﻿ Factoring Using the Distributive Property Worksheet | Problems & Solutions

# Factoring Using the Distributive Property Worksheet

Factoring Using the Distributive Property Worksheet
• Page 1
1.
Solve the equation.
4${x}^{4}$ - 36${x}^{2}$ = 0
 a. 3, 6 b. 0, 4, 9 c. 0, 3, 9 d. 0, 3, -3

#### Solution:

4x4 Ã¢â‚¬â€œ 36x2 = 0
[Given equation.]

4x2(x2 Ã¢â‚¬â€œ 9) = 0
[Factor out GCF.]

4x2(x2 Ã¢â‚¬â€œ 32) = 0

4x2(x Ã¢â‚¬â€œ 3)(x + 3) = 0
[Factor difference of squares.]

x = 0, 3, - 3
[By setting each variable factor equal to zero, you can find the solution for x.]

2.
Factor the GCF out of 9${x}^{4}$ + 18$x$.
 a. 9$x$(${x}^{3}$ + 3) b. 9$x$(${x}^{3}$ + 2) c. 3$x$(${x}^{3}$ + 6$x$) d. 3${x}^{2}$($x$ + 6)

#### Solution:

9x4 + 18x
[Given expression.]

9x4 = 3 · 3 · x · x · x · x
[Factor.]

18x = 3 · 3 · 2 · x
[Factor.]

The GCF = 3 · 3 · x = 9x
[It is the product of all the common factors.]

9x4 + 18x = 9x(x3 + 2)
[Use the distributive property to factor the greatest common factor out of the polynomial.]

3.
Factor the GCF out of 20${t}^{5}$ + 5${t}^{3}$.
 a. 5${t}^{2}$(4${t}^{3}$ + 1) b. 4${t}^{2}$(5${t}^{3}$ + 4) c. 5${t}^{3}$(4${t}^{2}$ + 1) d. 4${t}^{3}$(5${t}^{2}$ + 4)

#### Solution:

20t5 + 5t3
[Given expression.]

20t5 = 5 · 4 · t · t · t · t · t
[Factor.]

5t3 = 5 · t · t · t
[Factor.]

The GCF is 5t3.
[It is the product of all the common factors.]

20t5 + 5t3 = 5t3(4t2 + 1)
[Use the distributive property to factor the greatest common factor out of the polynomial.]

4.
Factor the GCF out of 6${a}^{5}$ + 12${a}^{3}$ - 3${a}^{2}$.
 a. 3${a}^{2}$(2${a}^{3}$ + 4$a$ - 1) b. 3${a}^{2}$(2${a}^{3}$ + 4$a$ + 1) c. 3${a}^{2}$(2${a}^{3}$ - 4$a$ - 1) d. 3${a}^{2}$(2${a}^{3}$ + 4$a$ - $a$)

#### Solution:

6a5 + 12a3 - 3a2
[Given expression.]

6a5 = 3(2)(a2)(a3)
[Factor.]

12a3 = 3(4)(a2)(a)
[Factor.]

3a2 = 3(a2)
[Factor.]

GCF = 3a2
[It is the product of all the common factors.]

6a5 + 12a3 - 3a2 = 3a2(2a3 + 4a - 1)
[Use the distributive property to factor the greatest common factor out of the polynomial.]

5.
Factor the GCF out of 16${d}^{6}$ - 4${d}^{2}$ + 2$d$.
 a. 2$d$(8${d}^{5}$ - 2$d$ - 1) b. 2$d$(8${d}^{5}$ - 2$d$ + 1) c. 2(8${d}^{5}$ - 2$d$ + 1) d. 2$d$(8${d}^{5}$ - 2$d$ + $d$)

#### Solution:

16d6 - 4d2 + 2d
[Given expression.]

16d6 = 2 × 8 × d × d2 × d3
[Factor.]

4d2 = 2 × 2 × d × d
[Factor.]

