Question

find all zeroes of the polynomial 2x4-2x3-7x2+3x+6 if two of its zeroes are-root3/2 and root3/2

Answer 1

(x-√3/2)(x+√3/2)=x²-3/2
g(x)=x²-3/2
p(x)=2x^4-2x³-7x²+3x+6
then q(x)=2x²-2x-4
⇒2x²-2x-4=2x²+2x-4x-4=2x(x+1)-4(x+1)=(2x-4)(x+1)
⇒2x-4=0         or               ⇒  x+1=0
⇒x=2               or                ⇒  x= -1

Answer 2

The zeroes are 2, -1, $\bf \frac{\sqrt3}{\sqrt2}$, and $\bf \frac{-\sqrt3}{\sqrt2}$.

Given:

The polynomial $2x^4-2x^3-7x^2 +3x+6$

Two of its zeroes are $\frac{\sqrt3}{\sqrt2}$ and $\frac{-\sqrt3}{\sqrt2}$

To Find:

The other roots of the polynomial

Solution:

As the two zeroes are given, we can find the factors of the polynomial which are $x-\frac{\sqrt3}{\sqrt2}$ and $x+\frac{\sqrt3}{\sqrt2}$.

Therefore

$[x+\frac{\sqrt3}{\sqrt2}][x-\frac{\sqrt3}{\sqrt2}]= x^2-1.5$

Dividing the polynomial by this expression we will get

$\frac{2x^4-2x^3-7x^2 +3x+6}{x^2-1.5} = \frac{2x^2[x^2-1.5]-2x^3-4x^2+3x+6}{x^2-1.5}\\\\ 2x^2+\frac{-2x^3-4x^2+3x+6}{x^2-1.5}= 2x^2+\frac{-2x[x^2-1.5]-4x^2+6}{x^2-1.5}\\\\2x^2-2x +\frac{-4x^2+6}{x^2-1.5}=2x^2-2x+\frac{-4[x^2-1.5]}{x^2-1.5}=2x^2-2x-4$

$2x^2-2x-4=2x^2-4x+2x-4\\\\2x^2-4x+2x-4=2x[x-2]+2[x-2]\\\\2x[x-2]+2[x-2]=2[x-2][x+1]$

Therefore the zeroes of the polynomial are 2, -1, $\bf \frac{\sqrt3}{\sqrt2}$, and $\bf \frac{-\sqrt3}{\sqrt2}$.

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