Question

f(y)=2y^3-y^2+2y-1/y^3-y^2+y-1 and g(y)=2y^2-3y+1/4y^2-4y+1. How much is f(y) x g(y)?

Answer 1

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Answer 2

Answer:

The required product is $4y^5-6y^4+10y^3-7y^2+\frac{5}{2}y+\frac{3}{2}-\frac{5}{4y}-\frac{3}{4y^2}-\frac{1}{y^3}$

Step-by-step explanation:

Given

$f(y)=2y^3-y^2+2y-\frac{1}{y^3} -y^2+y-1\\g(y)=2y^2-3y+\frac{1}{4y^2}-4y+1$

we have to find out the product of these two

$f(y)\times g(y)=(2y^3-y^2+2y-\frac{1}{y^3} -y^2+y-1)\times 2y^2-3y+\frac{1}{4y^2}-4y+1$

        =$(2y^3-2y^2+3y-\frac{1}{y^3}-1)\times (2y^2-y+\frac{1}{4y^2}+1)$

which gives

 $4y^5-2y^4+\frac{1}{2}y+2y^3-4y^4+2y^3-\frac{1}{2}-2y^2+6y^3-3y^2+\frac{3}{4y}+3y-\frac{2}{y}-\frac{1}{y^2}-\frac{1}{4y^5}-\frac{1}{y^3}-2y^2-y+\frac{1}{4y^2}+1$

=$4y^5-6y^4+10y^3-7y^2+\frac{5}{2}y+\frac{3}{2}-\frac{5}{4y}-\frac{3}{4y^2}-\frac{1}{y^3}$

which is required product