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.
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.
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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
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)
= 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.
3(x^2 + 2x + 4)
x^2 + 2x + 4 is prime. It doesn't factor nicely.