If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Algebra 1>Unit 13

Lesson 9: Strategy in factoring quadratics

# Factoring quadratics in any form

Tie together everything you learned about quadratic factorization in order to factor various quadratic expressions of any form.

#### What you need to know for this lesson

The following factoring methods will be used in this lesson:

#### What you will learn in this lesson

In this article, you will practice putting these methods together to completely factor quadratic expressions of any form.

## Intro: Review of factorization methods

MethodExampleWhen is it applicable?
Factoring out common factorsIf each term in the polynomial shares a common factor.
The sum-product patternIf the polynomial is of the form ${x}^{2}+bx+c$ and there are factors of $c$ that add up to $b$.
The grouping methodIf the polynomial is of the form $a{x}^{2}+bx+c$ and there are factors of $ac$ that add up to $b$.
Perfect square trinomialsIf the first and last terms are perfect squares and the middle term is twice the product of their square roots.
Difference of squaresIf the expression represents a difference of squares.

## Putting it all together

In practice, you'll rarely be told what type of factoring method(s) to use when encountering a problem. So it's important that you develop some sort of checklist to use to help make the factoring process easier.
Here's one example of such a checklist, in which a series of questions is asked in order to determine how to factor the quadratic polynomial.

Before starting any factoring problem, it is helpful to write your expression in standard form.
Once this is the case, you can proceed to the following list of questions:
Question 1: Is there a common factor?
If no, move onto Question 2. If yes, factor out the GCF and continue to Question 2.
Factoring out the GCF is a very important step in the factoring process, as it makes the numbers smaller. This, in turn, makes it easier to recognize patterns!
Question 2: Is there a difference of squares (i.e. ${x}^{2}-16$ or $25{x}^{2}-9$)?
If a difference of squares pattern occurs, factor using the pattern ${a}^{2}-{b}^{2}=\left(a+b\right)\left(a-b\right)$. If not, move on to Question 3.
Question 3: Is there a perfect square trinomial (i.e. ${x}^{2}-10x+25$ or $4{x}^{2}+12x+9$)?
If a perfect square trinomial is present, factor using the pattern ${a}^{2}±2ab+{b}^{2}=\left(a±b{\right)}^{2}$. If not, move on to Question 4.
Question 4:
a.) Is there an expression of the form ${x}^{2}+bx+c$?
If no, move on to Question 5. If yes, move on to b).
b.) Are there factors of $c$ that sum to $b$?
If yes, then factor using the sum-product pattern. Otherwise, the quadratic expression cannot be factored further.
Question 5: Are there factors of $ac$ that add up to $b$?
If you've gotten this far, the quadratic expression must be of the form $a{x}^{2}+bx+c$ where $a\ne 1$. If there are factors of $ac$ that add up to $b$, factor using the grouping method. If not, the quadratic expression cannot be factored further.
Following this checklist will help to ensure that you've factored the quadratic completely!
With this in mind, let's try a few examples.

## Example 1: Factoring $5{x}^{2}-80$‍

Notice that the expression is already in standard form. We can proceed to the checklist.
Question 1: Is there a common factor?
Yes. The GCF of $5{x}^{2}$ and $80$ is $5$. We can factor this out as follows:
$5{x}^{2}-80=5\left({x}^{2}-16\right)$
Question 2: Is there a difference of squares?
Yes. ${x}^{2}-16=\left(x{\right)}^{2}-\left(4{\right)}^{2}$. We can use the difference of squares pattern to continue factoring the polynomial as shown below.
$\begin{array}{rl}\phantom{5{x}^{2}-80}& =5\left(\left(x{\right)}^{2}-\left(4{\right)}^{2}\right)\\ \\ & =5\left(x+4\right)\left(x-4\right)\end{array}$
There are no more quadratics in the expression. We have completely factored the polynomial.
In conclusion, $5{x}^{2}-80=5\left(x+4\right)\left(x-4\right)$.

