Chapter 4.3, Problem 29WE

To determine

To multiply given expression assuming variable exponents represent positive integers

Answer to Problem 29WE

  $\left(y^4-3y^2+1\right)\left(y^2-2\right)=y^6-5y^4+7y^2-2$

Explanation of Solution

Given information : 

Given expression $\left(y^4-3y^2+1\right)\left(y^2-2\right)$

Concept used:

Basic rule of multiplication of trinomial and binomial

  $\left(a+b+c\right)\left(x+y\right)=a\left(x+y\right)+b\left(x+y\right)+c\left(x+y\right)$

Basic rule of multiplication of monomial and binomial

  $x\left(a+b\right)=\left(x\times a\right)+\left(x\times b\right)$

Rule of exponents

  $a^m.a^n=a^{m+n}$

Explanation: 

  $\begin{array}{l}\left(y^4-3y^2+1\right)\left(y^2-2\right)\\ =y^4\left(y^2-2\right)-3y^2\left(y^2-2\right)+1\left(y^2-2\right)\\ =\left(y^4\times y^2\right)-2y^4-\left(3y^2\times y^2\right)+\left(3y^2\times 2\right)+y^2-2\\ =y^{4+2}-2y^4-3y^{2+2}+6y^2+y^2-2\\ =y^6-2y^4-3y^4+7y^2-2\\ =y^6-5y^4+7y^2-2\end{array}$