Question

If $x^2$ + x + 1 =0 is a factors of 15$x^3$ + Ax + B , Find ( A, B)

• A. (15, 0)
• B. (0, 15)
• C. (-15, 0)
• D. (0, -15)

Answer 1

The correct option is D

(0, -15)

The roots of $x^2$ + x + 1 =0 are $\omega$ and ${\omega }^2$ where $\omega$ is a complex cube root of unity.
They are also the roots of 15$x^3$ + Ax + B =0
Let the third root be $\alpha$
Sum of the roots , $\alpha$ + $\omega$ + ${\omega }^2$ = $-\frac{\left(coefficientof{x}^2\right)}{\left(coefficientof{x}^3\right)}$
We know 1 + $\omega$ + ${\omega }^2$ = 0
$\Rightarrow$ $\alpha$ - 1 = 0
$\Rightarrow$ $\alpha$ = 1
$\therefore$ The roots are 1, $\omega$, ${\omega }^2$ . The equation , whose roots are 1, $\omega$, ${\omega }^2$ is $x^3$ - 1 = 0 or 15$x^3$ - 15 = 0. Comparing it with 15 $x^3$ + Ax + B , we get A = 0 and B = -15
We can solve this by substituting x = $\omega$ and x = ${\omega }^2$ in the equation. We will solve for A and B after that.