## Example 2: Factoring $4{x}^{2}+12x+9$‍

The quadratic expression is again in standard form. Let's start the checklist!
Question 1: Is there a common factor?
No. The terms $4{x}^{2}$, $12x$ and $9$ do not share a common factor. Next question.
Question 2: Is there a difference of squares?
No. There’s an $x$-term so this cannot be a difference of squares. Next question.
Question 3: Is there a perfect square trinomial?
Yes. The first term is a perfect square since $4{x}^{2}=\left(2x{\right)}^{2}$, and the last term is a perfect square since $9=\left(3{\right)}^{2}$. Also, the middle term is twice the product of the numbers that are squared since $12x=2\left(2x\right)\left(3\right)$.
We can use the perfect square trinomial pattern to factor the quadratic.
$\begin{array}{rl}& \phantom{=}4{x}^{2}+12x+9\\ \\ & =\left(2x{\right)}^{2}+2\left(2x\right)\left(3\right)+\left(3{\right)}^{2}\\ \\ & =\left(2x+3{\right)}^{2}\end{array}$
In conclusion, $4{x}^{2}+12x+9=\left(2x+3{\right)}^{2}$.

## Example 3: Factoring $12x-63+3{x}^{2}$‍

This quadratic expression is not currently in standard form. We can rewrite it as $3{x}^{2}+12x-63$ and then proceed through the checklist.
Question 1: Is there a common factor?
Yes. The GCF of $3{x}^{2}$, $12x$ and $63$ is $3$. We can factor this out as follows:
$3{x}^{2}+12x-63=3\left({x}^{2}+4x-21\right)$
Question 2: Is there a difference of squares?
No. Next question.
Question 3: Is there a perfect square trinomial?
No. Notice that $21$ is not a perfect square, so this cannot be a perfect square trinomial. Next question.
Question 4a: Is there an expression of the form ${x}^{2}+bx+c$?
Yes. The resulting quadratic, ${x}^{2}+4x-21$, is of this form.
Question 4b: Are there factors of $c$ that add up to $b$?
Yes. Specifically, there are factors of $-21$ that add up to $4$.
Since $7\cdot \left(-3\right)=-21$ and $7+\left(-3\right)=4$, we can continue to factor as follows:
$\begin{array}{rl}\phantom{3\left({x}^{2}+4x-21\right)}& =3\left({x}^{2}+4x-21\right)\\ \\ & =3\left(x+7\right)\left(x-3\right)\end{array}$
In conclusion, $3{x}^{2}+12x-63=3\left(x+7\right)\left(x-3\right)$.

## Example 4: Factoring $4{x}^{2}+18x-10$‍

Question 1: Is there a common factor?
Yes. The GCF of $4{x}^{2}$, $18x$ and $10$ is $2$. We can factor this out as follows:
$4{x}^{2}+18x-10=2\left(2{x}^{2}+9x-5\right)$
Question 2: Is there a difference of squares?
No. Next question.
Question 3: Is there a perfect square trinomial?
No. Next question.
Question 4a: Is there an expression of the form ${x}^{2}+bx+c$?
No. The leading coefficient on the quadratic factor is $2$. Next question.
Question 5: Are there factors of $ac$ that add up to $b$?
The resulting quadratic expression is $2{x}^{2}+9x-5$, and so we want to find factors of $2\cdot \left(-5\right)=-10$ that add up to $9$.
Since $\left(-1\right)\cdot 10=-10$ and $\left(-1\right)+10=9$, the answer is yes.
We can now write the middle term as $-1x+10x$ and use grouping to factor:

1) Factor $2{x}^{2}+4x-16$ completely.

2) Factor $3{x}^{2}-60x+300$ completely.

3) Factor $72{x}^{2}-2$ completely.

4) Factor $5{x}^{2}+5x+15$ completely.

5) Factor $8{x}^{2}-12x-8$ completely.

6) Factor $56-18x+{x}^{2}$ completely.

7) Factor $3{x}^{2}+27$ completely.

## Want to join the conversation?

• In number 5, where did the -4x+1x come from? -3x doesn't split up into -4x+1x and multiply out to -2. Help!

