# Chapter R - Section R.5 - Factoring Polynomials - R.5 Exercises: 100

$2(x+1)(4x-1)$

#### Work Step by Step

Factor out $2$ to obtain: $=2(4x^2+3x-1)$ RECALL: A quadratic trinomial $ax^2+bx+c$ may be factored if there are integers $d$ and $e$ such that $de = ac$ and $d+e =b$ If such integers exist, then $ax^2+bx + c = ax^2+dx + ex + c$, and may be factored by grouping. The trinomial above has $a=4$, $b=3$, and $c=-1$. Thus, $ac = 4(-1) = -4$ Note that $-4=4(-1)$ and $4+(-1)=3$ This means that $d=4$ and $e=-1$. Rewrite the middle term of the trinomial as $4x$ +($-x$) to obtain: $4x^2+3x-1 =4x^2+4x+(-x)-1$ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(4x^2+4x)+(-x-1) \\=4x(x+1) +(-1)(x+1)$ Factor out the GCF $x+1$ to obtain: $=(x+1)(4x-1)$ Therefore the completely factored form of the given expression is: $=2(x+1)(4x-1)$

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