How do you perform this problem? (2k²+4k-6) ÷ 2k-2/k-3?

Question

simplify please!!

Answer 1

First, factor (2k²+4k-6).
it becomes (2k-2)(k+3).

Dividing (2k-2)(k+3) by (2k-2)/(k-3) is the same as multiplying by the reciprocal [(k-3)/(2k-2)]

When you multiply these two terms together, you'll see that each (2k-2) from each term "cancel out", leaving you with (k+3)(k-3). depending on how you want your answer, (k+3)(k-3) is the same as k²-9.

Answer 2

(2k^2 + 4k - 6) divided by 2k - 2/ k-3
2 (k^2 + 2k - 3) divided by 2k - 2/ k-3
2 (k - 3) (k -1) divided by 2k-2 / k-3
so you can cross multiply ( sorry its hard to show that on the computer)

but the answer is 1 (you can cancel out k-3 on the numerator and denominator. and then you have 2 (k-1) over 2k-2
in which you can factor out a 2

so then you have 2 (k-1) over 2( k-1) and you get 1! HOPE THAT HELPS

Answer 3

2k^2+4k-6 / 2k-2/ k-3

(2k-2)(k+3) / 2k-2/k-3

2k-2 k+3 * k-3 / 2k-2

k+3/ k-3

Answer 4

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