Question

factor completely x^4 - 81y^4?

Answer 1

\[\Large\bf x^4-81y^4\]\[\Large\bf (x^2)^2-81(y^2)^2\]Is 81 is a perfect square?

Answer 2

9 x 9 give you 81 so yeah ?

Answer 3

\[\Large\bf (x^2)^2-9^2(y^2)^2\]\[\Large\bf (x^2)^2-(9y^2)^2\]So we can write it like that,
Any confusion how I got it to that form?

From there, we apply our rule for the `difference of squares`:\[\Large\bf\color{royalblue}{ a^2-b^2\quad=\quad (a-b)(a+b)}\]

Answer 4

i got it kinda, how would i plug it in the formula

Answer 5

I should have color-coded them a little better maybe.\[\Large\bf \color{orangered}{(x^2)}^2-\color{orangered}{(9y^2)}^2\quad=\quad \left[\color{orangered}{(x^2)}-\color{orangered}{(9y^2)}\right]\left[\color{orangered}{(x^2)}+\color{orangered}{(9y^2)}\right]\]

The orange represents our a and b.
So there is how we would complete the first step, applying the rule.

Answer 6

Lemme get rid of some of the extra brackets that we no longer need:\[\Large\bf (x^2-9y^2)(x^2+9y^2)\]

Answer 7

Is 9 a perfect square? :o

Answer 8

3

Answer 9

Ok good.
If we rewrite the first 9,
\[\Large\bf \color{royalblue}{(x^2-(3y)^2)}(x^2+9y^2)\]You might notice that again we have the difference of squares, so we want to apply the rule again.

Answer 10

ok i can seee that

Answer 11

\[\Large\bf \color{orangered}{x}^2-\color{orangered}{(3y)}^2\quad=\quad ?\]Understand how to apply the rule to this one?

Answer 12

nah..