Question
Using factor theorem, show that $g(x)$ is a factor of $p(x),$ when $p(x) = 3x^3 + x^2 - 20x + 12, g(x) = 3x - 2$
✓
Answer
By the factor theorem, $g(x) = 3x - 2$ will be a factor of $p(x)$ if $\text{p}\Big(\frac{2}{3}\Big)=0.$
Now, $\text{p}(\text{x}) = 3\text{x}^3 + \text{x}^2 - 20\text{x} + 12$
$\Rightarrow\text{p}\Big(\frac{2}{3}\Big)=3\Big(\frac{2}{3}\Big)^3+\Big(\frac{2}{3}\Big)^2-20\times\frac{2}{3}+12$
$=3\times\frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8+4-120+108}{9}$
$=\frac{0}{9}$
$=0$ Hence, $g(x) = 3x - 2$ is a factor of the given polynomial $p(x)$.
Now, $\text{p}(\text{x}) = 3\text{x}^3 + \text{x}^2 - 20\text{x} + 12$
$\Rightarrow\text{p}\Big(\frac{2}{3}\Big)=3\Big(\frac{2}{3}\Big)^3+\Big(\frac{2}{3}\Big)^2-20\times\frac{2}{3}+12$
$=3\times\frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12$
$=\frac{8+4-120+108}{9}$
$=\frac{0}{9}$
$=0$ Hence, $g(x) = 3x - 2$ is a factor of the given polynomial $p(x)$.
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