Question

List the possible rational roots and actual rational roots for the equation x^3+2x^2-5x-6=0

Answer 1

Answer:

$\boxed{\textsf{ The possible rational roots of equation are \textbf{ 2 , (-1) and (-3)}.}}$

Step-by-step explanation:

A cubic polynomial is given to us . And we need to find the rational roots . Rational roots means the roots which really exits and not imaginary . The given polynomial is :-

$\sf\implies p(x)= x^3+2x^2-5x-6$

On equating it with 0 , it becomes cubic equation. That is ,

$\sf\implies x^3+2x^2-5x-6=0$

Here we will factorise out the given equation and then equate it with 0 to find the roots . Here the constant term is 6 . So the possible factors is 6 is :-

$\sf\implies Factors\ of \ 6 \ = \ \pm 1 , \pm 2 , \ \pm 3 , \ \pm 6.$

So put them one by one checking out the factors .

Put x = 2 :-

$\sf\implies p(x)= x^3+2x^2-5x-6\\\\\sf\implies p(2)= 2^3 +2(2)^2 -5(2)-6\\\\\sf\implies p(2)= 8 +8 -10 -6 \\\\\sf\implies p(2)=16-16\\\\\sf\implies \boxed{\sf p(2)= 0 }$

This implies that (x-2) is a factor of p(x) . Now let's divide the polynomial by (x-2) .

$\rule{200}2$

x-2 ) x³ + 2x² - 5x -6 ( x² +4x +3

⠀⠀⠀ (-) x³(+) -2x²

_____________

⠀⠀⠀+4x² -5x - 6

⠀⠀⠀ (-) 4x² (+) -8x

_____________

⠀⠀⠀ 3x -6

⠀⠀⠀(+) 3x(+) - 6

_______________

⠀⠀⠀⠀ 0

Hence we can write , x³ +2x² -5x -6 as ,

$\sf\implies x^3+2x^2-5x-6 = (x-2)(x^2 +4x+3)$

Equating it with 0 .

$\sf\implies x^3+2x^2-5x-6 = 0\\\\\sf\implies (x-2)(x^2 +4x+3) = 0 \\\\\sf\implies (x-2)(x^2+3x+x+3)=0 \\\\\sf\implies (x-2)[ x(x+3)+1(x+3)]=0 \\\\\sf\implies (x-2)(x+1)(x+3)=0 \\\\\sf\implies\boxed{\pink{\frak { x = 2 , (-1) , (-3) }}}$