Question

What are the factor(s) for the expression 16x^2-72xy+81y^2 into a product of binomials?

Answer 1

Remember that (a-b)^2=a^2-2ab+b^2. Further, note that both the first term and last are perfect squares. Further, the equation fits the same form as above. This means we can say\[16x^2-72xy+81y^2=(4x-9y)^2\]

Answer 2

A more detailed breakdown might be\[16x^2-72xy+81y^2=(4x)^2-2(4x)(9y)+(9y)^2=(4x-9y)^2\]

Answer 3

Thanks, I'll try the solution you've referenced.