## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 6 - Section 6.2 - Factoring Trinomials of the Form x2+bx+c - Exercise Set: 87

#### Answer

Chapter 6 - Section 6.2 - Exercise Set: 87 (Answer) $z^2(x + 1) - 3z(x + 1) - 70(x + 1)$ = $(x + 1)(z - 10)(z + 7)$

#### Work Step by Step

Chapter 6 - Section 6.2 - Exercise Set: 87 (Solution) Factorize : $z^2(x + 1) - 3z(x + 1) - 70(x + 1)$ First step : Take out the GCF $(x + 1)$. $z^2(x + 1) - 3z(x + 1) - 70(x + 1)$ = $(x + 1)(z^2 - 3z - 70)$ Take $(z^2 - 3z - 70)$ to be $(z + \triangle)(z + \square)$ For this, we have to look for two numbers whose product is -70 and whose sum is -3. Factors of -70 $\Longleftrightarrow$ Sum of Factors 1,-70 $\Longleftrightarrow$ -69 (Incorrect sum) 2,-35 $\Longleftrightarrow$ -33 (Incorrect sum) 5,-14 $\Longleftrightarrow$ -9 (Incorrect sum) 7,-10 $\Longleftrightarrow$ -3 (Correct sum, so the two numbers are 7 and -10) … no need to consider more trials as 7 and -10 matched the criteria already Thus, $z^2(x + 1) - 3z(x + 1) - 70(x + 1)$ = $(x + 1)(z - 10)(z + 7)$

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