Question
Prove by factor theorem that
$(2x - 1)$ is a factor of $6x^3 - x^2 - 5x +2$
$(2x - 1)$ is a factor of $6x^3 - x^2 - 5x +2$
✓
Answer
$(2 x-1)$ is a factor of $6 x^3-x^2-5 x+2$
$2 x-1=0 \Rightarrow x=\frac{1}{2}$
Substituting this value, we get
$f \left(\frac{1}{2}\right)=6 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{1}{2}-5 \times \frac{1}{2}+2$
$=\frac{3}{4}-\frac{1}{4}-\frac{5}{2}+2$
$=\frac{1}{2}-\frac{5}{2}+2=-2+2=0$
Hence $(2 x-1)$ is a factor of $6 x^3-x^2-5 x+2$
$2 x-1=0 \Rightarrow x=\frac{1}{2}$
Substituting this value, we get
$f \left(\frac{1}{2}\right)=6 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{1}{2}-5 \times \frac{1}{2}+2$
$=\frac{3}{4}-\frac{1}{4}-\frac{5}{2}+2$
$=\frac{1}{2}-\frac{5}{2}+2=-2+2=0$
Hence $(2 x-1)$ is a factor of $6 x^3-x^2-5 x+2$
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