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: 97

Answer

$(2x+3)^2$

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 $a=4$, $b=12$, and $c=9$. Thus, $ac = 4(9) = 36$ 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: $4x^2+12x+9 =4x^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: $=(4x^2+6x)+(6x+9) \\=2x(2x+3) +3(2x+3)$ Factor out the GCF $2x+3$ to obtain: $=(2x+3)(2x+3) \\=(2x+3)^2$
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