Question

If the zeros of the polynomial $\left(x^4-6x^3-26x^2+138x-35\right)$ are $\left(2+\sqrt{3}\right)$ and $\left(2-\sqrt{3}\right)$, find the other zeros.

Answer 1

Since, it is given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are the zeroes of the polynomial $p\left(x\right)=x^4-6x^3-26x^2+138x-35$, therefore, $(x-2-\sqrt{3})$ and $(x-2+\sqrt{3})$ are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows: Now, consider the product of zeroes as follows:

$(x-2-\sqrt{3})(x-2+\sqrt{3})\left[\right(x-2)-\sqrt{3}]\left[\right(x-2)+\sqrt{3}]=(x-2{)}^2-(\sqrt{3}{)}^2(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)=[x^2+2^2-(2\times x\times 2\left)\right]-3(\because (a-b{)}^2=a^2+b^2-2ab)=x^2+4-4x-3=x^2-4x+1$

We now divide $x^4-6x^3-26x^2+138x-35$ by $(x^2-4x+1)$ as shown in the below image:

From the division, we observe that the quotient is $x^2-2x-35$ and the remainder is $0$.

Now, we factorize the quotient $x^2-2x-35$ by equating it to $0$ to find the other zeroes of the given polynomial:

$x^2-2x-35=0\Rightarrow x^2-7x+5x-35=0\Rightarrow x(x-7)+5(x-7)=0\Rightarrow (x+5)(x-7)=0\Rightarrow x=-5,x=7$

Hence, the other two zeroes of $x^4-6x^3-26x^2+138x-35$ are $-5,7$.