Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 5 - Polynomials and Factoring - Connecting: The Concepts - Exercises - Page 350: 10


$\left\{2, 8\right\}$

Work Step by Step

Subtract $10x$ to both sides: $x^2+16-10x=10x-10x \\x^2-10x+16=0$ RECALL: A trinomial of the form $x^2+bx+c$ can be factored if there are integers $d$ and $e$ such that $c=de$ and $b=d+e$. The trinomial's factored form will be: $x^2+bx+c=(x+d)(x+e)$ The trinomial above has $b=-10$ and $c=16$. Note that $16=-8(-2)$ and $-10= -8+(-2)$. This means that $d=-8$ and $e=-2$ Thus, the factored form of the trinomial is: $[x+(-8)][x+(-2)] = (x-8)(x-2)$ The given equation may be written as: $(x-8)(x-2)=0$ Use the Zero-Product Property by equating each factor to zero. Then, solve each equation to obtain: $\begin{array}{ccc} &x-8= 0 &\text{ or } &x-2=0 \\&x=8 &\text{ or } &x=2 \end{array}$ Thus, the solution set is $\left\{2, 8\right\}$.
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