Factor the polynomial completely: 16h^4 – 81?

Answer 1

Both terms are perfect squares: 16h^4 = (4h^2)^2, and 81=9^2, so you have

(4h^2 - 9)(4h^2 + 9)

The first term is also the difference of two squares, so you get

(2h - 3)(2h + 3)(4h^2 + 9)

This is a complete factorization if you're restricting yourself to real numbers, but if not, 4h^2 + 9 is also the difference of two squares, 4h^2 = (2h)^2 and -9=(3i)^2, so you have

(2h - 3)(2h + 3)(2h + 3i)(2h - 3i)

Answer 2

I've forgotten a lot about factoring, but,
16= 2^4
h^4 = h^4
81= 3^4

Can you factor out the 4th power somehow?
(2h)^4 - 3^4

Answer 3

James L is correct when factorised over the complex plane

(2h-3)(2h+3)(2h+3i)(2h-3i)

otherwise over the reals

(2h-3)(2h+3)(4h^2+9)

Answer 4

because the two 25 and 80 one are subtracted squares, it fairly is so basic it could truthfully be worked in one's head. The trick to ace-ing such exams is to get a sense for how those try questions are created via the tester, so which you would be able to truthfully artwork backward from what's supplied as a situation. what's being examined is extra how savvy one is, than how nicely one performs what could be gadget artwork.

Answer 5

16h^4 – 81
(4h^2+9)(4h^2-9)
(4h^2+9)(2h+3)(2h-3)