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 - Page R-8: 14



Work Step by Step

$15y^{2}+y-2$ Since the coefficient of $y^{2}$ is not equal to $1$, begin by multiplying the whole expression by $15$, which is the actual coefficient of $y^{2}$. Leave the product between $15$ and the second term expressed: $15(15y^{2}+y-2)=...$ $...=225y^{2}+15(y)-30$ Open two parentheses containing initially the square root of the second term, which is $15y$, followed by the sign of the second term, in the first parentheses, and the product of the signs of the second and third terms, on the second parentheses: $...=(15y+)(15y-)$ Find two numbers whose product is equal to the third term, $-30$ and whose sum is equal to the coefficient of the expression inside parentheses in the second term, $1$. These two numbers are $6$ and $-5$, because $(6)(-5)=-30$ and $6-5=1$. $...=(15y+6)(15y-5)$ The expression was affected initially by multiplying it by $15$. Divide it by $15$ to obtain the answer: $...=\dfrac{(15y+6)(15y-5)}{15}=...$ Substitute $15$ by $3\cdot5$ and divide the first parentheses by $3$ and the second parentheses by $5$: $...=\dfrac{(15y+6)(15y-5)}{3\cdot5}=...$ $...=\dfrac{(15y+6)}{3}\dfrac{(15y-5)}{5}=...$ $...=(5y+2)(3y-1)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.