which shows the factors of x^3-3x^2+4x-12?

Answer 1

x³ - 3x² + 4x - 12

Notice that the ratio of the first two coefficients (1:-3) is the same as the ratio of the last two (4:-12 = 1:-3). Whenever this happens (or when the ratio of the 1st and 3rd is the same as the ratio of the 2nd and 4th), you will have a common factor.

Factor x² out of the first two, and 4 out of the last two:

x²(x - 3) + 4(x - 3)

Now you can factor (x + 3) out of both:

(x - 3)(x² + 4)

Since x² + 4 is not factorable (over the real numbers), that's as far as you can go.

If you're looking for complex roots too, then x² + 4 factors to x + 2i and x - 2i:

(x + 3)(x - 2i)(x + 2i)

Answer 2

x^3-3x^2+4x-12 1 factor is x-3

x^2+4
(x-3)\x^3 - 3 x^2 + 4 x - 12
- ( x^3 - 3 x^2)
4 x - 12
4x -12
0

x^3-3x^2+4x-12=(x-3)(x^2+4) x^2+4 has no real factors
(x-3)(x+2i)(x-2i)

Answer 3

x^3-3x^2+4x-12

Group the first two terms and the last two terms together.

x^3-3x^2 + 4x-12

Factor an x^2 out of the first group and a 4 out of the second group.

x^2(x-3) + 4(x-3)

Factor (x-3) out of both terms and your answer is:

(x-3)(x^2+4)

Answer 4

Looks like factor by grouping

x^2(x-3)+4(x-3)=(x^2+4)(x-3)

Answer 5

x^3 - 3x^2 + 4x - 12
(x^2 + 4)(x - 3)

Answer 6

x^3-3x^2+4x-12 = (x-3)(x^2 + 4)

Now if you're not restricted to real factors..

x^3-3x^2+4x-12 = (x-3)(x+2i)(x-2i)

Answer 7

x^3-3x^2+4x-12
=x^2(x-3) +4(x-3)
=(x^2+4)(x-3)

Answer 8

x^2(x-3)+4(x-3)
(x^2+4)(x-3)

Answer 9

(x-3)(x^2+4)

Unfortunately, (x^2+4) can not be factored (using real numbers). It does, however, factor into (x+2i)(x-2i), using i (the square root of -1).

So the roots of the equation are 3 (the only real root) and 2i and -2i.

Answer 10

x^3-3x^2+4x-12
(x^3-3x^2)+(4x-12)
x^2(x-3)+4(x-3)
(x^2+4)(x-3)
(x-2i)(x+2i)(x-3)
x= 2i, -2i and 3