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.
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Yes. There is an easy solution for the difference of two squares, but the sum of two squares is a different matter.
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)