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Quick factorisation of polynomials
By Anonymous on Thursday, November 23, 2000 - 04:22 pm:

Hi,

Often we enounter the need to factorise, most times we need to do it as quick as possible. Are there are tricks that can be used to do this, so that factorisation can be done mentally and quickly?

For example.
2x3 - 3x2 - 4 = 0

By Arvan Pritchard (T708) on Thursday, November 23, 2000 - 04:42 pm:

One useful idea when all the coefficients are integers is to look for factors of the form (ax ± b), where a is a factor of the coefficient of the highest power of x, and b is a factor of the constant, and a,b have no common factors.

(ax-b) is a factor if (and only if) b/a is a root, and so we have 8 candidate roots (±½, ±1, ±2, ±4), and it's easy to check that x=2 is a root, so x-2 is a factor.

By Susan Langley (Sml30) on Thursday, November 23, 2000 - 04:50 pm:

Ok, no not really, as there is no general way to factorise polynomials of large roots at all easily. The general way to approach polynomially of order greater than 2 it to look for solutions and then factorise it by (x-the solution)
e.g. for your example:
2×23-3×22-4=0
so 2×x3-3×x2-4=0
goes to
(x-2)(2x2+x+2)=0

This method only really works in the equations you get in text books or similar, generally in real life you have non-constant coefficients and it is very hard to solve these polynomials at all except, possibly, by trial and error.

I presume you know the quadratic formula, that is probably the best we of solving quadratics until you get lots of practice and can just recognise things.

There is a cubic formula but it is very complicated.

Hope that helps

Susan

By Anonymous on Thursday, November 23, 2000 - 05:36 pm:

Isn't there something called Furbini's (Or Ruffini's?) rule to find the roots of a cubic polynomial?

By Anonymous on Thursday, November 23, 2000 - 06:56 pm:

Can anybody elaborate on how to factorise (find roots) for cubics? Just for fun?

By Michael Doré (Md285) on Thursday, November 23, 2000 - 10:06 pm:

OK, but like Susan says, the final formula is very complicated. I think the fact that there exists a closed form solution is of far more interest than the actual solution itself.

This derivation for the cubic formula is down to the Italian mathematician Del Ferro. I can show each of the steps - what I can't explain is the motivation behind them (i.e. how he knew that these steps would lead to a solution).

Let the cubic be:

ax3 + bx2 + cx + d = 0

Assume a is not 0 (otherwise we'd get a quadratic). Divide through by a and we obtain:

x3 + px2 + qx + r = 0

where p = b/a, q = c/a, r = d/a.

Now we can make a linear substitution to get rid of the x2 term. To this end let y = x + p/3:

(y - p/3)3 + p(y - p/3)2 + q(y - p/3) + r = 0

y3 + y2(-p + p) + y(p2/3 - 2p2/3 + q) + (p3/27 + p3/9 - pq/3 + r)

So the coefficient of y2 is zero and our equation becomes:

y3 + ly + m = 0

where l = q - p2/4 and m = 4p3/27 - pq/3 + r.

This is all straightforward so far. Now the crucial bit: let u,v be two numbers satisfying u + v = y. We have not yet uniquely determined u and v but we will do so later. We have:

(u + v)3 + l(u + v) + m = 0 (*)

So:

(u3 + v3) + (3uv + l)(u + v) + m = 0 (**)

Note that these two equations are satisfied if:

3uv = -l and u3 + v3 = -m.

In other words if these two equations are true then (**) is true, hence (*) is true, so u + v = y is indeed a root of the cubic. To find u,v use the relations:

3uv = -l and u3 + v3 = -m

Eliminate v to get a quadratic for u3. Then you can write down an expression for u. By taking the other root of the quadratic you can get an expression for v. Then use y = u + v to find the values of y and finally substitute back into x = y - p/3 to get the solution of the original cubic. Not nice.

Quartic equations also have a closed form solution - and the derivation of the formula is a good deal easier than the derivation of the cubic formula (but the derivation does rely on the cubic formula). It has been proved that all equations of degree greater than 4 do not have a general closed form formula, though I have no idea how to show this!

Yours,

Michael

By Anonymous on Friday, November 24, 2000 - 11:40 am:

Whoa! Looks great Michael!
Thanks.