# Chapter 5 - Polynomials and Factoring - 5.3 Factoring Trinomials of the Type ax2+bx+c - 5.3 Exercise Set - Page 326: 11

$(5a-3)(3a-1)$

#### Work Step by Step

Using the factoring of trinomials in the form $ax^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} 15a^2-14a+3 \end{array} has $ac= 15(3)=45$ and $b= -14 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ -9,-5 \right\}.$ Using these $2$ numbers to decompose the middle term of the trinomial expression above results to \begin{array}{l}\require{cancel} 15a^2-9a-5a+3 .\end{array} Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (15a^2-9a)-(5a-3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} 3a(5a-3)-(5a-3) .\end{array} Factoring the $GCF= (5a-3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (5a-3)(3a-1) .\end{array}

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