﻿ Factoring Trinomials Worksheet | Problems & Solutions

Factoring Trinomials Worksheet

Factoring Trinomials Worksheet
• Page 1
1.
Solve 2$x$2 + 3$x$ - 18 = $x$2 by factorizing.
 a. $x$ = -6 or $x$ = -3 b. $x$ = 6 or $x$ = 3 c. $x$ = -6 or $x$ = 3 d. None of the above

Solution:

2x2 + 3x - 18 = x2
[Original equation.]

x2 + 3x - 18 = 0
[Write the equation in the standard form.]

(x + 6) (x - 3) = 0
[Factorize the equation.]

x + 6 = 0 or x - 3 = 0
[Apply zero product property.]

x = -6 or x = 3

The values of x are -6, 3.

2.
Factor: $x$2 + 12$x$ + 35
 a. ($x$ - 7)($x$ - 5) b. ($x$ + 7)($x$ + 5) c. ($x$ - 8)($x$ + 5) d. ($x$ + 7)($x$ - 5)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = 12 and c = 35.

Find the numbers p and q whose product is 35 and sum is 12.

p and q    p + q
1,35            36
7, 5            12

The required values of p and q are 7 and 5.

So, the factors of the equation x2 + 12x + 35 are (x + 7)(x + 5).

3.
Factor: $x$2 - 3$x$ + 2
 a. ($x$ - 1)($x$ + 2) b. ($x$ - 1)($x$ - 2) c. ($x$ + 1)($x$ - 2) d. ($x$ + 1)($x$ + 2)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = -3 and c = 2.

Since c is positive, find the numbers p and q with the same sign, whose product is 2 and sum is -3.

p and q    p + q
-1, -2           - 3

The required values of p and q are -1 and -2.

So, the factors of the equation x2 - 3x + 2 are (x - 1)(x - 2).

4.
Factor: $x$2 + 3$x$ - 10
 a. ($x$ - 2)($x$ - 5) b. ($x$ + 2)($x$ - 5) c. ($x$ + 2)($x$ + 5) d. ($x$ - 2)($x$ + 5)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = 3 and c = -10.

Since c is negative, find the numbers p and q with different signs, whose product is -10 and sum is 3.

p and q    p + q
-1, 10         9
-2, 5            3

The required values of p and q are -2 and 5.

The factors of the equation x2 + 3x - 10 are (x - 2)(x + 5).

5.
Factor: $x$2 - 3$x$ - 18
 a. ($x$ - 6)($x$ + 3) b. ($x$ - 6)($x$ - 3) c. ($x$ + 6)($x$ - 3) d. ($x$ + 6)($x$ + 3)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = -3 and c = -18.

Since c is negative, find the numbers p and q with different signs, whose product is -18 and sum is -3.

p and q    p + q
-3, 6           3
-6, 3          -3

The required values of p and q are -6 and 3.

The factors of equation x2 - 3x - 18 are (x + 3)(x - 6).

6.
Factor: $x$2 - 4$x$ - 21
 a. ($x$ + 7)($x$ - 3) b. ($x$ + 7)($x$ + 3) c. ($x$ - 7)($x$ + 3) d. ($x$ - 7)($x$ - 3)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = -4 and c = -21.

Since c is negative, find the numbers p and q with different signs, whose product is -21 and sum is -4.

p and q    p + q
-3, 7           4
-7, 3           -4

The required values of p and q are -7 and 3.

So, the factors of equation x2 - 4x - 21 are (x + 3)(x - 7).

7.
Factor: $x$2 + 4$x$ - 12
 a. ($x$ - 2)($x$ - 6) b. ($x$ + 2)($x$ + 6) c. ($x$ + 2)($x$ - 6) d. ($x$ - 2)($x$ + 6)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = 4 and c = -12.

Since c is negative, find numbers p and q with different signs, whose product is -12 and sum is 4.

p and q     p + q
-6, 2            -4
-2, 6             4

The required values of p and q are -2 and 6.

So, the factors of the equation x2 + 4x - 12 are (x - 2)(x + 6).

8.
Factor: $x$2 - 7$x$ + 10
 a. ($x$ + 5)($x$ - 2) b. ($x$ + 5)($x$ + 2) c. ($x$ - 5)($x$ - 2) d. ($x$ - 5)($x$ + 2)

Solution:

The factors of a trinomial x2 + bx + c are in the form (x + p)(x + q), where b = p + q and c = pq.

Compare the equation with x2 + bx + c to get b and c values. So, b = -7 and c = 10.

Since c is positive, find the numbers p and q with the same sign, whose product is 10 and sum is -7.

p and q    p + q
-5, -2           -7

The required values of p and q are -5 and -2.

So, the factors of the equation x2 - 7x + 10 are (x - 5)(x - 2).

9.
Solve: $x$2 + 17$x$ + 70 = 0
 a. $x$ = -7 or $x$ = -10 b. $x$ = 7 or $x$ = 10 c. $x$ = 7 or $x$ = -10 d. $x$ = -7 or $x$ = 10

Solution:

x2 + 17x + 70 = 0
[Original equation.]

The factors are 7 and 10.
[7 + 10 = 17 and 7 x 10 = 70]

(x + 7) (x +10) = 0
[Factorize the equation.]

x + 7 = 0 or x + 10 = 0
[Apply zero product property.]

x = -7 or x = -10

The values of x are -7, -10.

10.
Solve: $x$2 - 19$x$ + 88 = 0
 a. $x$ = 8 or $x$ = 11 b. $x$ = -8 or $x$ = -11 c. $x$ = -8 or $x$ = 11 d. $x$ = 8 or $x$ = -11

Solution:

x2 - 19x + 88 = 0
[Original equation.]

The factors are -8 and -11.
[-8 + (-11) = -19 and (-8) x (-11) = 88]

(x - 8) (x - 11) = 0
[Factorize the equation.]

x - 8 = 0 or x - 11 = 0
[Apply zero product property.]

x = 8 or x = 11.

The values of x are 8, 11.