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

Answer:

The linear factors are:

$x - 2$

$x + 3$

$x + 1$

Step-by-step 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