENTER) to “zero” in on a zero. Do the same for each zero (use “Trace” + arrows to move closer to each zero).
Note that you won’t get exact answers for irrational zeros on the calculator. 
Note: If we are really having a difficult time factoring a quadratic trinomial, we could, as a last resort, use the Quadratic Formula, get the roots, and, if they are rational (integers or fractions), put the quadratic back in factored form. For example, if we used the Quadratic Formula and got roots \(\displaystyle \frac{4}{7}\) and –3, our factors would be \(\left( {7x+4} \right)\left( {x3} \right)\).
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try a Factoring problem. Click on Submit (the blue arrow to the right of the problem) and click on Factor to see the answer.
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
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On to Quadratic Inequalities – you are ready!