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

Hi, ShadowFax5! 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.

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.

What does "A sum of squares is never factorable over the real numbers," mean?

If you have a binomial like x^2+9, you have a sum of 2 squares. This binomial is not factorable using real numbers. Treat x^2+9 as x^2+0x+9. Try to find factors of 9 that add to 0. They don't exist. In higher level math, you learn that it can be factored using complex numbers.

One of the questions on the Unit Test was to expand (4-7y)(7+4y) and present the answer in standard form. I solved this easily but wanted to check it, as I always do in order to avoid a wrong answer. I expanded to 28-33y-28y^2, changed the order to -28y^2-33y+28, entered this (correct) answer, and then tried to check by factoring it in that order. I could imagine that there might be two numbers that would multiply to -784 and add up to 33, but they wouldn't involve ±7 and ±4. I assume it must be possible to factor -28y^2-33y+28 into (4-7y)(7+4y), but how? An inability to do so would seem disturbing, as it would suggest that you'd have to try all possible orders to know whether an expression could be factored or not – and actually I'm not sure how to factor this one in any order, though maybe I should be.

My preferred method of factoring expressions such as yours or those in the form ax^2+bx+c (with a>1) follows: 1. Multiply your a-value by c. (You get y^2-33y-784) 2. Attempt to factor as usual (This is quite tricky for expressions like yours with huge numbers, but it is easier than keeping the a coeffcient in.) If you find the two values, you should get (y+16)(y-49). 3. Divide your constants in the binomials by your original a-value and reduce as much as possible. So, divide 16 by -28 and divide 49 by -28, making sure to reduce the fractions and being aware of negative signs. 4. Take your denominators and multiply by your variable. You should have gotten (y+4/-7)(y+7/4). You will have gotten (-7y-4)(4y+7), which is the same to your original expression, with rearranged terms. Typically, you'll see basic equations on tests that don't involve such large numbers.

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?

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

Main content

Algebra 1

Course: Algebra 1 > Unit 13

Lesson 9: Strategy in factoring quadratics

Factoring quadratics in any form

Google Classroom

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

Method	Example	When is it applicable?
Factoring out common factors	$\begin{aligned}&\phantom{=}~6x^2+3x\\\\&=3x(2x+1)\\\\\end{aligned}$	If each term in the polynomial shares a common factor.
The sum-product pattern	$\begin{aligned}&\phantom{=}~x^2+7x+12\\\\&=(x+3)(x+4)\end{aligned}$	If the polynomial is of the form x, squared, plus, b, x, plus, c and there are factors of c that add up to b.
The grouping method	$\begin{aligned}&\phantom{=}~2x^2+7x+3\\\\&=2x^2+6x+1x+3\\\\&=2x(x+3)+1(x+3)\\\\&=(x+3)(2x+1)\\\\\\\end{aligned}$	If the polynomial is of the form a, x, squared, plus, b, x, plus, c and there are factors of a, c that add up to b.
Perfect square trinomials	$\begin{aligned}&\phantom{=}~x^2+10x+25\\\\&=(x+5)^2\end{aligned}$	If the first and last terms are perfect squares and the middle term is twice the product of their square roots.
Difference of squares	$\begin{aligned}&\phantom{=}~~x^2-9\\\\&=(x-3)(x+3)\end{aligned}$	If 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.

Factoring quadratic expressions

Before starting any factoring problem, it is helpful to write your expression in standard form.

[Why?]

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, squared, minus, 16 or 25, x, squared, minus, 9)?
If a difference of squares pattern occurs, factor using the pattern a, squared, minus, b, squared, equals, left parenthesis, a, plus, b, right parenthesis, left parenthesis, a, minus, b, right parenthesis. If not, move on to Question 3.

Question 3: Is there a perfect square trinomial (i.e. x, squared, minus, 10, x, plus, 25 or 4, x, squared, plus, 12, x, plus, 9)?
If a perfect square trinomial is present, factor using the pattern a, squared, plus minus, 2, a, b, plus, b, squared, equals, left parenthesis, a, plus minus, b, right parenthesis, squared. If not, move on to Question 4.

Question 4:

a.) Is there an expression of the form x, squared, plus, b, x, plus, 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 a, c that add up to b?
If you've gotten this far, the quadratic expression must be of the form a, x, squared, plus, b, x, plus, c where a, does not equal, 1. If there are factors of a, c 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!

[What does it mean to factor completely?]

With this in mind, let's try a few examples.

