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The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

Apr 21, 2018

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The polynomial f(x) has degree 3. If f(-1) = 15, f(0)= 0, f(1) = -5, and f(2) = 12, then what are the x-intercepts of the graph of f?

$$f(x)=ax^3\;\;so\;\; d=0\\ f(x)=ax^3+bx^2+cx\\~\\ f(-1)= -a+b-c=15\\ f((1)=a+b+c=-5\\ f(2)=8a+4b+2c=12$$

$$\begin{bmatrix} 1 && 1 && 1&&|&&-5\\ -1 && 1 && -1&&|&&15\\ 8 && 4 && 2 && | && 12 \end{bmatrix}\\~\\ *\text{row 2 replaced with row 2 + row 1}\\~\\ \begin{bmatrix} 1 && 1 && 1&&|&&-5\\ 0 && 2 && 0&&|&&10\\ 8 && 4 && 2 && | && 12\\ \end{bmatrix}\\~\\ *\text{row 3 replaced with row Row 3 - 8* row 1 }\\~\\ \begin{bmatrix} 1 && 1 && 1&&|&&-5\\ 0 && 2 && 0&&|&&10\\ 0 && -4 && -6 && | && 52\\ \end{bmatrix}\\~\\$$

2b=10

b=5

-4b-6c=52 becomes

-20-6c=52

-6c=72

c=-12

a+b+c=-5   becomes

a+5-12=-5

a=2

$$f(x)=2x^3+5x^2-12x\\ f(x)=x(2x^2+5x-12)\\$$

roots are  x=0 and

$$x = {-5 \pm \sqrt{25+96} \over 4}\\ x = {-5 \pm 11 \over 4}\\ x=0\qquad x=1.5\qquad x=-4$$

They are the x intercepts.

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Apr 22, 2018
edited by Melody  Apr 22, 2018