Question

find the value of m for which (2x - 1) close is a factor of
$8x {}^{4} + 4x {}^{3} - 16 {x}^{2} + 10x + m$
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Answer 1

Answer:

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Answer 2

Let me factorise the given polynomial as,

$\begin{aligned}&\sf{8x^4+4x^3-16x^2+10x+m}\\\\=\ &\sf{8x^4-4x^3+8x^3-4x^2-12x^2+6x+4x-2+m+2}\\\\=\ &\sf{4x^3(2x-1)+4x^2(2x-1)-6x(2x-1)+2(2x-1)+m+2}\\\\=\ &\sf{(2x-1)(4x^3+4x^2-6x+2)+(m+2)}\end{aligned}$

This implies $\sf{m+2}$ is the remainder on dividing the polynomial by $\sf{2x-1}$ but it should be zero, i.e.,

$\sf{m+2=0}\\\\\sf{\underline {\underline {m=-2}}}$