Question

Find the zeroes of the polynomial $x^2+\frac{1}{6}x-2$ and verify the relation between the coefficient and zeroes of the polynomial.

Answer 1

Step 1: Form the equation

The given polynomial is $x^2+\frac{1}{6}x-2$.

We know that the zeroes of a polynomial are evaluated by equating it with zero.

$\therefore p\left(x\right)=0$

$\Rightarrow x^2+\frac{1}{6}x-2=0$

Step 2: Solve to find the zeroes

$x^2+\frac{1}{6}x-2=0$

$\Rightarrow$ $6x^2+x-12=0$

$\Rightarrow$ $6x^2-8x+9x-12=0$

$\Rightarrow$$2x\left(3x-4\right)+3\left(x-4\right)=0$

$\Rightarrow$ $\left(3x-4\right)\left(2x+3\right)=0$

$\Rightarrow$ $x=\frac{4}{3},-\frac{3}{2}$

Step 3: Verify the relation between coefficient

Let $a$ and $b$ be the zeroes of the given polynomial, where $a=\frac{4}{3},b=-\frac{3}{2}$

A polynomial can also be written as,

$x^2-\left(a+b\right)x+ab=0$

$\Rightarrow x^2-\left(\frac{4}{3}-\frac{3}{2}\right)x+\left\{\frac{4}{3}\times \left(-\frac{3}{2}\right)\right\}=0$

$\Rightarrow$ $x^2-\frac{8-9}{6}x+\left(-2\right)=0$

$\Rightarrow$ $x^2+\frac{1}{6}x-2=0$

This is the original given polynomial, and thus the relation between the coefficient and zeroes of the polynomial has been verified.

Hence, the zeroes of this polynomial are $\frac{4}{3}$ and $-\frac{3}{2}$.