Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises: 27

Answer

$(a-3b)(a-6b)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To factor the quadratic expression $x^2+bx+c,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$. Then, express the factored form as $(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ In the given expression, $ a^2-9ab+18b^2 ,$ the value of $c$ is $ 18 $ and the value of $b$ is $ -9 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{1,18\}, \{2,9\}, \{3,6\}, \{-1,-18\}, \{-2,-9\}, \{-3,-6\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -3,-6 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} (a-3b)(a-6b) .\end{array}
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