Example 1: Factoring 5, x, squared, minus, 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, squared and 80 is 5. We can factor this out as follows:

5, x, squared, minus, 80, equals, 5, left parenthesis, x, squared, minus, 16, right parenthesis

Question 2: Is there a difference of squares?
Yes. x, squared, minus, 16, equals, left parenthesis, start color #11accd, x, end color #11accd, right parenthesis, squared, minus, left parenthesis, start color #1fab54, 4, end color #1fab54, right parenthesis, squared. We can use the difference of squares pattern to continue factoring the polynomial as shown below.

$\begin{aligned}\phantom{5x^2-80}&=5\left((\blueD {x})^2-(\greenD 4)^2\right)\\ \\ &=5(\blueD x+\greenD 4)(\blueD x-\greenD 4)\end{aligned}$

There are no more quadratics in the expression. We have completely factored the polynomial.

In conclusion, 5, x, squared, minus, 80, equals, 5, left parenthesis, x, plus, 4, right parenthesis, left parenthesis, x, minus, 4, right parenthesis.

Example 2: Factoring 4, x, squared, plus, 12, x, plus, 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, squared, 12, x 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, squared, equals, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, squared, and the last term is a perfect square since 9, equals, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis, squared. Also, the middle term is twice the product of the numbers that are squared since 12, x, equals, 2, left parenthesis, start color #11accd, 2, x, end color #11accd, right parenthesis, left parenthesis, start color #1fab54, 3, end color #1fab54, right parenthesis.

We can use the perfect square trinomial pattern to factor the quadratic.

$\begin{aligned}&\phantom{=}4x^2+12x+9\\\\&=(\blueD {2x})^2+2(\blueD{2x})(\greenD{3})+(\greenD{3})^2\\\\&=(\blueD{2x}+\greenD 3)^2\end{aligned}$

In conclusion, 4, x, squared, plus, 12, x, plus, 9, equals, left parenthesis, 2, x, plus, 3, right parenthesis, squared.

Example 3: Factoring 12, x, minus, 63, plus, 3, x, squared

This quadratic expression is not currently in standard form. We can rewrite it as 3, x, squared, plus, 12, x, minus, 63 and then proceed through the checklist.

Question 1: Is there a common factor?
Yes. The GCF of 3, x, squared, 12, x and 63 is 3. We can factor this out as follows:

3, x, squared, plus, 12, x, minus, 63, equals, 3, left parenthesis, x, squared, plus, 4, x, minus, 21, right parenthesis

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, squared, plus, b, x, plus, c?
Yes. The resulting quadratic, x, squared, plus, 4, x, minus, 21, is of this form.

Question 4b: Are there factors of c that add up to b?
Yes. Specifically, there are factors of minus, 21 that add up to 4.

Since 7, dot, left parenthesis, minus, 3, right parenthesis, equals, minus, 21 and 7, plus, left parenthesis, minus, 3, right parenthesis, equals, 4, we can continue to factor as follows:

$\begin{aligned}\phantom{3(x^2+4x-21)}&=3(x^2+4x-21)\\ \\ &=3(x+7)(x-3)\\ \end{aligned}$

In conclusion, 3, x, squared, plus, 12, x, minus, 63, equals, 3, left parenthesis, x, plus, 7, right parenthesis, left parenthesis, x, minus, 3, right parenthesis.

Example 4: Factoring 4, x, squared, plus, 18, x, minus, 10

Notice that this quadratic expression is already in standard form.

Question 1: Is there a common factor?
Yes. The GCF of 4, x, squared, 18, x and 10 is 2. We can factor this out as follows:

4, x, squared, plus, 18, x, minus, 10, equals, 2, left parenthesis, 2, x, squared, plus, 9, x, minus, 5, right parenthesis

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, squared, plus, b, x, plus, c?
No. The leading coefficient on the quadratic factor is 2. Next question.

Question 5: Are there factors of a, c that add up to b?
The resulting quadratic expression is 2, x, squared, plus, 9, x, minus, 5, and so we want to find factors of 2, dot, left parenthesis, minus, 5, right parenthesis, equals, minus, 10 that add up to 9.

Since left parenthesis, minus, 1, right parenthesis, dot, 10, equals, minus, 10 and left parenthesis, minus, 1, right parenthesis, plus, 10, equals, 9, the answer is yes.

