Find the zeros of the polynomial $6 x^{3} - 7 x^{2} - 11 x + 12$ , if $x - 1$ is a factor of the polynomial.

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Solution

Given
$(x - 1)$ is factor of $6 x^{3} - 7 x^{2} - 11 x + 12$
$x - 1) 6 x^{3} - 7 x^{2} - 11 x + 12 (6 x^{2} - x - 12 6 x^{3} - 6 x^{2}_- x^{2} - 11 x - x^{2} + x- 12 x + 12 - 12 x + 12__0$

$6 x^{2} - 7 x^{2} - 11 x - 112 = (x - 1) (6 x^{2} - x - 12)$
$= (x - 1) (6 x^{2} - 9 x + 8 x - 12)$
$= (x - 1) (3 x (2 x - 3) + 4 x (2 x - 3))$
$= (x - 1) (3 x + 4) (2 x - 3)$
$3 x + 4 = 0, 2 x - 3 = 0$
$x = \frac{- 4}{3}, x = \frac{3}{2}$
$∴$ Zero of polynomials are $\frac{- 4}{3}$ and $\frac{3}{2}$

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It is given that -1 is one of the zeros of the polynomial $x^{3} + 2 x^{2} - 11 x - 12$ . Find all the zeros of the given polynomial.

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