Question

Find all the zeroes of the polynomial $2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$ if you know the two zeroes are $\sqrt{2}$ and \[-\sqrt{2}\]

Answer 1

Hint: We all know if \[x=a\] is a root of a given polynomial then \[\left( x-a \right)\] will divide the given polynomial. Use the property to find the divisions of the polynomial and then divide the polynomial to convert it into a quadratic polynomial.

Complete step by step solution:
Given:
Polynomial: $2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2$
Roots: $\sqrt{2}$ and \[-\sqrt{2}\].
If $\sqrt{2}$ and \[-\sqrt{2}\] are roots of the given polynomial so factors are as
\[\left( x+\sqrt{2} \right)\left( x-\sqrt{2} \right)\]
\[\Rightarrow \left( {{x}^{2}}-2 \right)\]will divide the given polynomial

Now, $2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2=\left( {{x}^{2}}-2 \right)\left( 2{{x}^{2}}-3x+1 \right)$
To find the root of $2{{x}^{2}}-3x+1$ we will use factorisation,
$\Rightarrow 2{{x}^{2}}-2x-x+1=0$
$\Rightarrow 2x\left( x-1 \right)-1\left( x-1 \right)=0$
$\Rightarrow \left( 2x-1 \right)\left( x-1 \right)=0$
$\Rightarrow x=\dfrac{1}{2}\text{ or }x=1$
Hence, \[2{{x}^{4}}-3{{x}^{3}}-3{{x}^{2}}+6x-2=\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)\left( x-\dfrac{1}{2} \right)\left( x-1 \right)\]
Zeroes of the polynomials are \[\sqrt{2},-\sqrt{2},\dfrac{1}{2},1\]

Note: The degree of the polynomial determines the number of roots it will posses for eg. A polynomial of degree n will have n number of roots.
Example: \[{{x}^{3}}-6{{x}^{2}}+11x-6\] has degree \[=3\]
Hence it will have 3 roots which are \[x=1,x=2,x=3\]