How do you factor (x^2+5)(x^2+2x-3)?

1 Answer
May 20, 2015

(x^2+5)(x^2+2x-3) = (x^2+5)(x+3)(x-1)

The factors (x+3) and (x-1) were found by looking for two numbers a and b such that axxb = 3 and a-b = 2.
Then (x+a)(x-b) = x^2+(a-b)x-ab.

This is as far as you can break down the original expression into factors, unless you are allowed complex numbers as coefficients. You can tell that (x^2+5) has no smaller factors with real coefficients - which would be linear - because x^2+5 > 0 for all real values of x.

On the other hand, if you are allowed complex coefficients, you can factor as:

(x^2+5) = (x+isqrt(5))(x-isqrt(5)), where i = sqrt(-1), so

(x^2+5)(x^2+2x-3)

= (x+isqrt(5))(x-isqrt(5))(x+3)(x-1)