Question

Find all zeroes of the polynomial if $2$ of its zeroes are $\pm \sqrt{3}$.
$f\left(x\right)=x^4-3x^3-7x^2+12$

Answer 1

Since, it is given that $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of the polynomial $f\left(x\right)=x^4-3x^3-7x^2+9x+12$, therefore, $(x-\sqrt{3})$ and $(x+\sqrt{3})$ are also the zeroes of the given polynomial. Now, consider the product of zeroes as follows:

$(x-\sqrt{3})(x+\sqrt{3})=(x{)}^2-(\sqrt{3}{)}^2(\because a^2-b^2=(a+b\left)\right(a-b\left)\right)=x^2-3$

We now divide $x^4-3x^3-7x^2+9x+12$ by $(x^2-3)$ as shown in the above image:

From the division, we observe that the quotient is $x^2-3x-4$ 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:

$x^2-3x-4=0\Rightarrow x^2+x-4x-4=0\Rightarrow x(x+1)-4(x+1)=0\Rightarrow (x-4)(x+1)=0\Rightarrow (x-4)=0,(x+1)=0\Rightarrow x=4,x=-1$

Hence, the other two zeroes of $f\left(x\right)=x^4-3x^3-7x^2+9x+12$ are $-1,4$.