Question

Find the value of m if 2x-1 is a factor of 8x4+4x^3-16x^2+10x+m.​

Answer 1

Let p(x) = 8x4 + 4x3 − 16x2 + 10x + m

Given (2x − 1) is a factor of p(x) ⇒ p(1/2) = 0

Put x = 1/2 in p(x) ⇒ p(1/2)

= 8(1/2)4 + 4(1/2)3 − 16(1/2)2 + 10(1/2) + m 0 = 8(1/16) + 4(1/8) − 16(1/4) + 5 + m ⇒ (1/2) + (1/2) − 4 + 5 + m = 0 ⇒ 1 + 1 + m = 0

m= –2

Hope it helps you....

Answer 2

$let \: p(x) = 8 {x}^{4} + 4 {x}^{3} - 16 {x}^{2} + 10x + m$

$g(x)=2x−1 \\ </p><p> Now, g(x)=2x−1 \\ </p><p>=>2x−1=0 \\ </p><p>=>x= \frac{1}{2} </p><p></p><p></p><p>$

$If g(x) is a factor of p(x), then p( \frac{1}{2} ) must be 0 \\ </p><p>Thus, p( \frac{1}{2} )=0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: </p><p></p><p>$

$8 { \frac{1}{2} }^{4} + 4 { \frac{1}{2} }^{3} - 16 { \frac{1}{2} }^{2} + 10 \frac{1}{2} + m = 0$

$8 { \frac{1}{16} } + 4 { \frac{1}{8} } - 16 { \frac{1}{4} } + 10 \frac{1}{2} + m = 0$

$\frac{1}{2} + \frac{1}{2} - 4 + 5 \: + m = 0$

$2 + m = 0$

$m = - 2$