Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter R - Algebra Reference - R.2 Factoring - R.2 Exercises: 18

Answer

$4a^{2}+10a+6=2(2a+3)(a+1)$

Work Step by Step

$4a^{2}+10a+6$ Express $10a$ as $2(5a)$: $4a^{2}+10a+6=...$ $...=4a^{2}+2(5a)+6=...$ Open two parentheses including initially the square root of the first term, which is $2a$, followed by the sign of the second term, in the first parentheses and the product of the signs of the second and third term, in the second parentheses: $(2a+)(2a+)$ Find two numbers whose product is equal to the third term, $6$ and whose sum is equal to the coefficient of the expression inside parentheses in the second term, $5$. These two numbers are $3$ and $2$, because $(3)(2)=6$ and $3+2=5$ $...=(2a+3)(2a+2)=...$ Take out common factor $2$ from the second parentheses to finish the factoring process: $...=2(2a+3)(a+1)$
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