# Chapter 5 - Section 5.1 - Introduction to Polynomials and Polynomial Functions - Exercise Set - Page 325: 43

$22y^5+9y^4+7y^3-13y^2+3y-11$

#### Work Step by Step

Consider the given polynomial $(13y^5+9y^4-5y^2+3y+6)-(-9y^5-7y^3+8y^2+11)$ Need to drop the parentheses and will have to change the sign of each term of the second polynomial because of the presence of the negative outside. Thus, $(13y^5+9y^4-5y^2+3y+6)-(-9y^5-7y^3+8y^2+11)=13y^5+9y^4-5y^2+3y+6+9y^5+7y^3-8y^2-11$ or, $=13y^5+9y^5+9y^4+7y^3-5y^2+3y+6-8y^2-11$ or, $=22y^5+9y^4+7y^3-13y^2+3y-11$

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