Question

Find the zeroes of polynomial
2x^2-162x
2
−16​

Answer 1

Given :- Find the zeroes of polynomial 2x² -6x - 216 ?

Answer :-

put the given equation equal to zero .

→ 2x² - 6x - 216 = 0

taking common,

→ 2(x² - 3x - 108) = 0

→ x² - 3x - 108 = 0

now, splitting the middle term we get,

→ x² - 12x + 9x - 108 = 0

→ x(x - 12) + 9(x - 12) = 0

→ (x - 12)(x + 9) = 0

putting both equal to zero, we get,

→ x - 12 = 0,

→ x = 12

or,

→ x + 9 = 0

→ x = (-9) .

Hence, zeroes of the given polynomial are 12 and (-9) .

_______

if polynomial is 2x² - 16 ,

→ 2x² - 16 = 0

→ 2x² = 16

→ x² = 8

→ x = ±2√2

zeroes of polynomial are 2√2 and (-2√2) .

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Answer 2

SOLUTION

TO DETERMINE

The zeroes of the polynomial

$\sf{2 {x}^{2} - 16 }$

EVALUATION

Let p(x) be the given polynomial

$\sf{p(x) = 2 {x}^{2} - 16 }$

For Zero of the polynomial p(x) we have

$\sf{p(x) =0}$

$\sf{ \implies 2 {x}^{2} - 16 = 0 }$

$\sf{ \implies 2 {x}^{2} = 16 }$

$\sf{ \implies {x}^{2} = 8 }$

$\sf{ \implies x = \pm \sqrt{8} }$

$\sf{ \implies x = \pm \sqrt{2 \times 2 \times 2} }$

$\sf{ \implies x = \pm \: 2 \sqrt{2} }$

Hence the required zeroes of the polynomial are $\sf{2 \sqrt{2} \: \: \: and \: \: - 2 \sqrt{2} }$

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