Question

Find all the zeroes of $2x^4-3x^3-3x^2+6x-2$, if two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}.$

Answer 1

Step 1: Find a factor of the polynomial $p\left(x\right)=2x^4-3x^3-3x^2+6x-2$:

Here the zeroes of the polynomial are $\sqrt{2}$ and $-\sqrt{2}$.

$\begin{array}{rcl}& \Rightarrow & \left(x-\sqrt{2}\right)\left(x-\left(-\sqrt{2}\right)\right)=\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\\ & =& x^2-{\left(\sqrt{2}\right)}^2\\ & =& x^2-2\end{array}$

Thus, $x^2-2$ is a factor of $p\left(x\right)$.

Step 2: Find the quotient using long division method:

Let $g\left(x\right)=x^2-2$.

Divide the polynomial $p\left(x\right)$ by $g\left(x\right)$.

$\begin{array}{ccc}{x}^2& -& 2\end{array}\begin{array}{ccccc}2x^2& -& 3x& +& 1\end{array}\begin{array}{cccccc}2x^4& -3x^3-& 3x^2& +& 6x& -\end{array}\begin{array}{c}2\end{array}\begin{array}{cccccc}2x^4& -4x^2& & & & \end{array}\begin{array}{cccccc}-& & +& & & \end{array}\begin{array}{cccccc}-& 3x^3& +& x^2& +& 6x\\ -& 3x^3& +& 0x^2& +& 6x\\ +& & -& & -& \end{array}\begin{array}{cccc}& x^2& & +0x-2\\ & x^2& & +0x-2\\ & -& & ++\end{array}0$

Since the remainder is $0$, $g\left(x\right)=x^2-2$ is a factor of $p\left(x\right)=2x^4-3x^3-3x^2+6x-2$.

Step 3: Find the other zeroes:

Formula:

$\text{Dividend = Divisior ×Quotient +Remainder}$

$\Rightarrow 2x^4-3x^3-3x^2+6x-2=\left(x^2-2\right)\left(2x^2-3x+1\right)$.

Consider,

$$\begin{gathered} \Rightarrow 2x^2-3x+1=0 \\ \Rightarrow 2x^2-2x-x+1=0 \\ \Rightarrow 2x\left(x-1\right)-1\left(x-1\right)=0 \\ \Rightarrow \left(2x-1\right)\left(x-1\right)=0 \end{gathered}$$

Either $2x-1=0$ or $x-1=0$, that is, $x=\frac{1}{2}$ or $x=1$.

Therefore, the other zeroes are $1$ and $\frac{1}{2}$.