Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.2 - Factoring Trinomials of the Form x2+bx+c - Exercise Set: 90

Answer

$c=3$ $c=4$

Work Step by Step

If $y^2-4y+c=(y+p)(y+q)$, then $c=p\bullet q$ $p+q=-4$ Since $c$ must be positive and the sum of $p$ and $q$ is a negative value, that means $p$ and $q$ must both be negative. What two negative numbers add up to $-4$? $-1$ and $-3$ $-2$ and $-2$ These are the possible values of $p$ and $q$. Since $c=p\bullet q$, the product of the number combinations above will give us the values of $c$. $-1\bullet-3=3$ $-2\bullet-2=4$ Therefore, $c=3$ and $c=4$.
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