Question

The roots of the equation $x^3+8x^2+19x+12=0$ are

• A. $-4$
• B. $-1$
• C. $0$
• D. $-3$

Answer 1

Answer: (A), (B), (D)

$-3$

Let $f\left(x\right)=x^3+8x^2+19x+12$

We look for $x$ such that $f\left(x\right)=0$
Putting $x=1$, we get $f\left(1\right)=1^3+8(1{)}^2+19(1)+12=40$
$\Rightarrow x=1$, is not a root of $f\left(x\right)$

Putting $x=-1$, we get $f(-1)=(-1{)}^3+8(-1{)}^2+19(-1)+12=0$
$\therefore x=-1$, is a root of $f\left(x\right)$ or $(x+1)$ is a factor of $f\left(x\right)$.
$\therefore f\left(x\right)=(x+1)(x^2+7x+12)$
$\Rightarrow f\left(x\right)=(x+1)(x^2+(3+4)x+(3\left)\right(4\left)\right)$
$\Rightarrow f\left(x\right)=(x+1)(x+3)(x+4)$

Hence the roots of $f\left(x\right)=0\text{are}x=-1,-3,-4$