B....................................................
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The answer is: <u>2(k2−4k)(2c+5)</u>
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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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Y=6
i hope you understand