How do you factor polynomials into binomials?

Answer 1

First of all, you must bear in mind that not all polynomials fan be factored into binomials with real coefficients. For example, x^2 + 2x + 2 cannot be factorised because its roots are complex numbers.

There is no simple rule which will work for polynomials of order 3 or more. But here is one method.

Suppose the polynomial is of the order n, and is in the form

p(x) = a(n)*x^n + a(n-1)*x^(n-1) + ... + a(1)*x + a(0) = 0

where a(n), a(n-1), ... , a(1), a(0) are real constants, then

let f be some factor of a(n) and g a factor of a(0) - not necessarily positive nor prime.

If p(g/f) = 0 then (f*x - g) is a binomial factor of the polynomial.

You will need to try all combinations of +/-f and +/-g.

Divide p(x) by (f*x-g) and you will have a polynomial of degree (n-1). Continue finding a root and dividing the polynomial - thereby reducing its degree until you have only binomial terms.

Answer 2

There are a few methods - none of which are simple.

Suppose you have a polynomial. f(x). In order to factorise it you need to know a value for x, say x = a such that f(a) = 0. You can either find a graphically, by the remainder theorem or by using numerical methods. Then (x - a) is a factor of f(x).

Next, you calculate g(x) = f(x)/(x - a). This can be by long division or by writing g(x) as a polynomial with unknown coefficients and then comparing the coefficients of each power of x in f(x) and in (x - a)*g(x).

Repeat the procedure for g(x).

A polynomial need not have any real factors.

For example:

Suppose f(x) = x^3 - 4x^2 + x + 6

Then f(2) = 8 - 4*4 + 2 + 6 = 0 an so (x - 2) is a factor.

Since f(x) is of order 3, g(x) must be of order 2. Suppose g(x) = px^2 + qx + r

then (x - 2)*g(x) = px^3 + qx^2 + rx - 2px^2 - 2qx - 2r

= px^3 + qx^2 - 2px^2 + rx - 2qx - 2r

This is equal to f(x) = x^3 - 4x^2 + x + 6

Comparing coefficients of

x^3: p = 1

x^2: q - 2p = -4 => q - 2 = -4 => q = -2

x: r - 2q = 1 => r + 4 = 1 => r = -3

and the constant term gives a check: 2r = -6 => r = -3.

So you now have q(x) = 1x^2 - 2x - 3 = x^2 - 2x - 3

q(3) = 9 - 2*3 - 3 = 0 so (x - 3) is a factor.

etc

Therefore: f(x) = (x - 2)*(x - 3)*(x + 1)