Find the quotient (x^3 + 7x^2 + 7x - 15) divided by (x + 3).

Step-by-step solution

1. 1

   Predict the remainder before dividing. By the Remainder Theorem the remainder equals the dividend evaluated at $x=-3$: $(-27)+7(9)+7(-3)-15=-27+63-21-15=0$. A zero value means $x+3$ is a factor, so the division will come out exactly.

2. 2

   First step. $\dfrac{x^3}{x}=x^2$. Multiply back: $x^2(x+3)=x^3+3x^2$. Subtract: $7x^2-3x^2=4x^2$, bring down $7x$ to get $4x^2+7x$.

3. 3

   Second step. $\dfrac{4x^2}{x}=4x$. Multiply back: $4x(x+3)=4x^2+12x$. Subtract: $7x-12x=-5x$, bring down $-15$ to get $-5x-15$.

4. 4

   Third step. $\dfrac{-5x}{x}=-5$. Multiply back: $-5(x+3)=-5x-15$. Subtract: $0$, exactly as predicted in step 1.

5. 5

   Read the quotient and factor further. The quotient is $x^2+4x-5$, which factors as $(x+5)(x-1)$. So the full factorisation of the cubic is $(x+3)(x+5)(x-1)$, and its roots are $-3$, $-5$ and $1$.

6. 6

   Verify by expanding. $(x+3)(x^2+4x-5)=x^3+4x^2-5x+3x^2+12x-15=x^3+7x^2+7x-15$, which reproduces the dividend exactly.