Question
Using factor theorem, show that g(x) is a factor of p(x), when$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6,\ $ $\text{g}(\text{x})=\text{x}-\sqrt2$
✓
Answer
$\text{p}(\text{x})=7\text{x}^2-4\sqrt2\text{x}-6$By the factor theorem, (x - a) will be factor of p(x) if(a) = 0.
Here, $\text{f}(\sqrt{2})=7(\sqrt{2})^2-4\sqrt{2}\times\sqrt{2}-6$
$=14-8-6$
$=14-14=0$
$\therefore(\text{x}-\sqrt2)$ is a factor of $\big(7\text{x}^2-4\sqrt2\text{x}-6\big).$
Here, $\text{f}(\sqrt{2})=7(\sqrt{2})^2-4\sqrt{2}\times\sqrt{2}-6$
$=14-8-6$
$=14-14=0$
$\therefore(\text{x}-\sqrt2)$ is a factor of $\big(7\text{x}^2-4\sqrt2\text{x}-6\big).$
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