The same way you'd factor any other similar equation. You're going to run into the complication that the roots of that equation are not real numbers, though; unless you know what complex numbers are, you won't be able to do it.
Yes. There is an easy solution for the difference of two squares, but the sum of two squares is a different matter.
x2 + 4x + 4 = (x+2)2
(x + 2)(x + 2)
That factors to 4(x^2 + 4)
4x + 16 = 4 (x + 4)
4(x + 4)
4x2 + 8x + 4 = 4 (x2 + 2x + 1) = 4 (x + 1)2
If that's +14x + 3, the answer is (2x + 3)(4x + 1)
x - 4x + 4 = -3x + 4 which cannot be factorised.
4
4x2+13x+3 m*n=(4*3)=12 m+n=13 m=12 n=1 (4x/12)=(x/3) (4x+1)(x+3)
4(x^2 + x + 3)
4(x+y)^2
(2x+3)(2x+4)
The derivative of 2x2 + 4x + 8 is 4x+4.
4x2+4x-8 4(x2+x-2) and (4x+8)(x-1)
x(x-4)
x2 + 4x = x(x+4)
x2 + 4x = x(x + 4)
It is: 4x(1+4)
4X^2 - 8X factor out 4X 4X(X - 2)
4x2 + 8x + 4 = 4 (x2 + 2x + 1) = 4 (x + 1)2
3x2 + 4x factors to x(3x + 4)