Question

Which expressions are linear factors of the polynomial? Choose all that apply.

x3 + 2x2.5x-6
ox-2
O X+3
Ox-1
Ox+1

Answer 1

The linear factors are:

`x - 2`

`x + 3`

`x + 1`

Explanation:

Given

`x^3 + 2x^2 - 5x - 6`

Required

Select all linear factors of the polynomial

Solving for x - 2;

Equate x - 2 to 0 and solve for x

`x - 2 = 0`

`x = 2`

Substitute 2 for x in
`x^3 + 2x^2 - 5x - 6`

`2^3 + 2(2^2) - 5(2) - 6`

`= 8 + 8 - 10 - 6`

`= 0`

Because, we arrived at 0, then x - 2 is a factor

Solving for x + 3;

Equate x + 3 to 0 and solve for x

`x + 3 = 0`

`x = -3`

Substitute -3 for x in
`x^3 + 2x^2 - 5x - 6`

`(-3)^3 + 2(-3^2) - 5(-3) - 6`

`= -27 + 2(9) + 15 - 6`

`= -27 + 18 + 15 - 6`

`= 0`

Because, we arrived at 0, then x + 3 is a factor

Solving for x - 1;

Equate x - 1 to 0 and solve for x

`x - 1 = 0`

`x = 1`

Substitute 1 for x in
`x^3 + 2x^2 - 5x - 6`

`(1)^3 + 2(1^2) - 5(1) - 6`

`= 1 + 2(1) - 5 - 6`

`= 1 + 2 - 5 - 6`

`= -8`

Because, we did not arrived at 0, then x - 1 is not a factor

Solving for x + 1;

Equate x + 1 to 0 and solve for x

`x + 1 = 0`

`x = -1`

Substitute -1 for x in
`x^3 + 2x^2 - 5x - 6`

`(-1)^3 + 2(-1^2) - 5(-1) - 6`

`= -1 + 2(1) + 5 - 6`

`= -1 + 2 + 5 - 6`

`= 0`

Because, we arrived at 0, then x + 1 is a factor