We can now write the middle term as minus, 1, x, plus, 10, x and use grouping to factor:

\begin{aligned}&\phantom{=}~2(2x^2+9x-5)\\\\ &=2(2x^2-1x+10x-5)&&\small{\gray{\text{Split up middle term}}}\\ \\ &=2\left((2x^2-1x)+(10x-5)\right)&&\small{\gray{\text{Group terms}}}\\\\ &=2\left(x(2x-1)+5(2x-1)\right)&&\small{\gray{\text{Factor out GCFs}}}\\\\ &=2(2x-1)(x+5)&&\small{\gray{\text{Factor out $2x-1$}}} \end{aligned}

Check your understanding

1) Factor 2, x, squared, plus, 4, x, minus, 16 completely.

(Choice A)
left parenthesis, 2, x, plus, 4, right parenthesis, squared
(Choice B)
left parenthesis, 2, x, minus, 2, right parenthesis, left parenthesis, x, plus, 8, right parenthesis
(Choice C)
2, left parenthesis, x, plus, 2, right parenthesis, left parenthesis, x, minus, 4, right parenthesis
(Choice D)
2, left parenthesis, x, minus, 2, right parenthesis, left parenthesis, x, plus, 4, right parenthesis

[I need help!]

2) Factor 3, x, squared, minus, 60, x, plus, 300 completely.

[I need help!]

3) Factor 72, x, squared, minus, 2 completely.

[I need help!]

4) Factor 5, x, squared, plus, 5, x, plus, 15 completely.

(Choice A)
5, left parenthesis, x, squared, plus, x, plus, 3, right parenthesis
(Choice B)
left parenthesis, 5, x, plus, 1, right parenthesis, left parenthesis, x, plus, 3, right parenthesis
(Choice C)
5, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, plus, 1, right parenthesis

[I need help!]

5) Factor 8, x, squared, minus, 12, x, minus, 8 completely.

[I need help!]

6) Factor 56, minus, 18, x, plus, x, squared completely.

[I need help!]

7) Factor 3, x, squared, plus, 27 completely.

(Choice A)
3, left parenthesis, x, plus, 3, right parenthesis, squared
(Choice B)
3, left parenthesis, x, squared, plus, 9, right parenthesis
(Choice C)
3, left parenthesis, x, plus, 3, right parenthesis, left parenthesis, x, minus, 3, right parenthesis

[I need help!]

Want to join the conversation?

Sort by:

ShadowFax5
Posted 7 years ago. Direct link to ShadowFax5's post “In number 5, where did th...”
In number 5, where did the -4x+1x come from? -3x doesn't split up into -4x+1x and multiply out to -2. Help!
Button navigates to signup pageComment on ShadowFax5's post “In number 5, where did th...”
(22 votes)
Answer
- Beorn MacRupert
  Posted 7 years ago. Direct link to Beorn MacRupert's post “Hi, ShadowFax5! Be caref...”
  Hi, ShadowFax5!
  
  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!
  Comment on Beorn MacRupert's post “Hi, ShadowFax5! Be caref...”
  (37 votes)
Anastasia.Theys
Posted 6 years ago. Direct link to Anastasia.Theys's post “i don't understand how th...”
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 ?
Button navigates to signup pageButton navigates to signup page
(7 votes)
Answer
- Ellen Wight
  Posted 6 years ago. Direct link to Ellen Wight's post “3(x+3)^2 isn't equivalent...”
  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.
  Button navigates to signup page
  (11 votes)
Riley
Posted 3 months ago. Direct link to Riley's post “I'm struggling so hard wi...”
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.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Viktoria Chaucer
  Posted 3 months ago. Direct link to Viktoria Chaucer's post “Don't give up! Just remem...”
  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.
  Just remember that the five question method in this article works for ANY quadratic expressions.
  If u need any more help u can ask me :D
  
