## Algebra 1

Published by Prentice Hall

# Chapter 8 - Polynomials and Factoring - 8-4 Multiplying Special Cases - Practice and Problem-Solving Exercises - Page 496: 54

#### Answer

$(x+y)^{2} \ne x^{2} + y^{2}$ as this is untrue and can be proven by substitution of numbers into the equation.

#### Work Step by Step

To prove that $(x+y)^{2} \ne x^{2} + y^{2}$ We take numbers, for example, x=5 and y=6 and substitute them into the equation $(x+y)^{2} = x^{2} + y^{2}$ $(5+6)^{2} = 5^{2} + 6^{2}$ $(11)^{2}$ = 25 + 36 121 $\ne$ 61 Therefore we can prove that $(x+y)^{2} \ne x^{2} + y^{2}$

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