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 - Page 58: 122



Work Step by Step

The trinomial $x^2-6x+9$ is a perfect square trinomial since the square of half of the middle term's coefficient, which is $(\frac{-6}{2})^2=(-3)^2$, is equal to the third term of the trinomial. This means that the factored form of the trinomial is $(x-3)^2$. Thus, the given expression is equivalent to: $=7(x-3)^2+5(x-3)$ Factor out $x-3$ to obtain: $=(x-3)[7(x-3)+5] \\=(x-3)(7x-21+5) \\=(x-3)(7x-16)$
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