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Posted on 2003-10-29
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Does anyone knows how to factorize polynomials such as this?:

2x^3 -x^2 +8x -4

Thanks for your help!!!
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Question by:gothic130
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Assisted Solution

by:GwynforWeb
GwynforWeb earned 200 total points
ID: 9645849
Do you want this one to be factorised, or how to do for a general cubic. There is a method for a general cubic, it is very messy see http://mathworld.wolfram.com/CubicEquation.html
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by:GwynforWeb
ID: 9645887
There is a wonderful web page that will do the factorisation for you, see

http://www3.telus.net/thothworks/Quad3Deg.html
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SethHoyt earned 300 total points
ID: 9645904
For a cubic polynomial, the simplest way is to try to find one factor by finding a root of the equation:

2x^3 -x^2 +8x -4 = 0

A cubic must have at least one real root, but can have up to three. Every real root results in a factor of the form (x - a), where a is the root.

If all roots are real, the factorization is of the form:

2(x - a)(x - b)(x - c)

where I've factored out the 2 in the leading term, so that a, b and c are the three roots. This is actually the general form if we allow complex roots. But if we only allow real numbers for a, b and c, then only one of these is guaranteed to be real, and the remainder would not be factorable if the others are not real.

To find a root, you might try graphing the function to find where it crosses the x-axis, and try to guess the exact value. You can verify your guess by plugging it into the equation to see if you get zero, and adjusting if necessary. Another thing you can tell from a graph is the number of times the graph intersects the x-axis. If it only intersects once, there is only one real root and two real factors, one linear (like x - a) and one quadratic (contains x^2). If it intersects more than once, all roots are real, and there are three real factors (all linear).

After you find a root, you can divide the polynomial by the factor (x - a), where a is the root found. You can divide polynomials using synthetic division, which is basically an extension of long division for real numbers. After dividing, the quotient is the rest of the factorization, and would itself need to be factored if possible. But I assume you can do this part because the rest is quadradic, and its roots can be found using the quadratic formula, which you should know.


-Seth
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Expert Comment

by:SethHoyt
ID: 9645919
Oh, sorry I didn't see your post Gwyn...

But I think they are fairly orthogonal, so no harm done.
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by:gothic130
ID: 9646015
Thank you.
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by:SethHoyt
ID: 9646094
Here's another way to guess a root:

Let f(x) = 2x^3 -x^2 +8x -4

Find two values of x such that f(x) has opposite sign. Then a root must exist between those two values We can start with x=0 in general:

f(0) = -4

Now we need a positive value of f(x). Since the leading term's coefficent is positive, we know the value of f will eventually become positive as x increases. Trying x=1:

f(1) = 2 - 1 + 8 - 4 = 5

Thus, a root exists between 0 and 1. Let's try x = 1/2:

f(1/2) = 2/8 - 1/4 + 8/2 - 4 = 0

How fortunate! So 1/2 is a root, and that means that

(x - 1/2)

is a factor. Since we have an extra factor of 2, we can multiply that into this factor:

2(x - 1/2) = (2x - 1)

So (2x - 1) is our first factor. Now the rest of the factors can be found by dividing the f(x) by this.

The idea to dividing polynomials is to figure out what we'd need to multiply (2x - 1) by in order to get f(x).

We know the quotient must contain x^2 in order to get the 2x^3.

x^2 * (2x - 1) = 2x^3 - x^2

Remarkably, that gives us the first *two* terms; we got the second for free! We normally only expected to get the first term of f(x), but we were lucky here. Now we still need to get the rest of f(x), which is 8x - 4. To get this, we'd have to multiply (2x - 1) by 4. So adding 4 to our x^2, we get:

(x^2 + 4) * (2x - 1) = 2x^3 - x^2 + 8x - 4

Viola! The left side is the factorization of the right side. It cannot be reduced further, because (x^2 + 4) has no real factors. This can be verified by noting that x^2 + 4 >= 4, and never crosses the x-axis, so it has no real roots.

-Seth
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Expert Comment

by:SethHoyt
ID: 9646104
I really need to start refreshing before posting...

Anyway, thanks for the points.
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