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{{ printedBook.courseTrack.name }} {{ printedBook.name }} We want to find the zeros and sketch the graph of the given polynomial function. $f(x)=x_{3}−5x_{2}−4x+20 $ Let's do these things one at a time.

$x_{3}−5x_{2}−4x+20=0$

Factor

FactorOutFactor out $x_{2}$

$x_{2}(x−5)−4x+20=0$

FactorOutFactor out $-4$

$x_{2}(x−5)−4(x−5)=0$

FactorOutFactor out $(x−5)$

$(x−5)(x_{2}−4)=0$

FacDiffSquares$a_{2}−b_{2}=(a+b)(a−b)$

$(x−5)(x−2)(x+2)=0$

$(x−5)(x−2)(x+2)=0$

Solve using the Zero Product Property

ZeroProdPropUse the Zero Product Property

$x−5=0x−2=0x+2=0 (I)(II)(III) $

$x=5x−2=0x+2=0 $

$x=5x=2x+2=0 $

$x=5x=2x=-2 $

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$ | $x_{3}−5x_{2}−4x+20$ | $f(x)=x_{3}−5x_{2}−4x+20$ |
---|---|---|

$-3$ | $(-3)_{3}−5(-3)_{2}−4(-3)+20$ | $-40$ |

$0$ | $0_{3}−5(0)_{2}−4(0)+20$ | $20$ |

$4$ | $4_{3}−5(4)_{2}−4(4)+20$ | $-12$ |

$6$ | $6_{3}−5(6)_{2}−4(6)+20$ | $32$ |

The points $(-3,-40),$ $(0,20),$ $(4,-12),$ and $(6,32)$ are on the graph of the function. Now, we will determine the leading coefficient and degree of the polynomial function.
$f(x)=x_{3}−5x_{2}−4x+20⇕f(x)=1x_{3}−5x_{2}−4x+20 $
We can see now that the leading coefficient is $1,$ which is a positive number. Also, the degree is $3,$ which is an odd number. Therefore, the end behavior is **down** and **up**. With this in mind, let's plot the zeros and the obtained points, and graph the function.