Question

What is the completely factored form of p^4-16

Answer 1

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Answer 2

The Complete factored form of
`p^4-16` is
`\left(p^2+4\right)\left(p+2\right)\left(p-2\right)`

Explanation:

Consider the given polynomial
`p^4-16`

We have to write the given polynomial in factored form.

Consider the given polynomial
`p^4-16`

Rewrite 16 as
`4^2`, we get,

`=(p^2)^2-4^2`

`\mathrm{Apply\:exponent\:rule}:\quad \:a^(bc)=\left(a^b\right)^c`

Apply algebraic identity, we have,

`x^2-y^2=\left(x+y\right)\left(x-y\right)`

`\left(p^2\right)^2-4^2=\left(p^2+4\right)\left(p^2-4\right)`

Simplify, we have,

`=\left(p^2+4\right)\left(p^2-4\right)`

Again applying same identity, we have,

`\left(p^2\right)^2-4^2=\left(p^2+4\right)\left(p^2-4\right)`

`=\left(p^2+4\right)\left(p+2\right)\left(p-2\right)`

Thus, The Complete factored form of
`p^4-16` is
`\left(p^2+4\right)\left(p+2\right)\left(p-2\right)`