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.

• A. $\sqrt{5},\sqrt{5}+\sqrt{2},\sqrt{5}-\sqrt{2}$
• B. $\sqrt{5},\sqrt{5},\sqrt{5}-\sqrt{2}$
• C. $\sqrt{5},\sqrt{5}+\sqrt{2},\sqrt{5}$
• D. None of the above

Answer 1

The correct option is A $\sqrt{5},\sqrt{5}+\sqrt{2},\sqrt{5}-\sqrt{2}$

If $(x-\sqrt{5})$ is a factor, then we can write:

$x^3–3\sqrt{5}{x}^2+13x–3\sqrt{5}=(x–\sqrt{5})(x^2+bx+3)$

To determine the coefficient $b$, let's expand the product:

$(x–\sqrt{5})(x^2+bx+3)=x^3+b{x}^2+3x–\left(\sqrt{5}\right)x^2–\left(\sqrt{5}\right)bx–3\sqrt{5}$

$(x–\sqrt{5})(x^2+bx+3)=x^3+(b–\sqrt{5})x^2+(3–b\sqrt{5})x–3\sqrt{5}$

Comparing the right hand side to the original expression, we obtain

$b–\sqrt{5}=-3\sqrt{5}\Rightarrow b=-2\sqrt{5}$, or, with the same result:

$3–b\sqrt{5}=13\Rightarrow b\sqrt{5}=-10\Rightarrow b=-10/\sqrt{5}=-2\sqrt{5}\Rightarrow b=-2\sqrt{5}$

Therefore,

$x^3–3\sqrt{5}{x}^2+13x–3\sqrt{5}=(x–\sqrt{5})(x^2–2\sqrt{5}x+3)x^3–3\sqrt{5}{x}^2+13x–3\sqrt{5}=0(x–\sqrt{5})=0,(x^2–2\sqrt{5}x+3)=0x–\sqrt{5}=0\Rightarrow x=\sqrt{5}{x}^2–2\sqrt{5}x+3=0\Rightarrow x=\sqrt{5}\pm \sqrt{2}$

Hence, the zeros of the given expression are $\sqrt{5}+\sqrt{2},\sqrt{5}-\sqrt{2},\sqrt{5}$.