Question

check whether the polynomial Polynomial is

Answer 1

$ANSWER:\\ \\ 2y+1=0\\ =>2y= -1 \\ \\=>y=\frac{-1}{2}\:\:....(1)\\ \\ 4y^{3} + 4y^{2} - y-1 \:....(2) \\\\ Put\: (1) \: in (2)\\\\=>4\times(\frac{-1}{2})^{3}+4\times(\frac{-1}{2})^{2}- (\frac{-1}{2})-1\\\\=>\frac{-1}{2}+1+\frac{1}{2}-1\\ \\=0\\\\Hence,\: it \:is \:a\: multiple.$

Answer 2

Answer:yes

Step-by-step explanation:

2y+1=0

y=-1/2

p(y)=4*(-1/2)³+4*(-1/2)²-(-1/2)-1

4*-1/8+4*1/4+1/2-1

-1/2+1+1/2-1

0

So p(y) is a multiple of 2y+1.....

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