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jc

Guest

factoring polynomias

I was just wondering if somebody could help me factor the following problem:

16y^2-25x^2+10x-1

The teacher never explained how to factor with 2 coeficiants.

Thanks in Advance.
JC

MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: factoring polynomias

Bit of a puzzle, that one ...

I think we can tease apart the first two terms "16y^2-25x^2"

(4y-5x)(4y+5x) = 4y(4y+5x) -5x(4y+5x) = 16y^2 + 20xy -20xy -25x^2 = 16y^2  -25x^2

But we also need "+10x-1" ...

... making it (4y-5x+2)(4y+5x)  adds 8y+10x, so we need to wipe another -8y-1

What does (4y-5x+1)(4y+5x-1) get?

(4y-5x+1)(4y+5x-1) = 4y(4y+5x-1) -5x(4y+5x-1) +1(4y+5x-1) = 16y^2 + 20xy - 4y - 20xy -25x^2 +5x + 4y +5x -1

= 16y^2 -25x^2 +5x +5x -1 = 16y^2 -25x^2 +10x -1

It's a miracle ! I got it by a bit of lucky guessing! (unless I made a mistake)

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"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

JC

Guest

Re: factoring polynomias

Thanks alot for the help!

noobard

Member

Registered: 2009-06-07

Posts: 28

Re: factoring polynomias

ll tell u a simpler way
wen it comes to factorising polynomials
think first of a^2-b^2=(a+b)(a-b)
so keep 16y^2 aside for a while ....
and we can see that 25x^2 -10x+1 is a perfect square of 5x-1
so the problem is now 16y^2-(5x-1)^2which is nothin but
(4y-5x+1)(4y+5x-1)

;)

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Everything that has a begining has an EnD!!!