Question

Sandra graphed the system of equations that can be used to solve x^3-2x^2-11x+12=x^3-13x-12 What are the roots of the polynomial equation?
A.) –12, 12

B.) –4, 3

C.) –3, 4

D.) –1, 1

Answer 1

Answer: C.) –3, 4.

Step-by-step explanation: Given equation $x^3-2x^2-11x+12=x^3-13x-12$.

Subtracting x^3 from both sides, we get

$x^3-x^3-2x^2-11x+12=x^3-x^3-13x-12$.

$-2x^2-11x+12=-13x-12$.

Adding 13x and 12 on both sides, we get

$-2x^2-11x+12+13x+12=-13x-12+13x+12$.

$-2x^2+2x+24=0$.

Dividing whole equation by -2, we get

$x^2-x-12=0$.

Factoring quadratic, we get

$(x-4)(x+3)=0$.

x-4=0 and x+3=0

x=4 and x=-3.

Answer 2

First you need to solve the equation for x so it would become -2x^2+2x+24=0, when you factor it out it becomes -2(x^2-x-12)After factoring the inside it becomes -2(x-4)(x+3). Now (x-4) equals to 0 and (x+3) equals to 0 after solving for x you get the roots 4,-3