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).

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Solution

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

If p(x) is not a multiple of g(x), the remainder will be a non-zero value.

$p (\frac{- 1}{2}) = 2 (\frac{- 1}{2})^{3} - 11 (\frac{- 1}{2})^{2} - 4 (\frac{- 1}{2}) + 5 = - \frac{1}{4} - \frac{11}{4} + 2 + 5 = \frac{- 12}{4} + 7 = - 3 + 7 = 4 \neq 0$

Therefore p(x) is not a multiple of g(x)

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Similar questions

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

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5 and 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

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

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