Question
Using factor theorem, show that g(x) is a factor of p(x), when$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$ $\text{g}(\text{x})=\text{x}+\sqrt2$
✓
Answer
$\text{p}(\text{x})=2\sqrt2\text{x}^2+5\text{x}+\sqrt2,$By the factor theorem, (x - a) will be factor of p(x) if p(a) = 0.
Here, $\text{f}(-\sqrt{2})=2\sqrt2(-\sqrt2)^2+5(-\sqrt2)+\sqrt2$
$=2\sqrt2\times2-5\sqrt2+\sqrt2$
$=4\sqrt2-5\sqrt2+\sqrt2$
$=5\sqrt2-5\sqrt2=0.$
$\therefore(\text{x}+\sqrt2)$ is a factor of $2\sqrt2\text{x}^2+5\text{x}+\sqrt2.$
Here, $\text{f}(-\sqrt{2})=2\sqrt2(-\sqrt2)^2+5(-\sqrt2)+\sqrt2$
$=2\sqrt2\times2-5\sqrt2+\sqrt2$
$=4\sqrt2-5\sqrt2+\sqrt2$
$=5\sqrt2-5\sqrt2=0.$
$\therefore(\text{x}+\sqrt2)$ is a factor of $2\sqrt2\text{x}^2+5\text{x}+\sqrt2.$
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