Question
Prove by factor theorem that
$(3x-2)$ is a factor of $18x^3 - 3x^2 + 6x -12$
$(3x-2)$ is a factor of $18x^3 - 3x^2 + 6x -12$
✓
Answer
$(3 x-2)$ is a factor of $18 x^3-3 x^2+6 x-12$
$3 x-2=0 \Rightarrow x=\frac{2}{3}$
Substituting this value, we get
$f \left(\frac{2}{3}\right)=18 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)-3 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)+6 \times\left(\frac{2}{3}\right)-8 $
$=\frac{16}{3}-\frac{4}{3}+4-8$
$=4+4-8$
$=0$
Hence $(3 x-2)$ Is a factor of $18 x^3-3 x^2+6 x-8$.
$3 x-2=0 \Rightarrow x=\frac{2}{3}$
Substituting this value, we get
$f \left(\frac{2}{3}\right)=18 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)-3 \times\left(\frac{2}{3}\right) \times\left(\frac{2}{3}\right)+6 \times\left(\frac{2}{3}\right)-8 $
$=\frac{16}{3}-\frac{4}{3}+4-8$
$=4+4-8$
$=0$
Hence $(3 x-2)$ Is a factor of $18 x^3-3 x^2+6 x-8$.
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