Question

Given that $x-\sqrt{5}$ is a factor of the cubic polynomial $x^3-3\sqrt{5}{x}^2+13x-3\sqrt{5}$ , find all the zeroes of the polynomial .

Answer 1

We know that if is a zero of a polynomial, and then is a factor of .

It is given that $x-\sqrt{5}$ is a factor of = $x^3-3\sqrt{5}{x}^2+13x-3\sqrt{5}$.

Now, we divide $x^3-3\sqrt{5}{x}^2+13x-3\sqrt{5}$ by $x-\sqrt{5}$ to find the other zeroes of .

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∴ Quotient = $x^2-2\sqrt{5}x+3$ and remainder = 0.

By using division algorithm, we have $f\left(x\right)=g\left(x\right)\times q\left(x\right)+r\left(x\right)$.

$$\begin{gathered} f\left(x\right)=\left(x-\sqrt{5}\right)\left(x^2-2\sqrt{5}x+3\right)+0 \\ =\left(x-\sqrt{5}\right)\left[x-\left(\sqrt{5}+\sqrt{2}\right)\right]\left[x-\left(\sqrt{5}-\sqrt{2}\right)\right] \end{gathered}$$

Hence, the zeroes of the given polynomial are $\sqrt{5},\sqrt{5}+\sqrt{2}\mathrm{and}\sqrt{5}-\sqrt{2}$.