Question

Find all zeros of a polynomial ($2x^4-9x^3+5x^2+3x-1$) if two of its zeros are ($2+\sqrt{3}$) and ($2-\sqrt{3}$).

Answer 1

Since, it is given that $2+\sqrt{3}$ and $2-\sqrt{3}$ are the zeroes of the polynomial $f\left(x\right)=2x^4-9x^3+5x^2+3x-1$, therefore, $(x-(2-\sqrt{3}\left)\right)$ and $(x-(2+\sqrt{3}\left)\right)$ are the factors of the given polynomial.

Now, consider the product of zeroes as follows:

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

We now divide $2x^4-9x^3+5x^2+3x-1$ by $(x^2-4x+1)$ as shown in the above image:

From the division, we observe that the quotient is $2x^2-x-1$ and the remainder is $0$.

So we factorize the quotient $2x^2-x-1$ as follows:

$2x^2-x-1=2x^2-2x+x-1=2x(x-1)+1(x-1)=(2x+1)(x-1)$

Hence, the other zeroes of $f\left(x\right)=2x^4-9x^3+5x^2+3x-1$ are $-\frac{1}{2},1$.