  Hope this helps.
  Comment on Viktoria Chaucer's post “Don't give up! Just remem...”
  (4 votes)
601044
Posted 3 years ago. Direct link to 601044's post “Is a 4th dimensional grap...”
Is a 4th dimensional graph a thing?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Silenyx: YT
  Posted 3 years ago. Direct link to Silenyx: YT's post “It is, but very hard too ...”
  It is, but very hard too graph and not encountered that much (if at all) in real life.
  Button navigates to signup page
  (2 votes)
Lexi Tischman
Posted 5 years ago. Direct link to Lexi Tischman's post “can there be no way to fa...”
can there be no way to factor a problem?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- David Severin
  Posted 5 years ago. Direct link to David Severin's post “Yes some problems are un-...”
  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.
  Comment on David Severin's post “Yes some problems are un-...”
  (0 votes)
Soshi_Garelik
Posted 4 years ago. Direct link to Soshi_Garelik's post “is there any classes that...”
is there any classes that explain this minus the exponents?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “Quadratics have an expone...”
  Quadratics have an exponent of 2. So, I'm not aware of any lessons what would explain it with out using an exponent.
  Button navigates to signup page
  (1 vote)
Ella
Posted 7 years ago. Direct link to Ella's post “Is there a perfect square...”
Is there a perfect square trinomial?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “Yes... there is a perfect...”
  Yes... there is a perfect square trinomial. You can find lessons on it at these links:
  1) Multiplying to create a perfect square trinomial:
  -- https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-special-products-of-polynomials/v/pattern-for-squaring-simple-binomials
  -- https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-special-products-of-polynomials/v/square-a-binomial
  2) Factoring a perfect square trinomial: https://www.khanacademy.org/math/algebra-home/alg-polynomials/alg-factoring-quadratics-perfect-squares/v/factoring-perfect-square-trinomials
  Comment on Kim Seidel's post “Yes... there is a perfect...”
  (1 vote)
Michael Byers
Posted 8 months ago. Direct link to Michael Byers's post “What does "A sum of squar...”
What does "A sum of squares is never factorable over the real numbers," mean?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 8 months ago. Direct link to Kim Seidel's post “If you have a binomial li...”
  If you have a binomial like x^2+9, you have a sum of 2 squares. This binomial is not factorable using real numbers.
  Treat x^2+9 as x^2+0x+9. Try to find factors of 9 that add to 0. They don't exist.
  
  In higher level math, you learn that it can be factored using complex numbers.
  Button navigates to signup page
  (3 votes)
TheSnowyOwl9455
Posted 3 months ago. Direct link to TheSnowyOwl9455's post “So, question 6 says to fa...”
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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 3 months ago. Direct link to Kim Seidel's post “The polynomial has no com...”
  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
  Button navigates to signup page
  (5 votes)
Roy McCoy
Posted a year ago. Direct link to Roy McCoy's post “One of the questions on t...”
One of the questions on the Unit Test was to expand (4-7y)(7+4y) and present the answer in standard form. I solved this easily but wanted to check it, as I always do in order to avoid a wrong answer. I expanded to 28-33y-28y^2, changed the order to -28y^2-33y+28, entered this (correct) answer, and then tried to check by factoring it in that order. I could imagine that there might be two numbers that would multiply to -784 and add up to 33, but they wouldn't involve ±7 and ±4. I assume it must be possible to factor -28y^2-33y+28 into (4-7y)(7+4y), but how? An inability to do so would seem disturbing, as it would suggest that you'd have to try all possible orders to know whether an expression could be factored or not – and actually I'm not sure how to factor this one in any order, though maybe I should be.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- APTeamOfficial
  Posted a year ago. Direct link to APTeamOfficial's post “My preferred method of fa...”
  My preferred method of factoring expressions such as yours or those in the form ax^2+bx+c (with a>1) follows:
  
  1. Multiply your a-value by c. (You get y^2-33y-784)
  2. Attempt to factor as usual (This is quite tricky for expressions like yours with huge numbers, but it is easier than keeping the a coeffcient in.) If you find the two values, you should get (y+16)(y-49).
  3. Divide your constants in the binomials by your original a-value and reduce as much as possible. So, divide 16 by -28 and divide 49 by -28, making sure to reduce the fractions and being aware of negative signs.
  4. Take your denominators and multiply by your variable. You should have gotten (y+4/-7)(y+7/4).
  
  You will have gotten (-7y-4)(4y+7), which is the same to your original expression, with rearranged terms. Typically, you'll see basic equations on tests that don't involve such large numbers.
  Comment on APTeamOfficial's post “My preferred method of fa...”
  (2 votes)

Algebra 1

Course: Algebra 1 > Unit 13

What you need to know for this lesson

What you will learn in this lesson

Intro: Review of factorization methods

Putting it all together

Factoring quadratic expressions

Example 1: Factoring 5x2−805x^2-805x2−805, x, squared, minus, 80

Example 2: Factoring 4x2+12x+94x^2+12x+94x2+12x+94, x, squared, plus, 12, x, plus, 9

Example 3: Factoring 12x−63+3x212x-63+3x^212x−63+3x212, x, minus, 63, plus, 3, x, squared

Example 4: Factoring 4x2+18x−104x^2+18x-104x2+18x−104, x, squared, plus, 18, x, minus, 10

Check your understanding

Want to join the conversation?

Example 1: Factoring 5, x, squared, minus, 80

Example 2: Factoring 4, x, squared, plus, 12, x, plus, 9

Example 3: Factoring 12, x, minus, 63, plus, 3, x, squared

Example 4: Factoring 4, x, squared, plus, 18, x, minus, 10