SOLUTION: I got two questions the first is a quadratic expression: 9k^2+66k+21 Next is a quadratic equation: 3k^2-39k=-120

Question 775824: I got two questions the first is a quadratic expression:
9k^2+66k+21
Next is a quadratic equation:
3k^2-39k=-120
Answer by tanjo3(60) (Show Source): You can put this solution on YOUR website!
(1) Factor out a 3 from the expression.
3(3k^2 + 22k + 7)

(2) Observe that the expression is in the form ak^2 + bk + c. Here we can apply the quadratic forumla.

k = , -7
(3) Place you zeros back into the expression.
3[(3k + 1)(k + 7)]

-----------------------------------

(1) Move all terms to one side.
3k^2 - 39k + 120 = 0
(2) Factor out a 3 from the expression.
3(k^2 - 13k + 40) = 0

(3) Observe that the expression is in the form ak^2 + bk + c. Find two number which add up b and multiply to c.
-5 + -8 = 13; -5 � -8 = 40
(4) Substitute these numbers in 3[(k + ___) (k + ___)]
3[(k - 5) (k - 8)]
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