Question

please give it answer , please fast ​

Answer 1

We don't need to divide since there is 'remainder theorem' here.

If we use division algorithm here, we have

q(x)=r(x)Q(x)+R

"Since r(x) is linear, remainder is a constant. Therefore R is a constant."

We found out that no matter the value we put in, R doesn't change.

So we put in -1/2 (α here) as the value of x in order to make r(α)=0

q(α)=0Q(α)+R

∴q(α)=R

q(-1/2)=R

$q(\alpha)=(2\times \frac{-1}{8}) -(11\times \frac{1}{4} )-(4\times \frac{1}{2} )+5$

$=(-\frac{1}{4} )-\frac{11}{4} -\frac{16}{4} +\frac{20}{4} =\frac{20-16-11-1}{4}$

$=\frac{20-16-11-1}{4}=\frac{-8}{4}$

$=-2$

Since it has a remainder of -2, r(x) doesn't divide q(x).

Therefore, it is not a multiple.

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