Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 7 - Exponential and Logarithmic Functions - 7-1 Exploring Exponential Models - Practice and Problem-Solving Exercises - Page 441: 55

Answer

$(3x - 1)(x + 4)$

Work Step by Step

To factor a quadratic polynomial in the form $ax^2 + bx + c$, we look at factors of $(a)(c)$ such that, when added together, equal the $b$ term. For the expression $3x^2 + 11x - 4$, we look for the factors that, when multiplied together, will equal $(a)(c)$, which is $(3)(-4)$ or $-12$, but when adding the factors together will equal $b$ or $11$. We need for one factor to be negative and the other factor to be positive, but the positive number must have the greater absolute value. A negative number multiplied with a positive number equal a negative number, but when added together, can make a positive or negative number, depending on the sign of the factor with the greater absolute value. Let's look at some possible factors: $(a)(c)$ = $(12)(-1)$ $b = 11$ $(a)(c)$ = $(4)(-3)$ $b = 1$ $(a)(c)$ = $(6)(-2)$ $b = 4$ The first pair works. We will use that pair to split the middle term: $3x^2 + 12x - x - 4$ Now, we can factor by grouping. We group the first two terms together and the second two terms together: $(3x^2 + 12x) + (-x - 4)$ We see that $3x$ is a common factor for the first group, and $-1$ is a common factor for the second group, so let's factor those out: $3x(x + 4) - (x + 4)$ We see that $x + 4$ is common to both groups, so we put that binomial in parentheses. The other binomial will be $3x - 1$, which is composed of the coefficients in front of the binomials. We now have the two factors: $(3x - 1)(x + 4)$
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