Question

Question 1 (Essay Worth 10 points)

Answer 1

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The answer is: 2(k2−4k)(2c+5)

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Step:

* Consider 2ck2+5k2−8ck−20k. Do the grouping 2ck2+5k2−8ck−20k=(2ck2+5k2) +(−8ck−20k), and factor out k2 in the first and −4k in the second group.

* Factor out the common term 2c+5 by using the distributive property.

* Rewrite the complete factored expression.

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