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: 98

Answer

$(3x-2)^2$

Work Step by Step

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 given trinomial has $a=9$, $b=-12$, and $c=4$. Thus, $ac = 9(4) = 36$ Note that $36=-6(-6)$ and $(-6)+(-6)=-12$ This means that $d=-6$ and $e=-6$. Rewrite the middle term of the trinomial as $-6x$ +($-6x$) to obtain: $9x^2-12x+4 =9x^2-6x+(-6x)+9 $ Group the first two terms together and the last two terms together. Then, factor out the GCF in each group to obtain: $=(9x^2-6x)+(-6x+4) \\=3x(3x-2) +(-2)(3x-2)$ Factor out the GCF $3x-2$ to obtain: $=(3x-2)(3x-2) \\=(3x-2)^2$
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