Question

If two zeroes of the polynomial $x^4-6x^3-26x^2+138x-35$ are $2\pm \sqrt{3},$ find other zeroes.

Answer 1

Interpret the given data and find the factor of the given polynomial.
Since this is a polynomial equation of degree 4, hence there will be total 4 roots,
Let $f\left(x\right)=x^4-6x^3-26x^2+138x-35$
Since $2+\sqrt{3}\text{and}2-\sqrt{3}$ are zeroes of given polynomial $f\left(x\right)$.
$\therefore [x-(2+\sqrt{3}\left)\right][x-(2-\sqrt{3}\left)\right]=0$
$(x-2-\sqrt{3})(x-2+\sqrt{3})=0$
On multiplying the above equation, we get,
$x^2-4x+1,$ this is a factor of a given polynomial $f\left(x\right)$.

To find unknown factor, divide the given polynomial by known factor.

Now, if we will divide $f\left(x\right)$ by $g\left(x\right)$, the quotient will also be a factor of $f\left(x\right)$ and the remainder will be $O$.

$x^2-2x-35$
$x^2-4x+1\sqrt{x^4-6x^3-26x^2+138x-35}$

$\frac{x^4-4x^3+x^2}{-2x^3-27x^2+138x-35}$

$\frac{-2x^3+8x^2-2x}{-35x^2+140x-35}$

$\frac{-35x^2+140x-35}{0}$

Find the other zeroes by factorizing the factor.
So, $x^4-6x^3-26x^2+138x-35=(x^2-4x+1)(x^2-2x-35)$
Now, on further factorizing $(x^2-2x-35)$ we get,
$x^2-(7-5)x-35=x^2-7x+5x+35=0$
$x(x-7)+5(x-7)=0$
So, its zeroes are given by :
$x=-5$ and $x=7$.
Therefore, all four zeroes of given polynomial equation are : $2+\sqrt{3},2-\sqrt{3},-5\text{and}7.$