Question

1. for each of the following cubic equations one root is given. determine the other roots of each cubic.

a) $x^3 + 3x^2 - 6x - 8 = 0$ has a root at $x = 2$.
b) $x^3 + 2x^2 - 21x + 18 = 0$ has a root at $x = 3$.
c) $x^3 + 4x^2 + 7x + 6 = 0$ has a root at $x = -2$.
d) $2x^3 + 9x^2 + 3x - 4 = 0$ has a root at $x = -4$.

Explanation:

Part (a)

Step1: Factor the cubic using the given root

Since \( x = 2 \) is a root, \( (x - 2) \) is a factor. We perform polynomial division or use synthetic division. Let's use synthetic division for \( x^3 + 3x^2 - 6x - 8 \) with root \( 2 \):
\[

$$\begin{array}{r|rrrr} 2 & 1 & 3 & -6 & -8 \\ & & 2 & 10 & 8 \\ \hline & 1 & 5 & 4 & 0 \\ \end{array}$$

\]
So the cubic factors as \( (x - 2)(x^2 + 5x + 4) \).

Step2: Factor the quadratic

Factor \( x^2 + 5x + 4 \): \( x^2 + 5x + 4=(x + 1)(x + 4) \).

Step3: Find the roots

Set each factor to zero: \( x - 2 = 0 \) (given), \( x + 1 = 0 \) (so \( x=-1 \)), \( x + 4 = 0 \) (so \( x=-4 \)).

Step1: Factor the cubic using the given root

Since \( x = 3 \) is a root, \( (x - 3) \) is a factor. Use synthetic division for \( x^3 + 2x^2 - 21x + 18 \) with root \( 3 \):
\[

$$\begin{array}{r|rrrr} 3 & 1 & 2 & -21 & 18 \\ & & 3 & 15 & -18 \\ \hline & 1 & 5 & -6 & 0 \\ \end{array}$$

\]
So the cubic factors as \( (x - 3)(x^2 + 5x - 6) \).

Step2: Factor the quadratic

Factor \( x^2 + 5x - 6 \): \( x^2 + 5x - 6=(x + 6)(x - 1) \).

Step3: Find the roots

Set each factor to zero: \( x - 3 = 0 \) (given), \( x + 6 = 0 \) (so \( x=-6 \)), \( x - 1 = 0 \) (so \( x=1 \)).

Step1: Factor the cubic using the given root

Since \( x = -2 \) is a root, \( (x + 2) \) is a factor. Use synthetic division for \( x^3 + 4x^2 + 7x + 6 \) with root \( -2 \):
\[

$$\begin{array}{r|rrrr} -2 & 1 & 4 & 7 & 6 \\ & & -2 & -4 & -6 \\ \hline & 1 & 2 & 3 & 0 \\ \end{array}$$

\]
So the cubic factors as \( (x + 2)(x^2 + 2x + 3) \).

Step2: Solve the quadratic

For \( x^2 + 2x + 3 = 0 \), use the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 2 \), \( c = 3 \):
\[
x=\frac{-2\pm\sqrt{4 - 12}}{2}=\frac{-2\pm\sqrt{-8}}{2}=\frac{-2\pm2i\sqrt{2}}{2}=-1\pm i\sqrt{2}
\]

Answer:

The other roots are \( \boldsymbol{x=-1} \) and \( \boldsymbol{x=-4} \).

Part (b)