2d = 2 × d
[Factor.]

The GCF = 2d.
[It is the product of all the common factors.]

16d6 - 4d2 + 2d = 2d(8d5 - 2d + 1)
[Use the distributive property to factor the greatest common factor out of the polynomial.]

6.
Factor the expression completely.
25${x}^{3}$ + 10${x}^{2}$
 a. 5${x}^{2}$(5$x$ + $x$) b. 5${x}^{2}$(5${x}^{2}$ + 1) c. 5${x}^{2}$(5$x$ + 2) d. 5${x}^{2}$(5$x$ - 2)

#### Solution:

25x3 + 10x2
[Given expression.]

25x3 = 5 × 5 × x2 × x
[Factor.]

10x2 = 5 × 2 × x2
[Factor.]

25x3 + 10x2 = 5x2(5x + 2)
[GCF = 5x2.]

This polynomial is factored completely. Because 5x + 2 cannot be factored using integer co-efficient.

7.
Factor the expression completely.
- 4${w}^{4}$ + 28${w}^{3}$
 a. 4${w}^{3}$($w$ - 7) b. - 4${w}^{3}$($w$ + 7) c. - 4${w}^{3}$($w$ - 7) d. - 4${w}^{3}$($w$ - 1)

#### Solution:

- 4w4 + 28w3
[Given expression.]

4w4 = 4 × w3 × w
[Factor.]

28w3 = 4 × 7 × w3
[Factor.]

- 4w4 + 28w3 = - 4 × w3 × w + 4 × 7 × w3

= - 4w3(w - 7)
[Take the common factor out.]

8.
Factor the expression completely.
3${d}^{4}$ + 3${d}^{3}$ - 60${d}^{2}$
 a. 3${d}^{2}$($d$ - 4)($d$ + 5) b. 3${d}^{2}$($d$ - 4)($d$ - 5) c. 3${d}^{2}$($d$ + 4)($d$ + 5) d. 3${d}^{2}$($d$ + 4)($d$ -5)

#### Solution:

3d4 + 3d3 - 60d2
[Given.]

3d4 = 3 × d2 × d2
[Factor.]

3d3 = 3 × d2 × d
[Factor.]

60d2 = 3 × 20 × d2
[Factor.]

The GCF is 3d2.

3d4 + 3d3 - 60d2 = 3d2(d2 + d - 20)
[Factor out GCF.]

= 3d2(d - 4)(d + 5)
[Factor the trinomial.]

9.
Factor the expression completely.
- 27${z}^{4}$ + 3${z}^{2}$
 a. - 3${z}^{2}$ b. - 3${z}^{2}$ c. 3${z}^{2}$(3$z$ - 1)(3$z$ + 1) d. - 3${z}^{2}$(3$z$ - 1)(3$z$ + 1)

#### Solution:

- 27z4 + 3z2
[Given.]

27z4 = 3 × 9 × z2 × z2
[Factor.]

3z2 = 3 × z2
[Factor.]

The GCF is 3z2.

- 27z4 + 3z2 = - 3z2(9z2 - 1)
[Factor out GCF.]

= - 3z2(3z + 1)(3z - 1)
[Use (a2 - b2) = (a + b)(a - b).]

10.
Factor the expression completely ${x}^{4}$ + ${x}^{3}$ - 16$x$ -16.
 a. (${x}^{3}$ + 16)($x$ - 1) b. (${x}^{3}$ + 16)($x$ + 1) c. (${x}^{3}$ - 1)($x$ + 16) d. (${x}^{3}$ - 16)($x$ + 1)

#### Solution:

x4 + x3 -16x -16
[Given.]

= (x4 + x3) - (16x + 16)
[Group terms.]

= x3(x + 1) - 16(x + 1)
[Factor each group.]

= (x3 - 16)(x + 1)
[Use distributive property.]

Therefore, x4 + x3 - 16x - 16 = (x3 - 16)(x + 1)