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## Algebra 1 (Eureka Math/EngageNY)

### Unit 4: Lesson 5

Topic A: Lessons 3-4: Factoring by grouping- Intro to grouping
- Factoring by grouping
- Factoring quadratics by grouping
- Factoring quadratics: leading coefficient ≠ 1
- Factor quadratics by grouping
- Strategy in factoring quadratics (part 1 of 2)
- Strategy in factoring quadratics (part 2 of 2)
- Factoring quadratics in any form

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# Factoring quadratics in any form

CCSS.Math: , ,

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.

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 tob).

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!

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:

**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.

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.

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:

**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:

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:

**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:

## Check your understanding

## 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!(22 votes)
- 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!(35 votes)

- 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 ?(7 votes)- 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.(11 votes)

- Is a 4th dimensional graph a thing?(4 votes)
- It is, but very hard too graph and not encountered that much (if at all) in real life.(2 votes)

- can there be no way to factor a problem?(4 votes)
- 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.(0 votes)

- is there any classes that explain this minus the exponents?(3 votes)
- Quadratics have an exponent of 2. So, I'm not aware of any lessons what would explain it with out using an exponent.(1 vote)

- 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.(2 votes)
- 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.(2 votes)

- Is there a perfect square trinomial?(2 votes)
- 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(1 vote)

- Towards the end of the article, it says, "A sum of squares is never factorable", why is this the case?(1 vote)
- Starting with the difference of perfect squares, the middle term cancels out when you have (x + 3)(x - 3) = x^2 - 3x + 3x - 9 = x^2 - 9, however in you have the sum of perfect squares, both factors have to be the same (both negative or both positive) and in both cases you end up with a middle term (x+3)(x+3) = x^2 + 6x + 9 and (x-3)(x-3) = x^2 - 6x + 9(3 votes)

- If there is a question like x2 + 18x + 81 how would i solve that?(1 vote)
- Why couldn't the answer to question number 7 be the third option? I don't understand it especially when the third question (which is similar to question number 7) you get the same answer that you would expect from such a question. Why the discrepancy?(0 votes)
- The difference between the 2 questions is the sign of the second term.

Question 3 is about the DIFFERENCE of 2 squares, ( 72x^2 - 2).

Question 7 is about the SUM of 2 squares ( 3x^2 + 27 ).

If you chose the third option for question 7, when you checked it by multiplying it out, you would have gotten 3x^2 - 27 --- not what the question was asking for.

As it says in the help area, the SUM of 2 squares is NEVER factorable over the real numbers.(6 votes)