Question

81x^4 – 1

Answer 1

This is a difference of two squares:
\[a ^{2}-b ^{2} = (a+b)(a-b)\]
can you now factorize with this in mind?

Answer 2

how do i do that @gitahimart

Answer 3

\[81x ^{4}-1 = (9x ^{2})^{2} -1^{2}\]
get it now?

Answer 4

\[81x^{4}-1\]
\[=(9x^{2})^{2}-(1)^{2}\]
now use the above form

Answer 5

so it would be a+9 a-9?? right

Answer 6

@broderick365 u should separate the factors with parenthesis
and a should be 9x^2
get it?

Answer 7

nope: it would be:
\[(9x ^{2}-1)(9x ^{2}+1)\]
but can you see a term that you can factorize further?

Answer 8

9

Answer 9

Hint: the \[9x ^{2}-1\] can further be factorized to?

Answer 10

3z^2-1

Answer 11

@broderick365 that's just one factor, look at the formula at the very top

Answer 12

\[9x ^{2}-1 = (3x-1)(3x+1)\]
so in totality:
\[81x ^{4}-1 = (3x-1)(3x+1)(9x+1)\]
That's the naswer, try and follow, see if you can attempt another