Could you give me the prime factorizations? 64x^2 -176xy + 121y^2

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Could you give me the prime factorizations? 64x^2 -176xy + 121y^2

Mathematics
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Perfect square trinomial. (8x - 11y)^2
I didn't want the answer, I wanted to work it out on my own, I need them wrote out. Could you do that.
just the numbers though, i can do the letters, I just never can get the right numbers with it.

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The first thing I check for is to see if the trinomial is a perfect square. The first and last terms of the trinomial are perfect squares. That satisfies one of the conditions required for the quadratic to be a square. The other condition is that the middle term is twice the product of the square roots of the first and last terms. We compare 176xy with 2(8x)(11y). They are equal, and the trinomial is a square. The only thing left is determine if you add or subtract 11y from 8x. Since the middle term is minus, you subtract.
I need these without the letters it looks like this: 176xy= 8*22*x*y
That's right.
I need the one for 121y^2 I have 64x^2=8*8*x*x I do not have a 8 for the 121y^2 so how is it worked out.
121y^2 = (11y)^2.
what is the common factor then?
Here you work it out for me, maybe. So I can see your work. Completely factor the following expression: Show all work, prime factorizations. 64x^2-176xy +121y^2
64x^(2)-176xy+121y^(2) For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (7744) that add up to b (-176). a=64, b=-176, c=121 For a polynomial of the form ax^(2)+bx+c, find two factors of a*c (7744) that add up to b (-176).In this problem -(11)/(8)*-(11)/(8)=(121)/(64) (which is (c)/(a)) and -(11)/(8)-(11)/(8)=-(176)/(64) (which is (b)/(a)) , so insert -(11)/(8) as the right hand term of one factor and -(11)/(8) as the right-hand term of the other factor. (x-(11)/(8)*y)(x-(11)/(8)*y) Remove the fraction by multiplying the first term of the factor by the denominator of the second term. (8x-11y)(8x-11y) Combine the two common factors of (8x-11y) by adding the exponents. (8x-11y)^(2)
To get a commom factor, you can split the middle term and rewrite the trinomial as\[64x^{2} - 88xy - 88xy + 121y^{2}, \]and group:\[= (64x^{2} - 88xy) + (-88xy + 121y^{2}.\]Factor each binomial:\[= 8x(8x - 11y) - 11y(8x - 11y). \]Now we have the common factor (8x - 11y), and we can factor it out to get\[(8x - 11y)(8x - 11y) = (8x - 11y)^{2}.\]

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