Question

Find zeroes of polynomial 2x^2 -16

Answer 1

Step-by-step explanation:

$2x {}^{2} - 16 = 0 \\ = > 2x {}^{2} = 16 \\ = > x { }^{2} = \frac{16}{2} \\ = > x {}^{2} = 8 \\ = > x = \sqrt{8}$

Answer 2

The zeroes of the polynomial are 2✓2 and -2✓2.

Solution: The zeroes of the polynomial are values at which the polynomial f(x) has value equal to 0.

A quadratic polynomial f(x) can have at most two zeroes since the highest power of x in the polynomial is 2.

Let f(x) = $2 {x}^{2} - 16$

For finding the zeroes of this polynomial, we equate the polynomial with 0.

Therefore,

$2 {x}^{2} - 16 = 0$

$= > 2 {x}^{2} = 16$

$= > {x}^{2} = \frac{16}{2}$

$= > {x}^{2} = 8$

$= > x = \sqrt{8}$

$= > x = 2 \sqrt{2} \: or \: x = - 2 \sqrt{2}$

Here, both 2✓2 and -2✓2 when put in place of x in the polynomial makes the value of the polynomial 0 and are hence called the zeroes of the polynomial.