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

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

defazajx
2021-05-09
Answered

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

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

You can still ask an expert for help

insonsipthinye

Answered 2021-05-11
Author has **83** answers

Step 1

Given:

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

First check the solution of cubic equation by putting the different value of x.

If x=1

$P\left(1\right)=4{\left(1\right)}^{3}-7\left(1\right)+3=7-7=0$

So, x = 1 is the solution of given cubic equation.

$4{x}^{3}-7x+3$ is divided by x-1

$\frac{4{x}^{3}-7x+3}{x-1}=4{x}^{2}+4x-3$

Step 2

Factorize$4{x}^{2}+4x-3$

$4{x}^{2}-2x+6x-3=0$

2x(2x-1)+3(2x-1)=0

(2x-1)(2x+3)=0

(2x-1)=0 or (2x+3)=0

2x=1 or 2x=-3

$x=\frac{1}{2}$ or $x=-\frac{3}{2}$

So,

Zeroes of polynomial are: x=1,$x=\frac{1}{2},x=-\frac{3}{2}$

Given:

First check the solution of cubic equation by putting the different value of x.

If x=1

So, x = 1 is the solution of given cubic equation.

Step 2

Factorize

2x(2x-1)+3(2x-1)=0

(2x-1)(2x+3)=0

(2x-1)=0 or (2x+3)=0

2x=1 or 2x=-3

So,

Zeroes of polynomial are: x=1,

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for which the attempted solution is to convert the sine terms into complex natural exponents (engineering notation using j as imaginary unit) as

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the next step in the solution moves the$\frac{1}{2j}$ term outside of the integral to form

$\frac{-{A}^{2}}{4\tau}{\int}_{0}^{\tau}({e}^{j\mathrm{\Omega}t}-{e}^{-j\mathrm{\Omega}t})\cdot ({e}^{j\mathrm{\Omega}(t-\lambda )}-{e}^{-j\mathrm{\Omega}(t-\lambda )})dt$

I'm struggling to understand how$\frac{1}{2j}\to \frac{-1}{4}$ when being moved out of the integral.

for which the attempted solution is to convert the sine terms into complex natural exponents (engineering notation using j as imaginary unit) as

the next step in the solution moves the

I'm struggling to understand how

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