Question

given that x-root 5 is a factor of the cubic polynomial x^3-3root5x^2+13x-3root5,find all the zeros of the polynomial

Answer 1

hope that helps dear friend! you just have to divide givenp(x) with the factor and then solve the resulting quotient with factorisation or quadratic formula.

Answer 2

Answer:

$\text{The other two zeroes are }\sqrt5\pm\sqrt2$

Step-by-step explanation:

Given that $x-\sqrt5$ is the factor of cubic polynomial $x^3-3\sqrt5x^2+13x-3\sqrt5$.

we have to find all the zeroes of the polynomial.

As given $x-\sqrt5$ is the factor

Hence, by using synthetic division as shown in attachment the quotient can be written as

$x^2-2\sqrt5x+3$

In order to find zeroes put the quotient equals to 0.

$x^2-2\sqrt5x+3=0$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\=\frac{2\sqrt5\pm \sqrt{20-4(1)(3)}}{2(1)}\\\\=\frac{2\sqrt5\pm \sqrt8}{2}=\sqrt5\pm\sqrt2$

$\text{The other two zeroes are}\sqrt5\pm\sqrt2$