Question

"we divide 3y4 – y3 + 12y2 + 2 by 3y2 – 1 then remainder is.​

Answer 1

make me abrailiest.

Thanq.

Answer 2

The remainder would be $\frac{19\sqrt{3}-1}{3\sqrt{3}}$

Step-by-step explanation:

First finding the value of y

$3y^2-1 = 0$

$3y^2=1$

$y^2 =\frac{1}{3}$

$y =\sqrt{\frac{1}{3}}$

Substituting the value of y in the expression, we obtain

$3y^4-y^3+12y^2 + 2$

$3(\sqrt{\frac{1}{3}})^4-(\sqrt{\frac{1}{3}})^3+12(\sqrt{\frac{1}{3}})^2 + 2$

$3(\frac{1}{9}})-(\frac{1}{3}\sqrt{\frac{1}{3}})+12(\frac{1}{3}}) + 2$

$(\frac{1}{3}})-(\frac{1}{3}\sqrt{\frac{1}{3}})+4+ 2$

$(\frac{1}{3}})-(\frac{1}{3\sqrt{3}})+6$

$\frac{\sqrt{3}-1+ 18\sqrt{3}}{3\sqrt{3}}$

$\frac{19\sqrt{3}-1}{3\sqrt{3}}$