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Old 08-15-2012, 12:20 AM   #1
kKFB1BxX

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Default Toni's maths thread
I'm stuck.

I'm to factorise z^2 + 14zy + 49y^2

and the answer is (z + 7y)^2

and is a perfect square.

I can see that z is the square root of z^2 and that 7y is the square root of 49y^2
but I don't understand how I'm meant to look at it and know that it's a perfect square.
I suppose if I could do that, I would know how to factorise it.

given that the perfect square factoring formula (?)
is x^2 + 2xy + y^2 = (x + y)^2

what am I looking for/at?
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Old 08-15-2012, 12:25 AM   #2
Ndptbudd

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You are on the right track. You just need to recognise that 2 times 7 is 14, and see that it fits into the standard formula you have quoted.
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Old 08-15-2012, 12:34 AM   #3
kKFB1BxX

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ok ok I think I'm seeing it Morrie
its 2 x 1 (for the co-efficient of z^2) x the 7y (being the square root of 49y^2) = 14zy

if I had (6a + 7b)^2
it would be 36a^ + 84yz + 49y^2
so how do I look at 84 in relation to it being a perfect square?
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Old 08-15-2012, 12:37 AM   #4
kKFB1BxX

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ahhhh. I need to double the 6 x 7 because of the 2xy!
I got it. I think
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Old 08-15-2012, 12:41 AM   #5
kKFB1BxX

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*thinking out loud*
If I halve the middle term, and its product of the square root of the first and third terms, then its a perfect square

gunna go and see if I can find a site to practise on.
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Old 08-15-2012, 12:42 AM   #6
xquFzpNw

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ok ok I think I'm seeing it Morrie
its 2 x 1 (for the co-efficient of z^2) x the 7y (being the square root of 49y^2) = 14zy

if I had (6a + 7b)^2
it would be 36a^ + 84yz + 49y^2
so how do I look at 84 in relation to it being a perfect square?
Mind your unknowns there Purps, you've turned a b into a y and added a z

It's simple enough I think to see that 36 and 49 are both squares, yeah? That's the first clue. Then just see if/how the square roots of those can fit in with the middle term.

84ab = 2*42ab

= 2*6*7*ab

=2*6a*7b
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Old 08-15-2012, 12:52 AM   #7
kKFB1BxX

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84ab = 2*42ab

= 2*6*7*ab

=2*6a*7b

I get that
off to practise
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Old 08-15-2012, 05:39 AM   #8
RorieSorNearop

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http://www.helpalgebra.com/articles/...gthesquare.htm
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