Question

If $\sqrt{3}$ and $-\sqrt{3}$ are the zeros of p(x)= $x^4+4x^3-8x^2-12x+15$, then find the remaining zeroes of $p\left(x\right)$.

Answer 1

Since, it is given that $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of the polynomial $f\left(x\right)=x^4+4x^3-8x^2-12x+15$, 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 $x^4+4x^3-8x^2-12x+15$ by $(x^2-3)$ as shown in the above image:

From the division, we observe that the quotient is $x^2+4x-5$ and the remainder is $0$.

Now, we factorize the quotient $x^2+4x-5$ as shown below:

$x^2+4x-5=x^2-x+5x-5=x(x-1)+5(x-1)=(x+5)(x-1)$

Either $x+5=0$ or $x-1=0$

Hence, the remaining zeroes of $f\left(x\right)=x^4+4x^3-8x^2-12x+15$ are $-5,1$.