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Can someone explain how to write squares as a trinomial like:

(7u-3)squared

and...

How to decide whether each trinomial is a perfect square.
(If it is, factor it):

64usquared + 72uv + 81vsquared

CAN SOMEONE PLEASE EXPLAIN AS SIMPLE AS POSSIBLE!

Thanks!

2007-11-26 11:08:15 · 2 answers · asked by wavves 4 in Science & Mathematics Mathematics

2 answers

In general if you have (a + b)² this becomes:
a² + 2ab + b²

To expand:
(7u - 3)²

a = 7u, b = -3

a² = 49u²
2ab = -42u
b² = 9

So the final expanded equation is:
49u² - 42u + 9

To reverse the process, start by taking the square roots of the first and last numbers:
64u² --> (8u)²
81v² --> (9v)²

Now see if the middle can be 2(8u)(9v)... in this case it can't. You should be seeing 144uv, but instead you have 72uv. Can't factor it this way.

2007-11-26 11:18:38 · answer #1 · answered by Puzzling 7 · 0 0

(7u - 3)^2...
The simple way is to remember that
if you have
(a + b)^2, it will equal a^2 + 2ab + b^2.
If you have
(a - b)^2, it will equal a^2 - 2ab + b^2.

(7u - 3)^2 is in the form of (a - b)^2,
so 49u^2 - 2*7u*3 + 9 = 49u^2 - 42u + 9.


How you know that
64u^2 + 72uv + 81v^2 is a perfect square?

This is obviously in the form (a+b)^2 IF it is a perfect square.
(8u + 9v)^2. But the middle term is 2*a*b and that is
2*8u*9v and that doesn't match the middle term. So I don't think it is a perfect square. To make it a perfect square the middle term must be 144uv.

2007-11-26 19:17:36 · answer #2 · answered by Axis Flip 3 · 0 0

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