Question

If one of the zeros of $f\left(x\right)=x^3+13x^2+32x+20$ is $-2$ then all its zeros are

A. $2,13,11$

B. $-10,-1,-2$

C. $4,-2,-10$

D. $-2,5,10$

• A. 4, -2, -10
• B. 2, 13, 11
• C. -10, -1, -2
• D. -2, 5, 10

Answer 1

The correct option is B. $-10,-1,-2$

Since, $-2$ is a zero of $f\left(x\right)$, therefore, $(x+2)$ is a factor of $f\left(x\right)$.

Thus,

$x^3+13x^2+32x+20=(x+2)(x^2+11x+10)$

Using Middle Term Splitting, we get,

$(x^2+11x+10)=x^2+1x+10x+10$

$=x(x+1)+10(x+1)$

$=(x+1)(x+10)$

Thus, $x^3+13x^2+32x+20=(x+2)(x+1)(x+10)$

Hence, zeroes of $f\left(x\right)$ are $-2,-1,-10$