Question

If two zeroes of the polynomial \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\] are $ 2\pm \sqrt{3} $ . Find the other zeroes.

Answer 1

Hint: We first try to express the polynomial in its factorised form. We know that if a is root of polynomial $ p\left( x \right) $ , then $ \left( x-a \right) $ is a root of the polynomial. So, we form a polynomial factor with the given roots $ 2\pm \sqrt{3} $ . We form its polynomial and divide \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\] to find the other polynomial factor. We solve that function to find its roots.

Complete step by step answer:
The given polynomial has a variable with max power of 4 as \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\].
The number of roots for the polynomial is 4. We have been given two out of those 4 roots as $ 2\pm \sqrt{3} $ .
We know that if a is root of polynomial $ p\left( x \right) $ , then $ \left( x-a \right) $ is a root of the polynomial.
So, for \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\] the roots of the polynomials are $ \left\{ x-\left( 2\pm \sqrt{3} \right) \right\} $ .
We can express the given polynomial as the multiplication of the factorised roots form where
\[\begin{align}
  & {{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0 \\
 & \Rightarrow \left\{ x-\left( 2+\sqrt{3} \right) \right\}\left\{ x-\left( 2-\sqrt{3} \right) \right\}g\left( x \right)=0 \\
\end{align}\]
Here, $ g\left( x \right) $ gives us the other two roots of the polynomial. The highest power of the polynomial $ g\left( x \right) $ is 2 as the rest of the roots already have power 2.
We can find the polynomial $ g\left( x \right) $ by dividing polynomial \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35\] with multiplied form of \[\left\{ x-\left( 2+\sqrt{3} \right) \right\}\left\{ x-\left( 2-\sqrt{3} \right) \right\}\]. The multiplied form is
\[\left\{ x-\left( 2+\sqrt{3} \right) \right\}\left\{ x-\left( 2-\sqrt{3} \right) \right\}={{\left( x-2 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}={{x}^{2}}-4x+4-3={{x}^{2}}-4x+1\].
We get $ g\left( x \right) $ as $ g\left( x \right)=\dfrac{{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35}{{{x}^{2}}-4x+1}={{x}^{2}}-2x-35 $ .
We need to find the roots of the polynomial $ {{x}^{2}}-2x-35 $ .
\[\begin{align}
  & {{x}^{2}}-2x-35=0 \\
 & \Rightarrow {{x}^{2}}-7x+5x-35=0 \\
 & \Rightarrow \left( x-7 \right)\left( x+5 \right)=0 \\
\end{align}\]
The other roots are $ x=-5,7 $ .
Therefore, the other zeroes of the polynomial \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\] is $ x=-5,7 $ .

Note:
The given polynomial’s power decides the power of its polynomial factors. The addition of the highest order of the factors is equal to the highest order of the main polynomial.