Find the value of m so that 2x - 1 is a facto | vedclass

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(B) Let $p(x) = 8x^4 + 4x^3 - 16x^2 + 10x + m$.Since $(2x - 1)$ is a factor of $p(x)$,by the factor theorem,$p(\frac{1}{2}) = 0$.Substituting $x = \fr

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(B) Let $p(x) = 8x^4 + 4x^3 - 16x^2 + 10x + m$.Since $(2x - 1)$ is a factor of $p(x)$,by the factor theorem,$p(\frac{1}{2}) = 0$.Substituting $x = \frac{1}{2}$ into the polynomial:$8(\frac{1}{2})^4 + 4(\frac{1}{2})^3 - 16(\frac{1}{2})^2 + 10(\frac{1}{2}) + m = 0$$8(\frac{1}{16}) + 4(\frac{1}{8}) - 16(\frac{1}{4}) + 5 + m = 0$$\frac{1}{2} + \frac{1}{2} - 4 + 5 + m = 0$$1 - 4 + 5 + m = 0$$2 + m = 0$Therefore,$m = -2$.

Find the value of $m$ so that $2x - 1$ is a factor of $8x^4 + 4x^3 - 16x^2 + 10x + m$.

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