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Nickie_S
 4 years ago
Could you give me the prime factorizations?
64x^2 176xy + 121y^2
Nickie_S
 4 years ago
Could you give me the prime factorizations? 64x^2 176xy + 121y^2

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abtrehearn
 4 years ago
Best ResponseYou've already chosen the best response.1Perfect square trinomial. (8x  11y)^2

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0I didn't want the answer, I wanted to work it out on my own, I need them wrote out. Could you do that.

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0just the numbers though, i can do the letters, I just never can get the right numbers with it.

abtrehearn
 4 years ago
Best ResponseYou've already chosen the best response.1The 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.

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0I need these without the letters it looks like this: 176xy= 8*22*x*y

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0what is the common factor then?

Nickie_S
 4 years ago
Best ResponseYou've already chosen the best response.0Here you work it out for me, maybe. So I can see your work. Completely factor the following expression: Show all work, prime factorizations. 64x^2176xy +121y^2

heromiles
 4 years ago
Best ResponseYou've already chosen the best response.064x^(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 righthand 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. (8x11y)(8x11y) Combine the two common factors of (8x11y) by adding the exponents. (8x11y)^(2)

abtrehearn
 4 years ago
Best ResponseYou've already chosen the best response.1To 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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