Question

if x minus root 5 is a factor of the cubic polynomial x cube minus 3 root 5 x square + 13 x minus 3 root 5 find all the zeros of the polynomial​

Answer 1

GIVEN :

If $x-\sqrt{5}$ is a factor of the cubic polynomial $x^3-3\sqrt{5}x^2+ 13 x-3\sqrt{5}$

TO FIND :

The zeroes of the given polynomial​.

SOLUTION :

Given cubic polynomial is $x^3-3\sqrt{5}x^2+ 13 x-3\sqrt{5}$

Also given that $x-\sqrt{5}$ is a factor for the given polynomial.

Let $p(x)=x^3-3\sqrt{5}x^2+ 13 x-3\sqrt{5}$

Since $x-\sqrt{5}$ is a factor for the given polynomial

It must satisfy the given polynomial.

ie., p(x)=0 for $x=\sqrt{5}$

Put $x=\sqrt{5}$ in p(x) we get

$p(\sqrt{5})=(\sqrt{5})^3-3\sqrt{5}(\sqrt{5})^2+ 13(\sqrt{5})-3\sqrt{5}$

$p(\sqrt{5})=(\sqrt{5})^3-3\sqrt{5}(\sqrt{5})^2+ 13(\sqrt{5})-3\sqrt{5}$

$=5\sqrt{5}-3(5)\sqrt{5}+13\sqrt{5}-3\sqrt{5}$

$=5\sqrt{5}-15\sqrt{5}+13\sqrt{5}-3\sqrt{5}$

$=18\sqrt{5}-18\sqrt{5}$

= 0

$p(\sqrt{5})=0$

∴ $x=\sqrt{5}$ is a zero for the given polynomial.

By using Synthetic Division we can find other zeroes :

Since $x=\sqrt{5}$ is a zero for the given polynomial we have,

$\sqrt{5}$ _| 1    $-3\sqrt{5}$     13    $-3\sqrt{5}$

        0     $\sqrt{5}$        -10   $3\sqrt{5}$

        _______________________________________

        1      $-2\sqrt{5}$     3      0

Now we have the quadratic equation

$x^2-2\sqrt{5}x+3=0$  

For Quadratic equation $ax^2+bx+c=0$ then

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-(-2\sqrt{5})\pm \sqrt{(-2\sqrt{5})^2-4(1)(3)}}{2(1)}$

$=\frac{2\sqrt{5}\pm \sqrt{(-2)^2(\sqrt{5})^2-12}}{2}$

$=\frac{2\sqrt{5}\pm \sqrt{4(5)-12}}{2}$

$=\frac{2\sqrt{5}\pm \sqrt{20-12}}{2}$

$=\frac{2\sqrt{5}\pm \sqrt{8}}{2}$

$=\frac{2\sqrt{5}\pm 2\sqrt{2}}{2}$

$=\frac{2(\sqrt{5}\pm \sqrt{2})}{2}$

$=\sqrt{5}\pm \sqrt{2}$

∴ $x=\sqrt{5}+\sqrt{2}$ and $x=\sqrt{5}-\sqrt{2}$

∴ the other zeroes are $\sqrt{5}+\sqrt{2}$ and $\sqrt{5}-\sqrt{2}$ for the given cubic polynomial $x^3-3\sqrt{5}x^2+ 13 x-3\sqrt{5}$