## factorization

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### factorization

I have these to factorize:

1. 5-2x-x^2
2. 16x^2-y^2-4y-4

This is what I did:
(1) 5-2x-x^2 = -(x^2+2x-5) = -(x+1)^2+6 = -(x+1-6^.5)(x+1+6^.5).

(2) 16x^2-y^2-4y-4 = 4(4x^2-y-1)-y^2 = 4((2x+1)(2x-1)-y) = (4x-y)(4x+y)-4(y+1) = (4x-y-2)(4x+y+2).

For #1, how can -(x+1)^2+6 become -(x+1-6^.5)(x+1+6^.5)?
For #2, how can 16x^2-y^2-4y-4 become (4x-y-2)(4x+y+2)?
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joyrider

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Location: Melbourne, Oz (Australia)

I don't understand the questions.

Alvaro
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Okay then...

Firstly, I don't understand what the steps are to make -x^2-2x+5 into
-(x+1-6^.5)(x+1+6^.5). The reason is because when I get to -(x+1)^2+6 I can't use the difference of two squares rule (the rule where a^2-b^2 = (a+b)(a-b)) as 6 isn't negative.

Secondly, I don't understand how to get 16x^2-y^2-4y-4 into (4x-y-2)(4x+y+2). One of my experiments brought me to this: 16x^2-y^2-4y-4 = (4x-y)(4x+y)-4y-4 = (4x-y+2)(4x+y-2)-4y, with the stray -4y at the end. In order to factor that all I used was the difference of two squares rule and I know that (4x-y+2)(4x+y-2) = 16x^2-y^2-4y-4 through expansion.
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joyrider

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Joined: Mon Jul 12, 2004 2:39 am
Location: Melbourne, Oz (Australia)

joyrider wrote:Okay then...

Firstly, I don't understand what the steps are to make -x^2-2x+5 into
-(x+1-6^.5)(x+1+6^.5). The reason is because when I get to -(x+1)^2+6 I can't use the difference of two squares rule (the rule where a^2-b^2 = (a+b)(a-b)) as 6 isn't negative.

But -(x+1)^2 is negative: (6 - (x+1)^2)

Secondly, I don't understand how to get 16x^2-y^2-4y-4 into (4x-y-2)(4x+y+2). One of my experiments brought me to this: 16x^2-y^2-4y-4 = (4x-y)(4x+y)-4y-4 = (4x-y+2)(4x+y-2)-4y, with the stray -4y at the end. In order to factor that all I used was the difference of two squares rule and I know that (4x-y+2)(4x+y-2) = 16x^2-y^2-4y-4 through expansion.

-y^2-4y-4 is equal to -(y+2)^2
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MXP

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Thanks for that!
Nothing in this world is good or bad, only your reaction to it.

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joyrider

Posts: 92
Joined: Mon Jul 12, 2004 2:39 am
Location: Melbourne, Oz (Australia)