Question
Show that: $2x - 3$ is a factor of $x + 2x^3 - 9x^2 + 12.$
✓
Answer
Let $p(x) = x + 2x^3 - 9x^2 + 12, g(x) = 2x - 3 g(x) = 2x - 3 = 0$
gives $\text{x}=\frac{3}{2}$ g(x) will be a factor of p(x) if $\text{p}\Big(\frac{3}{2}\Big)=0$ (Factor theorem)
Now, $\text{p}\Big(\frac{3}{2}\Big)=\frac{3}{2}+2\Big(\frac{3}{2}\Big)^3-9\Big(\frac{3}{2}\Big)^2+12$
$=\frac{3}{2}+2\Big(\frac{27}{8}\Big)-9\Big(\frac{9}{4}\Big)+12$
$=\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12$
$=\frac{6+27-81+48}{4}=\frac{0}{4}=0$
Since, $\text{p}\Big(\frac{3}{2}\Big)=0,$
So $g(x)$ is a factor of $p(x)$.
gives $\text{x}=\frac{3}{2}$ g(x) will be a factor of p(x) if $\text{p}\Big(\frac{3}{2}\Big)=0$ (Factor theorem)
Now, $\text{p}\Big(\frac{3}{2}\Big)=\frac{3}{2}+2\Big(\frac{3}{2}\Big)^3-9\Big(\frac{3}{2}\Big)^2+12$
$=\frac{3}{2}+2\Big(\frac{27}{8}\Big)-9\Big(\frac{9}{4}\Big)+12$
$=\frac{3}{2}+\frac{27}{4}-\frac{81}{4}+12$
$=\frac{6+27-81+48}{4}=\frac{0}{4}=0$
Since, $\text{p}\Big(\frac{3}{2}\Big)=0,$
So $g(x)$ is a factor of $p(x)$.
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