Be careful there: it does not have to multiply out to -2. You've missed a step. It has to multiply out to -4, since the leading term is 2 (different, thus, than 1) and it has to be multiplied by the last term -2, remember? Take a look at how I solved it:
8x^2-12x-8
Factor out 4:
4(2x-3x-2)
Leading term 2 times last term -2 = -4 and middle term -3. Find:
A+B = -3 that is -4+1
A*B = -4 that is -4*1
Group it:
4[(2x^2-4x)+(1x-2)]
Factor out the CGF of the groups:
4[2x(x-2)+1(x-2)]
Regroup it by factoring out the (x-2):
4(2x+1)(x-2)

There you have it.
Hope it helps and if someone find any error, please comment!
Cheers!
• i don't understand how the answer to the last question (#7) is 3(x^2 + 9)
isn't x^2 + 9 a sum of squares so shouldn't it be 3(x+3)^2 ?
• 3(x+3)^2 isn't equivalent to x^2+9. The x^2+9 does look like a sum of squares at first, but to be that it would need to factor to (x+3)(x+3); (x+3)(x+3) FOILed is x^2+6x+9, not x^2+9.
• can there be no way to factor a problem?
• Yes some problems are un-factorable in the real domain. If you look at b^2 - 4ac, if this is positive you have 2 factors, if it is 0 you have one factor, and if it is negative, is does not have factors in real domain. The first crosses x axis at two places, the second has the vertex on the x-axis, and the third has the vertex above if open upward or below if open downward and never crosses the x axis.
• I'm struggling so hard with this. I watched every video, read this whole article, thought I understood everything, spent 5 hours studying and doing all the practices until I got 100% on them, but once I read this article and tried to do the practice questions without just being told which method to use upfront I suddenly just can't do anything beyond factoring out the GCF. I could only get 1/7 practice questions on this article right and I'm so frustrated I just want to cry. Does anyone have advice because I just want to give up.
• Don't give up! Just remember that once you have tried factoring out the GCF, u need to check these things:

1. Is there a difference of squares?

2. Is there a perfect square trinomial?

3. a.) Is there an expression of the form x^2 + bx + c?
b.) If so, R there factors of c that sum to b?

4. If you've gotten this far, the quadratic expression must be of the form ax^2 + bx + c where a≠1. If there are factors of ac that add up to b, factor using the grouping method.

If there is anything specific u don't understand then I recommend going to previous videos and watching them thoroughly - they should help and I often go back to them bc it is hard to keep up with everything and remember it all.
If u need any more help u can ask me :D

Hope this helps.
• Is a 4th dimensional graph a thing?
• It is, but very hard too graph and not encountered that much (if at all) in real life.
• So, question 6 says to factor 56-18x+x^2 completely. After placing in standard form: x^2 -18X +56, the next step should be to find the GCF. 2( x^2 -9x + 28). After this, there is no a+b=-9 and a * b = 28. Factors of 28 are 1,28, 2, 14, 4, 7. So none of them work. However, apparently, when I asked for the help, they just skipped right past the GCF that they had taught us to try first and went right on to the sum/product step. Does anyone else feel confused about this?
(1 vote)
• The polynomial has no common factor other than 1. In order for there to have been a common factor of 2, the problem would have been: 2x^2-18x+56.

Yes, you should always look for a GCF. But all terms need to be evenly divisible by the value you pick. x^2 does not divide evenly by 2 in your problem, so the GCF=1 and there is no need to factor out a GCF.

Hope this help.s
• Just to confirm

Is (x-4)^2 = (4-x)^2
• Yes, you are correct here is why:

(x-4)^2 = x^2 + 2(x)(-4) + 16 = x^2 - 8x + 16
(4-x)^2 = x^2 + 2(-x)(4) + 16 = x^2 - 8x + 16

Notice, however, that in the second expression it is -x and 4 while in the first it is x and -4.
• x^2 (a+b)x + ab is the general rule for factorising, basically signifying that you find 2 numbers a and b which add to the coefficient of x and multiply to the constant term, to factorise to (x + a)(x + b) = 0.
x^2 + 2bx + b^2 (not x^2 + 2abx + b^2) is the expansion of (x + b)^2, a perfect square; this is a certain situation in factorising where it's a perfect square. It is useful to know for factoring but is most important in the completing the square method.