Question

If two zero of the polynomial $p\left(x\right)=2x^4-3x^3-3x^2+6x-2$ are$\sqrt{-2}$ and $\sqrt{2}$, find its other two zeroes.

Answer 1

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

$(x-\sqrt{2})(x+\sqrt{2})=(x{)}^2-(\sqrt{2}{)}^2(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)=x^2-2$

We now divide $2x^4-3x^3-3x^2+6x-2$ by $(x^2-2)$ as shown in the above image:

From the division, we observe that the quotient is $2x^2-3x+1$ and the remainder is $0$.

Now, we factorize the quotient $2x^2-3x+1$ by equating it to $0$ to find the other zeroes of the given polynomial:

$2x^2-3x+1=0\Rightarrow 2x^2-2x-x+1=0\Rightarrow 2x(x-1)-1(x-1)=0\Rightarrow (2x-1)(x-1)=0\Rightarrow (2x-1)=0,(x-1)=0\Rightarrow x=\frac{1}{2},x=1$

Hence, the other two zeroes of $p\left(x\right)=2x^4-3x^3-3x^2+6x-2$ are $\frac{1}{2},1$.