Let $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) = 2 x + 1$ , show that $g (x)$ is not a factor of $p (x)$

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Solution

$\Rightarrow p (x) = 2 x^{7} - 11 x^{2} - 4 x + 5$
$g (x) = 2 x + 1$
Now, if $\frac{p (x)}{g (x)}$ is not divisible or we get a remainder on dividing, we can say $g (x)$ is not a factor of $g (x)$
So, $Q (x) = x^{2} - 5 x$
$R (x) = x + 5$
since we get the remainder, non zero, we can say $g (x)$ is not a factor of $p (x) .$
Hence, solved.

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Similar questions

If $p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5$ and $g (x) = 2 x + 1$ , show that g(x) is not multiple of p(x).

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

Using factor theorem, show that g(x) is a factor of p(x), when
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