Answer:
The Complete factored form of [tex]p^4-16[/tex] is [tex]\left(p^2+4\right)\left(p+2\right)\left(p-2\right)[/tex]
Step-by-step explanation:
Consider the given polynomial [tex]p^4-16[/tex]
We have to write the given polynomial in factored form.
Consider the given polynomial [tex]p^4-16[/tex]
Rewrite 16 as [tex]4^2[/tex], we get,
[tex]=(p^2)^2-4^2[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c[/tex]
Apply algebraic identity, we have,
[tex]x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex]
[tex]\left(p^2\right)^2-4^2=\left(p^2+4\right)\left(p^2-4\right)[/tex]
Simplify, we have,
[tex]=\left(p^2+4\right)\left(p^2-4\right)[/tex]
Again applying same identity, we have,
[tex]\left(p^2\right)^2-4^2=\left(p^2+4\right)\left(p^2-4\right)[/tex]
[tex]=\left(p^2+4\right)\left(p+2\right)\left(p-2\right)[/tex]
Thus, The Complete factored form of [tex]p^4-16[/tex] is [tex]\left(p^2+4\right)\left(p+2\right)\left(p-2\right)[/tex]
