Factor the following quadratic expressions.
a. 2x^2 + 10x + 12
b. 6x^2 + 5x − 6

  a.  2(x +2)(x +3)

  b.  (2x +3)(3x -2)

Step-by-step explanation:

a. First the common factor of 2 can be removed:

  2(x^2 +5x +6)

Now, we're looking for factors of 6 that sum to 5

  6 = 1·6 = 2·3 . . . . . the latter pair has a sum of 5

These constants go into the binomial factors directly:

  2(x +2)(x +3)

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b. There are no common factors among the coefficients. We need to find factors of ac = (6)(-6) = -36 that have a sum of b = 5.

  -36 = -1·36 = -2·18 = -3·12 = -4·9 = -6·6

The factor pair (-4, 9) has a sum of +5, so that is the one we're looking for. We can rewrite the middle term using these numbers, then factor by grouping.

  6x² -4x +9x -6

  2x(3x -2) +3(3x -2) . . . . . . factor the first pair of terms and the last pair

  (2x +3)(3x -2) . . . . . . . . . . factor out the common factor of 3x-2

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)