Question

16x^4-y^4 factorise easy method​

Answer 1

Answer:

16x^4-y^4

=(4x^2)^2 - (y^2)^2

={(2x)^2}^2-(y^2)^2

Here applies the identity of a^2-b^2 = (a+b)(a-b) where:

a^2={(2x)^2}^2 and b={y^2}^2

So, a=(2x)^2 and b=y^2

Then, {(2x)^2 + y^2} {(2x)^2 - y^2}

Therefore,

16x^4-y^4 will be factorized as {(2x)^2 + y^2} {(2x)^2 - y^2}

Answer 2

Answer:

{(2x)^2 + y^2} {(2x)^2 - y^2}

Step-by-step explanation:

16x^4-y^4

=(4x^2)^2 - (y^2)^2

={(2x)^2}^2-(y^2)^2

Here applies the identity of a^2-b^2 = (a+b)(a-b) where:

a^2={(2x)^2}^2 and b={y^2}^2

So, a=(2x)^2 and b=y^2

Then, {(2x)^2 + y^2} {(2x)^2 - y^2}.

Therefore,

16x^4-y^4 will be factorized as {(2x)^2 + y^2} {(2x)^2 - y^2}.

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