Write the cubic polynomial function f(x) in expanded form with zeros 4, -2, and 1, given that f(-1)=-12

tun1ju2k1ki
2022-10-02
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Piper Pruitt

Answered 2022-10-03
Author has **9** answers

Given that $$f(-1)=-12$$

The cubic polynomial function is

$$f(x)=k(x-a)(x-b)(x-c)$$

where $$a=-4,b=-2,c=1\phantom{\rule{0ex}{0ex}}f(x)=k(x+4)(x+2)(x-1)\phantom{\rule{0ex}{0ex}}f(-1)=k(-1+4)(-1+2)(-1-1)\phantom{\rule{0ex}{0ex}}-12=k(3)(1)(-2)\phantom{\rule{0ex}{0ex}}-12=-6k\phantom{\rule{0ex}{0ex}}k=2$$

put the value in equation

$$f(x)=2[(x+4)(x+2)(x-1)]\phantom{\rule{0ex}{0ex}}=2[({x}^{2}+2x+4x+8)(x-1)]\phantom{\rule{0ex}{0ex}}=2[{x}^{3}+5{x}^{2}+2x-8]\phantom{\rule{0ex}{0ex}}f(x)=2{x}^{3}+10{x}^{2}+4x-16$$

The cubic polynomial function is

$$f(x)=k(x-a)(x-b)(x-c)$$

where $$a=-4,b=-2,c=1\phantom{\rule{0ex}{0ex}}f(x)=k(x+4)(x+2)(x-1)\phantom{\rule{0ex}{0ex}}f(-1)=k(-1+4)(-1+2)(-1-1)\phantom{\rule{0ex}{0ex}}-12=k(3)(1)(-2)\phantom{\rule{0ex}{0ex}}-12=-6k\phantom{\rule{0ex}{0ex}}k=2$$

put the value in equation

$$f(x)=2[(x+4)(x+2)(x-1)]\phantom{\rule{0ex}{0ex}}=2[({x}^{2}+2x+4x+8)(x-1)]\phantom{\rule{0ex}{0ex}}=2[{x}^{3}+5{x}^{2}+2x-8]\phantom{\rule{0ex}{0ex}}f(x)=2{x}^{3}+10{x}^{2}+4x-16$$

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