Question

Find the value of m if 2y³+my²+11y+m+3 is exadly divisible by (2y-1)​

Answer 1

2y3 + my2 + 11y + m + 3 is exactly divisible but (2y-1) is

-7

Answer 2

$\large\underline{\sf{Solution-}}$

Given that:

$\sf p(y) = 2y^3+my^2+11y+m+3$

We are also given that, p(y) is divisible by 2y - 1.

So, 2y - 1 is a factor of p(y).

$\sf\longmapsto 2y-1=0$

$\sf\longmapsto y=\dfrac{1}{2}$

So, 1/2 is a zero of p(y).

$\sf\longmapsto p(1/2) =0$

$\sf\longmapsto p(y) = 2y^3+my^2+11y+m+3$

$\sf\longmapsto p\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{2}\right)^3+m\left(\dfrac{1}{2}\right)^2+11\left(\dfrac{1}{2}\right)+m+3$

So,

$\sf\longmapsto p\left(\dfrac{1}{2}\right) = 2\left(\dfrac{1}{8}\right)+m\left(\dfrac{1}{4}\right)+11\left(\dfrac{1}{2}\right)+m+3$

$\sf\longmapsto p\left(\dfrac{1}{2}\right) = \dfrac{1}{4}+\dfrac{m}{4}+\dfrac{11}{2}+m+3$

$\sf\longmapsto p\left(\dfrac{1}{2}\right) =\dfrac{1}{4}+\dfrac{m}{4}+\dfrac{22}{4}+\dfrac{4m}{4}+\dfrac{12}{4}$

$\sf\longmapsto 0 =\dfrac{1+m+22+4m+12}{4}$

$\sf\longmapsto 0 =\dfrac{5m+35}{4}$

$\sf\longmapsto 5(m+7)=4\times0$

$\sf\longmapsto 5(m+7)=0$

$\sf\longmapsto m+7=0$

$\longmapsto\bf m=-7$

Therefore, the value of m is -7.