Question

is x-√5 is a factor of x^3-3√5x^2-5x+15√5

Answer 1

(√5)^3-3√5(√5)^2-5(√5)+15√5
5√5-15√5-5√5+15√5
=0
Yes x-√5 is a factor ofx^3-3√5x^2-5x+15√5

Answer 2

SOLUTION :

Let P(x) = $\bf x^{3}-3\sqrt{5}x^{2}-5x+15\sqrt{5}$

and

g(x) = $\bf x-\sqrt{5}$

⇒ $\bf x=\sqrt{5}$

now,

$\bf P(\sqrt{5} ) = (\sqrt{5})^{3}-3\sqrt{5}(\sqrt{5})^{2}-5(\sqrt{5})+15\sqrt{5}$

          $\bf = 5\sqrt{5}-3\times 5\sqrt{5}-5\sqrt{5}+15\sqrt{5}$

          $\bf = \cancel{ 5\sqrt{5} } -15\sqrt{5} \cancel{ -5\sqrt{5} } +15\sqrt{5}$

         $\bf = \cancel{-15\sqrt{5}} \cancel{ +15\sqrt{5}}$

          $\bf =0$

Hence,

g(x) is the factor of P(x)