Question

What polynomial has quotient x2 + 6x – 4 when divided by x + 3?​

Answer 1

Answer:

$(x^{3} + 9\, x^{2} + 14\, x - 12)$.

Step-by-step explanation:

Let $p(x)$ denote the polynomial in question.

The question states that dividing $p(x)$ by $(x + 3)$ gives the quotient $(x^{2} + 6\, x - 4)$ (with no remainder.) In other words:

$\displaystyle \frac{p(x)}{x + 3} = x^{2} + 6\, x - 4$.

Multiply both sides by $(x + 3)$ to find an expression for the polynomial in question, $p(x)$:

$p(x) = (x^{2} + 6\, x - 4)\, (x + 3)$.

Expand this expression to obtain:

$\begin{aligned}p(x) &= (x^{2} + 6\, x - 4) \, (x + 3) \\ &= x\, (x^{2} + 6\, x - 4) \\ & \quad + 3\, (x^{2} + 6\, x - 4) \\ &= x^{3} + 6\, x^{2} - 4\, x \\ &\quad + 3\, x^{2} + 18\, x - 12 \\ &= x^{3} + 9\, x^{2} + 14\, x - 12\end{aligned}$.