Khadija Wells

2021-03-20

Find all rational zeros of the polynomial, and write the polynomial in factored form.

$P\left(x\right)=6{x}^{4}-7{x}^{3}-12{x}^{2}+3x+2$

davonliefI

Skilled2021-03-22Added 79 answers

Given

$P\left(x\right)=6{x}^{4}-7{x}^{3}-12{x}^{2}+3x+2$

Answer

$P\left(x\right)=6{x}^{4}-7{x}^{3}-12{x}^{2}+3x+2$

use long division to find zeros of p(x)

from the long division, we get

x=-1, x=2 and$6{x}^{2}-x-1=0$

$x=\frac{-(-1)\pm \sqrt{{(-1)}^{2}-4\left(6\right)(-1)}}{2\left(6\right)}$

$=\frac{1\pm \sqrt{1+24}}{12}$

$=\frac{1\pm \sqrt{25}}{12}$

$=\frac{1\pm 5}{12}$

$x=\frac{6}{12}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x=\frac{-4}{12}$

$x=\frac{1}{2}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}x=-\frac{1}{3}$

Therefore the zeros of p(x) is x =-1,$-\frac{1}{3},\frac{1}{2}$ and 2

Factor of p(x),

$p\left(x\right)=(x+1)(x+\frac{1}{3})(x-\frac{1}{2})(x-2)$

Answer

use long division to find zeros of p(x)

from the long division, we get

x=-1, x=2 and

Therefore the zeros of p(x) is x =-1,

Factor of p(x),

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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