How to factor 3x^2+6x+12
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How to factor 3x^2+6x+12

[From: ] [author: ] [Date: 12-05-13] [Hit: ]
a = 3. b = 6,x^2 + 2x + 4 is prime.It doesnt factor nicely.......
First divide through by the common 3 to get:

3(x^2 + 2x + 4)

Since there are no factors of 4 that add up to 2, the trinomial can't be factored to integer and you're done.


You can confirm that by using the discriminant (the b^2 - 4ac term) from the quadratic equation.

Use 1 for a, 2 for b, and 4 for c:

2^2 - (4 * 1 * 4) = 4 - 16 = -12

Since the discriminant is negative and the quadratic formula takes the square root of it, there are no real roots and the trinomial can't be factored to integers.

-
3x^2 + 6x + 12
a = 3. b = 6, and c = 12
x
= [-b +/- sqrt (b^2 - 4ac)]/2a
= [-6 +/- sqrt (36 - 144)/6
= (-6 +/- sqrt (-108))/6
= (-6 +/- 6sqrt (-3))/6
= -1 +/- i sqrt 3

-
3x² + 6x + 12
= 3(x² + 2x + 6)
= 3 (-i x + √3 - i) (i x + √3 + i)

-
3x^2+6x+12 =
3(x^2 + 2x + 4)

x^2 + 2x + 4 is prime. It doesn't factor nicely.
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