Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 5 - Section 5.6 - A General Factoring Strategy - Exercise Set - Page 380: 96

Answer

False

Work Step by Step

The trinomial $4x^2+8x+3$ has the discriminant $8^2-4(4)(3)=16$, so its roots are: $$\begin{align*} x&=\dfrac{-8\pm\sqrt{16}}{2(4)}=\dfrac{-8\pm 4}{8}\\ x_1&=-\dfrac{12}{8}=-\dfrac{3}{2}\\ x_2&=-\dfrac{4}{8}=-\dfrac{1}{2}. \end{align*}$$ We factor the trinomial: $$4x^2+8x+3=4\left(x+\dfrac{3}{2}\right)\left(x+\dfrac{1}{2}\right)=(2x+3)(2x+1).$$ The trinomial $4x^2+8x+1$ has the discriminant $8^2-4(4)(1)=48$, so its roots are: $$\begin{align*} x&=\dfrac{-8\pm\sqrt{48}}{2(4)}=\dfrac{-8\pm 4\sqrt 3}{8}=\dfrac{-2\pm \sqrt 3}{2}\\ x_1&=\dfrac{-2-\sqrt 3}{2}\\ x_2&=\dfrac{-2+\sqrt 3}{2}. \end{align*}$$ We factor the trinomial: $$4x^2+8x+1=4\left(x+\dfrac{2+\sqrt 3}{2}\right)\left(x+\dfrac{2-\sqrt 3}{2}\right)=(2x+2+\sqrt 3)(2x+2-\sqrt 3).$$ Therefore the given statement is FALSE.
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