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# Obtain all the zeroes of $f(x) = {x^3} + 13{x^2} + 32x + 20$ if one of the zeroes is ${\text{ - 2}}$ .

Last updated date: 15th Jul 2024
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Hint – With the help of one of the root make the equation quadratic by taking the common $x + 2$ . Then solve the quadratic equation.

Given , $f(x) = {x^3} + 13{x^2} + 32x + 20 = 0\,\,\,\,\,\,\,\,\,\,\,\,({\text{i}})$
One of the root of the equation is ${\text{ - 2}}$ (Given)
So , ${\text{ }}f( - 2) = 0$
Hence,
$x + 2{\text{ }}$ is one of the factors of the equation .
So we can write $f(x)$ as ,
$f(x) = (x + 2)({x^2} + 11x + 10)$
Then we can solve quadratic equation present above by splitting the middle term so,
$f(x) = (x + 2)({x^2} + x + 10x + 10) \\ f(x) = (x + 2)(x(x + 1) + 10(x + 1)) \\ f(x) = (x + 2)(x + 10)(x + 1) \\$
From (i) we get ,
$(x + 2)(x + 10)(x + 1) = 0$
So ,
$x = - 2,x = - 10,x = - 1$
Therefore the roots of $x$ are ${\text{ - 1, - 2, - 10}}$.
Note: – Whenever you cave been asked to find the roots of cubic equation, then try to get any one of the factor by hit and trial method or any other method , by the way here in this question one of the root is given, therefore we got one of the factor of the equation and then after taking the factor out, the equation will be quadratic, solve it then get two more roots. Number of roots of the equation will be equal to the power of the highest degree term present in the equation.