Question

find zero of polynomial: a) (x + 2)^2 - (x-2)^2 b) 3x^2 + 2x I'll mark u brainliest.

Answer 1

SOLUTIONS:

a) (x + 2)² - (x - 2)²

Given ,

p(x) = (x + 2)² - (x - 2)²

Simplifying using

a² - b² = (a + b)(a - b)

→ p(x) = (x + 2 + x - 2)[x + 2 - (x - 2)]

→ p(x) = 2x(x + 2 - x + 2)

→ p(x) = 2x(4)

→ p(x) = 8x

Putting p(x) = 0

→ 8x = 0

→ x = 0/8

→ x = 0

b) 3x² + 2x

Given,

p(x) = 3x² + 2x

Taking common & putting p(x) = 0

→ x(3x + 2) = 0

→ x = 0/3x + 2

→ x = 0

Answer 2

Find zero of polynomials:

a) (x + 2)^2 - (x-2)^2

it is in the form a² - b² if we take (x+2) as 'a' and (x-2) as 'b'

So,

a² - b² = (a+b)(a-b)

(x + 2)² - (x-2)²

= [ (x + 2)+(x - 2) ] [ (x + 2) - (x - 2) ]

= [ x + 2 + x - 2 ] [ x + 2 - x + 2) ]

= (2x) (4)

= 8x

Now,

→ 8x = 0

→ x = 0 ÷ 8 = 0

Hence,

$\boxed{\bf{x=0}}$

$\rule{200}2$

b) 3x² + 2x

We can take x as common from both the terms.

So,

= 3x² + 2x

= x(3x+2)

now,

→ x(3x+2) = 0

→ x = 0 ÷ (3x+2)

→ x = 0