Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 7 - Cumulative Review Exercises - Page 565: 3


$x$ can be either $-6$ or $3$.

Work Step by Step

To solve the equation, let us first subtract $18$ from both sides of the equation to set the equation to zero: $$x^2 + 3x - 18 = 0$$ We can now factor this trinomial. We need to find the factors of $-18$, the constant term, that will add up to the coefficient of the middle term, $3$. Let's look at the factors for $-18$: $-18$ and $1$ or $18$ and $-1$ $-9$ and $2$ or $9$ and $-2$ $-6$ and $3$ or $6$ and $-3$ It looks like $6$ and $-3$ will work. Let's set up the factorization: $$(x + 6)(x - 3) = 0$$ According to the zero product rule, if either of the factors equals zero, then the product of the factors is zero. We can now set each of the factors equal to zero and solve for $x$: $$x + 6 = 0$$ Subtract $6$ from each side to solve for $x$: $$x = -6$$ Let's set the other factor equal to zero: $$x - 3 = 0$$ Add $3$ to each side of the equation to solve for $x$: $$x = 3$$ $x$ can be either $-6$ or $3$.
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