Question

What are the roots of the polynomial equation mc012-1.jpg? Use a graphing calculator and a system of equations.

–6, 6

–4, 1, 3

–3, –1, 4

1, 3

Answer 1

at_answer_text_math

The required roots are 1, -4, 3

option (2) is correct

at_explanation_text_math

Given equation is

$x^3-7x=6x-12$

$x^3-13x+12=0$

Sum of the coefficients = 1 -13 +12 =0

$\therefore$ (x-1) is a factor

Now,

$x^3-13x+12=(x-1)(x^2+px-12)$

Equating coefficients of x on both sides we get

$\implies\:-13=-12-p$

$\implies\:-13+12=-p$

$\implies\:-1=-p$

$\implies\:p=1$

Another factor is

$x^2+x-12$

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

$\:x^3-13x+12=(x-1)(x+4)(x-3)$

$\therefore$ The required roots are 1, -4, 3