Chapter 5 - Section 5.2 - Factoring Trinomials - 5.2 Exercises - Page 337: 55

$13y(y+1)(y-4)$

Work Step by Step

$\bf{\text{Solution Outline:}}$ First, factor the GCF of the terms. Then, 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 $b$. Express the factored form as $GCF(x+m_1)(x+m_2).$ $\bf{\text{Solution Details:}}$ Factoring the $GCF= 13y ,$ the given expression, $13y^3+39y^2-52y ,$ is equivalent to \begin{array}{l}\require{cancel} 13y(y^2+3y-4) .\end{array} In the expression above, the value of $c$ is $-4$ and the value of $b$ is $3 .$ The possible pairs of integers whose product is $ac$ are \begin{array}{l}\require{cancel} \{1,-4\}, \{2,-2\}, \{-1,4\}, \{-2,2\} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ 1,-4 \}.$ Hence, the factored form of the given expression is \begin{array}{l}\require{cancel} 13y(y+1)(y-4) .\end{array}

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