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timbeckman
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Best method to factor like a quadratic 06 Feb 2010, 19:32
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If I have an equation as follows:
a^4 + 100a^2 + 2304 = 0

and I need to factor like a quadratic to:
(a^2+36)(a^2+64) = 0

What is the easiest method to come up with the integers of 36 & 64?
I realize the sum of the numbers must = 100, and the multiple equal 2304, but without multiple trial & errors how is 36 & 64 determined?



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Best method to factor like a quadratic 06 Feb 2010, 19:41
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timbeckman
If I have an equation as follows:
a^4 + 100a^2 + 2304 = 0

and I need to factor like a quadratic to:
(a^2+36)(a^2+64) = 0

What is the easiest method to come up with the integers of 36 & 64?
I realize the sum of the numbers must = 100, and the multiple equal 2304, but without multiple trial & errors how is 36 & 64 determined?
Show more

In the above equation, what value of "a" will yield 0? If a < 0 then squaring that number will give you a positive number. Am I missing something?
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timbeckman
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Best method to factor like a quadratic 07 Feb 2010, 11:17
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Good point...I'm sorry it should have read:
a^4 - 100a^2 + 2304 = 0
(a^2-36)(a^2-64) = 0
...I realize the sum of the numbers must = -100...
Same question still exists?
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esledge
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Best method to factor like a quadratic 07 Feb 2010, 12:58
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Hi Tim,

I start by creating a factor tree for 2304. I don't necessarily go down to the prime factors every time, though that's OK. For example, I get 2304 = (4 * 4 * 12 * 12) or (2^8 * 3^2).

Then I look at these numbers and think, "How do I mix and match these factors to get two numbers that sum to/differ by 100?" While there's still a bit of trial and error involved, having the factors in mind and starting in the middle (i.e. the square root of 2304 = (2^4)(3^1) = 48) can help.

First try: 48 and 48. Sum of 96, difference of 0. But 96 is close to 100....so swap a 3 and a 2 between the numbers for a small change. That is, 48/3 * 2 and 48/2 * 3.

Second try: 32 and 72. Sum of 104, difference of 40. 104 is still close to 100, but we went too far. Go back to 48 and 48 and swap a 3 and a 4 instead. That is, 48/4 * 3 and 48/3 *4.

Third try: 36 and 64. Sum of 100, difference of 28. Bingo.

The sign of the coefficients plays a role, of course. Since +2304 was positive in the quadratic, the factors will have the same sign. That tells us to focus on getting the sum of the factors to be 100. If it had been -2304, the factors would have opposite signs and we would pay attention to the difference as we try to match the middle coefficient.



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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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