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Question

Simplify the factor completely: 4(x3+1)(x2−9)

A
4(x+1)(x2x+1)(x+3)(x+3)
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B
4(x+1)(x2x+1)(x3)(x+3)
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C
4(x+1)(x2x+1)(x+3)2
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D
(x1)(x2x+1)(x3)(x+3)
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Solution

The correct option is B 4(x+1)(x2x+1)(x3)(x+3)
4(x3+1)(x29)
We know that x3+y3=(x+y)(x2x+1)
x2y2=(xy)(x+y)
So, 4(x3+1)(x232)=4(x+1)(x2x+1)(x3)(x+3)

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