stevemilw
Hello, im revising eigenvalues in matrices and ive come across a part where i need to factorise the following quadratic equation; let lander = x
36-36x+11x^2-x^3 = 0
I know that the answer is (x-2)(-x^2+9x-18), but i dont know how he got to it.
Ive look through google but i cant find any information on quadratics of this type.

Mentor
Quadratic suggests polynomial of 2nd degree, that is one with the highest power 2. Here you have x3, so it is a cubic polynomial.

stevemilw
aaah i see, i know it = 0 however, thats probably why i havent been able to find search results, ill have a look again, thank you

stevemilw
hhhm, im still not finding any decent material. I will continue this in the morning.

stevemilw
can someone please go through how this is done, i cant find an easy way to do it, or any help or google
Thanks

Count Iblis
You can try to use the Rational Roots theorem. If x = p/q with p and q integers that don't have any factors in common, then p will divide the constant term and q will divide the coefficent of the highest power of x.

In this case we have:

f(x) = 0

with

f(x) = x^3 - 11 x^2 + 36 x - 36

A rational zero of f thus has to be an integer that divides 36. They can thus be:

x= ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36

Since were too lazy to try out all these 18 candidates, we're going to use the following trick. Let's substitute for x some random integer r such that f(r) is some ugly number with only a few divisors. How will that help?

Consider the polynomial g(y) defined as:

g(y) = f(r+y)

We apply the Rational Roots theorem to g(y). We note that to do that you don't need to expand out f(r+y), all you need are the coefficients of y^3 and the constant term of f(1+y). The constant term is the value for y = 0, which is f(r) = ugly number with only a few divisors. The coefficient of y^3 is the same as the coefficient of x^3, so the rational roots of g(y) are those few divisors of the ugly number f(r)

Since g(y) = f(r+y), adding r to these candidates gives the possible rational roots for f.

E.g., we have that f(5) = -6. Tis means that the possible rational roots of g(y) = f(5+y) are:

y = ±1, ±2, ±3, ±6

The possible rational roots of f are thus:

x = 5+y = -1, 2, 3, 4, 6, 7, 8, 11

But the rational roots also have to be divisors of 36, therefore we can strike out any items in the list that are not divisors of 36, leaving us with:

x = -1, 2, 3, 4, 6,

If we try f(-1), we see that f(-1) = -84. If we then take r = -1 and play the same game as above, we see that the list of candidates is reduced to:

x = 2, 3, 6

These are in fact the three roots of f.

stevemilw
okay, thanks for the reply, but how does knowing the roots help me in factoring the expression.
What i am trying to do is but it into two brackets.
Sorry if i sound stupid, but im not seeing as too how this helps. X can be treated as any value when i put it into brackets, so why am i finding the roots? that was all way too complicated for me im afraid.