College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter R - Section R.5 - Factoring Polynomials - R.5 Exercises: 99

Answer

$2(x+1)(3x+1)$

Work Step by Step

Factor out $2$ to obtain: $=2(3x^2+4x+1)$ 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 $a=3$, $b=4$, and $c=1$. Thus, $ac = 3(1) = 3$ Note that $3=3(1)$ and $3+1=4$ This means that $d=3$ and $e=1$. Rewrite the middle term of the trinomial as $3x$ +($x$) to obtain: $3x^2+4x+1 =3x^2+3x+x+1 $ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(3x^2+3x)+(x+1) \\=3x(x+1) +(1)(x+1)$ Factor out the GCF $x+1$ to obtain: $=(x+1)(3x+1)$ Therefore the completely factored form of the given expression is: $=2(x+1)(3x+1)$
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