Question

factorise 16x power 4 -81

Answer 1

hope it helps.............

Answer 2

Factors are $\bf \red{ 16 {x}^{4} - 81 = (4 {x}^{2} + 9)(2x + 3)(2x - 3)}$

Given:

• $16 {x}^{4} - 81$

To find:

• Factorise the polynomial.

Solution:

Identity to be used:

$\bf {a}^{2} - {b}^{2} = (a + b)(a - b)$

Step 1:

Rewrite the polynomial.

$16 {x}^{4} - 81 = ( {4 {x}^{2} )}^{2} - ( {9)}^{2}$

apply Identity.

As

$a= 4x^2\:and\:b=9$

$( {4 {x}^{2} )}^{2} - ( {9)}^{2} = (4 {x}^{2} + 9)(4 {x}^{2} - 9)$

Step 2:

Rewrite the second term again.

$(4 {x}^{2} + 9)(4 {x}^{2} - 9) =(4 {x}^{2} + 9)( {(2x)}^{2} - ( {3)}^{2} )$

Apply the identity again; as

$a= 2x\:and\:b=3$

$(4 {x}^{2} + 9)( {(2x)}^{2} - ( {3)}^{2} ) =(4 {x}^{2} + 9)(2x + 3)(2x - 3)$

Thus,

Factors are $\bf \: 16 {x}^{4} - 81 = (4 {x}^{2} + 9)(2x + 3)(2x - 3)$

Learn more:

1) factorise 225x^2-9x^2-36y^2+36xy

https://brainly.in/question/29246923

2) positive real root of 81x^2-1=0

https://brainly.in/question/15078788