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Factoring quadratic equations

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Factoring quadratic equations [#permalink] New post 02 Apr 2012, 18:06
Dear all,

I'm having some trouble with factoring quadratic equations, especially if the numbers get higher / more complex.

More specifically, I'm struggling with this quadratic equation:

t^2+4t-672=0

this comes down to

(t-24)(t+28)=0

But I don't know how to arrive at this solution in the most efficient way.

Would really appreciate if someone could help me out here.

Thanks a lot!
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Re: Factoring quadratic equations [#permalink] New post 02 Apr 2012, 18:31
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This thread was quite helpful because it has a lot of examples.
http://www.purplemath.com/modules/factquad.htm

If x1 and x2 are the solutions to the quadratic equation and working backwards:
(x-x1)*(x-x2) = x^2 - (x1+x2) x + x1*x2
so you can identify the terms of the developed form with the sum and product of the solutions.

In your specific example, I would decompose 672 knowing you want to have a product of 2 terms
of opposite signs which adds up to -4

672=6*(112)=6*4*28=24*28
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Re: Factoring quadratic equations [#permalink] New post 03 Apr 2012, 00:06
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jfk wrote:
Dear all,

I'm having some trouble with factoring quadratic equations, especially if the numbers get higher / more complex.

More specifically, I'm struggling with this quadratic equation:

t^2+4t-672=0

this comes down to

(t-24)(t+28)=0

But I don't know how to arrive at this solution in the most efficient way.

Would really appreciate if someone could help me out here.

Thanks a lot!


Solving and Factoring Quadratics:
http://www.purplemath.com/modules/solvquad.htm
http://www.purplemath.com/modules/factquad.htm

Hope it helps.
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Re: Factoring quadratic equations [#permalink] New post 03 Apr 2012, 00:17
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jfk wrote:
Dear all,

I'm having some trouble with factoring quadratic equations, especially if the numbers get higher / more complex.

More specifically, I'm struggling with this quadratic equation:

t^2+4t-672=0

this comes down to

(t-24)(t+28)=0

But I don't know how to arrive at this solution in the most efficient way.

Would really appreciate if someone could help me out here.

Thanks a lot!


First, check out this post: hard-factoring-question-109006.html#p870223
It discusses how to factor harder quadratic equations.

Coming to this question, you have -672 so the two factors which multiply to give 672 will have opposite signs i.e. one of them will be positive and the other will be negative. Also, their difference will be 4 i.e. very small.
Try and split 672 into its factors. We see that 672 is divisible by 4.
672 = 4*168 = 4*4*42 = 4*4*2*3*7 (We try to split the number till we get manageable factors)

Now you have to try various combinations of these factors till you get a pair with a difference of 4. Notice that the difference is very small so we need to get 2 factors which are kind of equal to each other. Say, one factor can be 7*4 and another can be 6*4. 28 and 24 actually do have a difference of 4 so we have got our 2 factors.

t^2 + 28t - 24t - 672 = 0
(t-24)(t+28) = 0
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Re: Factoring quadratic equations [#permalink] New post 03 Apr 2012, 00:31
Thanks a lot everyone! Very helpful comments!
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Re: Factoring quadratic equations [#permalink] New post 05 Apr 2012, 19:31
Do you need to solve this problem with paper and pen or can you use a calculator, program, or website?

If you need an analytical solution, or pen-and-paper, another way to attack this type of problem would be to use the Quadratic Formula (QF); there is a lot of good material about the QF on Wikipedia or MathWorld. This approach is easier if you have a calculator, but it allows for exact, algebraic, solutions.

On the other hand, if you cannot solve the problem easily, you might not have a choice but to use the QF.
And if you are allowed to use a program or web applet, then you can solve the problem numerically.

For example, here is a free online applet that solves a Quadratic Equation:

Quadratic Equation Solver

Whether you use a web program, a full-blown computer program, or a pocket calculator, once you have numeric solutions root1 and root2, you could then re-state the original equation in the following form:

(t - root1)(t - root2)

If the roots are nice round numbers, that is nice but doesn't always happen. The solutions are what they are.
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Re: Factoring quadratic equations [#permalink] New post 05 Apr 2012, 19:42
Quote:
Do you need to solve this problem with paper and pen or can you use a calculator, program, or website?


I'm pretty sure GMAC would frown upon any of the later....
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Re: Factoring quadratic equations [#permalink] New post 05 Apr 2012, 19:55
nsspaz151 wrote:
I'm pretty sure GMAC would frown upon any of the later....


True. I forgot the overall context of these forums. I looked at the question by itself and just gave a general answer.

In any case, if a solution cannot be found quickly by inspection, I suppose the QF is the only way to go. Unless a person wants to make a bunch of guesses and hope they get lucky.
Re: Factoring quadratic equations   [#permalink] 05 Apr 2012, 19:55
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