That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: (-3 plus or minus 2i times the square root of 3) divided by 3.
x = -1 + 1.1547005383792515i
x = -1 - 1.1547005383792515i
where i is the square root of negative one.
If that's 3x2 - 7x + 4, the answer is (3x - 4)(x - 1)
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3x2-2x-21 = (3x+7)(x-3)
6x(3x2 - x + 4)
-(x + 2)(3x + 4)
3x2 + 27x +60
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}
3x2-9x-30 =3x2+6x-15x-30 =3x(x+2)-15(x+2) =(3x-15)(x+2) OR 3(x - 5)(x+2)
While it is possible to factor 3x2 from both of these and get 3x2(4 - 1), it's a lot easier to subtract 3x2 from 12x2 and get 9x2
factor out a 3x 3x(x-3)=3x2-9x
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42
the answer is 3x2
21x2 + 27x - 30 = 3*(7x2 + 9x - 10) = 3*(7x2 - 5x + 14x - 10) =3*[x*(7x - 5) + 2*(7x - 5)] = 3*(7x - 5)*(x + 2)
x3 + 9x2 + 27x + 27 Given the numbers in the equation, we can likely bet on (x + 3) being a factor. Let's try it with artificial division: 3 * 1 = 3 9 - 3 = 6 3 * 6 = 18 27 - 18 = 9 3 * 9 = 27 27 - 27 = 0 Bingo. So let's carry it out in long division:                       x2 + 6x + 9                    _____________________ x + 3 ) x3 + 9x2 + 27x + 27                        x3 + 3x2                                        6x2 + 27x                                        6x2 + 18x                                                                9x + 27                                                                9x + 27                                                                                    0 So we have: x3 + 9x2 + 27x + 27 = (x + 3)(x2 + 6x + 9) Which we can now factor further with relative ease: = (x + 3)(x + 3)(x + 3) = (x + 3)3
-8x + 3x2 - 3
27x + 3 use distibutive law to factor
You can take a 3x2 out of both of those, leaving 1 and 3x.