## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter P - Section P.7 - Equations - Exercise Set - Page 106: 57

#### Answer

$\left\{3, 5\right\}$

#### Work Step by Step

Move all terms to the left side. Note that when a term moves to the other side of the equation, the sign changes to its opposite. $x^2-8x+15=0$ With a leading coefficient of $1$, factor the trinomial by looking for factors of the constant term $(15)$ whose sum is equal to the coefficient of the middle term $(-8)$. Note that $15=(-5)(-3)$ and $-5+(-3) = -8$. This means that the factors of the trinomial are $x-5$ and $x-3$. Thus, the factored form of the trinomial is: $(x-5)(x-3)=0$ Equate each factor to zero then solve each equation to obtain: $x-5 = 0 \text{ o } x-3=0 \\x=5 \text{ or } x=3$ The solution set of the given equation is $\left\{3, 5\right\}$.

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