Obtain all zeros of the polynomial $f (x) = x^{4} + x^{3} - 14 x^{2} - 19 x - 6,$ if two of its zeros are $- 2$ and $- 1$ .

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Solution

let other two roots are $a, b$
$a b \times 2 = - 6; a b = - 3; b = - \frac{3}{a}$
sum of roots $= 1$
$a + b = - 1 + 3 a - \frac{3}{a} = 2$
$a^{2} - 2 a - 3 = 0 a^{2} - 3 a + a - 3 + 0 (a - 3) (a + 1) = 0$
Roots are of the polynomial are $f (x) = 2 x^{4} + x^{3} - 14 x^{2} - 19 x - 6$ are $3$ , $- 1$ , $- 2$ and $- 1$

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