When factoring what do you do when it is impossible to get a number that multiplies or adds properly?

Answer 1

You write that it cannot be factored.

Answer 2

I suppose that you are talking about factoring polynomials, for instance
x^2 - 4 x - 18

This one is not easy to factor; for -18 you have 2 x -9 or 3 x -6, but in both cases the sum is not equal to -4.

There is a factoring possible, but it is not easy to just "see" it. That is why other methods have been invented to solve equations. The methods are: "completing the square" and the "quadratic formula". I will show them briefly, but if you learn factoring now you will probably learn these techniques next.

Completing the square: ( x^2 - 4 x - 18 = 0 )
If you expand the square
(x - 2)^2 = x^2 - 4 x + 4
you have almost the same formula, except for the last number (4 instead of -18). Since this is 22 too high, I conclude
(x - 2)^2 = 22
Therefore, x - 2 = + or - the square root of 22, and
x = 2 +- sqrt(22)
This shows that the factoring was
x^2 - 4 x - 18 = (x - 2 - V22) (x - 2 + V22)

Quadratic formula: ( x^2 - 4 x - 18 = 0 )
The general pattern is a x^2 + b x + c = 0, so here a = 1, b = -4, c = -18.
First we calculate the "discriminant"
D = b^2 - 4 a c = (-4)^2 - 4x1x(-18)
= 16 - (-72) = 88
Next, we determine
x = (-b +- sqrt D) / 2a
= (4 +- V88) / 2
= 2 +- V22

Answer 3

It won't happen with the ones you're learning to factor. It's only when you get to polynomials of degree 5 or higher that factoring cannot always be performed with radical expressions involving complex numbers.


Doug

Answer 4

Please, clarify your question. If you are talking about numbers, then, if these cant be factorized, they are prime numbers, like someone else said

If you are talking about polynomials, then the problem is much more difficult

Ana

Answer 5

then it's called a "Prime" if it can't be factored so you don't have anything to do with it.

Answer 6

the quadratic formula might help

Answer 7

just leave the question and try other.