Question

Ok, I can not find the prime factorization for my 81 in this equation.

16a^2 - 72ab+81b^2. At the moment I am only looking for the numbers that go for the 81.

The other ones I have narrowed it down to
16=4*4
72=4*2*9

(I know this is not including the exponents, I have done that part, got stuck on how the 81 will fit in there.) Can you help me please?

Answer 1

3^4=81

Answer 2

to solve this polynomial why do u need to fctorise 81?????

Answer 3

u hv to factorise this polynomial ?????

Answer 4

well I need to find the prime factorization to solve. Do you want the direction that came with the problem I will type them, give me a minute.

Answer 5

(4a-9b)^2

Answer 6

16a^2 - 72ab+81b^2

This can be expressed as follows:
= (4a)^2 - 2(4a)(9b) + (9b)^2

We know that x^2 - 2xy + y^2 = (x - y)^2
so
= (4a - 9b)^2

16a^2 - 72ab+81b^2 = (4a - 9b)(4a - 9b)

Answer 7

Factor the expression 16a^2-72ab+81b^2 into a product of binomials.

Answer 8

that is what i have done.....
(4a - 9b) is a binomial because it has two terms......

Answer 9

because the 16 is = to 4*4 the same number which would then mean that number split, then you put it in with the 2nd number then the 3rd number. Ok, I get it!