Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 3 - Polynomial and Rational Functions - 3.3 Zeros of Polynomial Functions - 3.3 Exercises - Page 337: 60

Answer

$$\color{blue}{\bf{f(x)= -2x^3-8x^2 }}$$

Work Step by Step

We are given three zeros of a polynomial and the value of the function for a given ${x}$: Our zeros are $\bf{ -4}$, and $\bf{0 }$ of multiplicity 2, so let's make them into factors of our polynomial. If $\bf{ -4 }$ is a zero then: $( -4 +n)=0$ $n= 4 $ so our factor is $\bf{(x +4 )}$ because when $x=\bf{ -4 }$, $(x+4 )=0$ If $\bf{ 0 }$ is a zero then: $( 0 +n)=0$ $n= 0 $ so our factor is $\bf{(x )}$ because when $x=\bf{ 0 }$, $(x +0 )=0$ Since $0$ is a zero of multiplicity 2, $(x)$ is a factor twice: $(x)^2$ or $(x)(x)$ Now we have three factors of our 3rd degree polynomial: $\bf{(x +4 )(x )(x )}$, which, multiplied by some unknown factor $a$, make up our function: $f(x)=a(x +4 )(x )(x )$ If $f( -1 )= -6 $ then ${(-1+4 )( -1 )( -1 )}$ times some number $a$, equals $ -6 $ or: $f( -1 )= -6 =a(-1+4 )( -1 )( -1 ) $ $ -6 =a(3 )( -1 )( -1 ) $ $ -6 =a(3 ) $ $-\dfrac{6}{3}=a$ $\bf{a=-2 }$ Now that we have the value of $a$, we can find $f(x)$ $f(x)=-2 (x +4 )(x )(x ) $ $f(x)=-2 (x +4 )(x^2) $ $f(x)=-2 (x^3+4x^2) $ $$\color{blue}{\bf{f(x)= -2x^3-8x^2 }}$$
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