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# Factoring by Grouping

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The volume of a rectangular prism is $3x^5 - 27x^4 - 2x^2 + 18x$ . What are the lengths of the prism's sides?

### Guidance

In the Factoring when the Leading Coefficient Doesn't Equal 1 concept, we introduced factoring by grouping. We will expand this idea to other polynomials here.

#### Example A

Factor $x^4+7x^3-8x-56$ by grouping.

Solution: First, group the first two and last two terms together. Pull out any common factors.

$\underbrace{x^4+7x^3}_{x^3{\color{red}(x+7)}}\underbrace{-8x-56}_{-8{\color{red}(x+7)}}$

Notice what is inside the parenthesis is the same . This should always happen when factoring by grouping. Pull out this common factor.

$& x^3(x+7)-8(x+7)\\& (x+7)(x^3-8)$

Look at the factors. Can they be factored any further? Yes. The second factor is a difference of cubes. Use the formula.

$& (x+7)(x^3-8)\\& (x+7)(x-2)(x^2+2x+4)$

#### Example B

Factor $x^3+5x^2-x-5$ by grouping.

Solution: Follow the steps from above.

$& x^3+5x^2-x-5\\& x^2(x+5)-1(x+5)\\& (x+5)(x^2-1)$

Look to see if we can factor either factor further. Yes, the second factor is a difference of squares.

$& (x+5)(x^2-1)\\&(x+5)(x-1)(x+1)$

#### Example C

Find all real-number solutions of $2x^3-3x^2+8x-12 = 0$ .

Solution: Follow the steps from Example A.

$2x^3-3x^2+8x-12 &= 0\\x^2(2x-3)+4(2x-3) &= 0\\(2x-3)(x^2+4) &= 0$

Now, determine if you can factor further. No, $x^2+4$ is a sum of squares and not factorable. Setting the first factor equal to zero, we get $x = \frac{3}{2}$ .

Intro Problem Revisit

We need to factor $3x^5 - 27x^4 - 2x^2 + 18x$ to find the lengths of the prism's sides.

First, pull out the common factor. $x(3x^4 - 27x^3 - 2x + 18)$

Next, factor $(3x^4 - 27x^3 - 2x + 18)$ by grouping the first two and last two terms together.

$(3x^4 - 27x^3 - 2x + 18)\\= (3x^4 - 27x^3) + (- 2x + 18)\\= 3x^3(x - 9) - 2(x - 9)$

Now pull out the common factor.

$3x^3(x - 9) - 2(x - 9)\\= (3x^3 - 2)(x - 9)$

The expression can't be factored further, so $3x^5 - 27x^4 - 2x^2 + 18x = x(3x^3 - 2)(x - 9)$ and the lengths of the sides of the rectangular prism are $x$ , $3x^3 - 2$ , and $x - 9$ .

### Guided Practice

Factor the following polynomials by grouping.

1. $x^3+7x^2-2x-14$

2. $2x^4-5x^3+2x-5$

3. Find all the real-number solutions of $4x^3-8x^2-x+2 = 0$ .

Each of these problems is done in the same way: Group the first two and last two terms together, pull out any common factors, what is inside the parenthesis is the same, factor it out, then determine if either factor can be factored further.

1. $& x^3+7x^2-2x-14\\& x^2(x+7)-2(x+7)\\& (x+7)(x^2-2)$

$x^2-2$ is not a difference of squares because 2 is not a square number. Therefore, this cannot be factored further.

2. $& 2x^4-5x^3+2x-5\\& x^3(2x-5)+1(2x-5)\\& (2x-5)(x^3+1) \quad \ \ \text{Sum of cubes, factor further}.\\& (2x-5)(x+1)(x^2+x+1)$

3. Factor by grouping.

$4x^3-8x^2-x+2 &= 0\\4x^2(x-2)-1(x-2) &= 0\\(x-2)(4x^2-1) &= 0\\(x-2)(2x-1)(2x+1) &= 0\\x & = 2, \frac{1}{2}, -\frac{1}{2}$

### Practice

Factor the following polynomials using factoring by grouping. Factor each polynomial completely.

1. $x^3-4x^2+3x-12$
2. $x^3+6x^2-9x-54$
3. $3x^3-4x^2+15x-20$
4. $2x^4-3x^3-16x+24$
5. $4x^3+4x^2-25x-25$
6. $4x^3+18x^2-10x-45$
7. $24x^4-40x^3+81x-135$
8. $15x^3+6x^2-10x-4$
9. $4x^3+5x^2-100x-125$
10. $3x^3-2x^2+12x-8$

Find all the real-number solutions of the polynomials below.

1. $9x^3-54x^2-4x+24 = 0$
2. $x^4+3x^3-27x-81 = 0$
3. $x^3-2x^2-4x+8 = 0$
4. Challenge Find ALL the solutions of $x^6-9x^4-x^2+9 = 0$ .
5. Challenge Find ALL the solutions of $x^3+3x^2+16x+48 = 0$ .