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Question

# Simplify the expression: (p3+11p2+28p)(p+4)

A
(p+7)
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B
p(p+7)
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C
(p+8)
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D
p(p+8)
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Solution

## The correct option is B p(p+7)Given: (p3+11p2+28p)(p+4) Taking 'p' common from the numerator, we get p(p2+11p+28)(p+4) ..(i) Now, comparing the expression p2+11p+28 with the identity x2+(a+b)x+ab we note that, (a+b)=11 and ab=28. So, (7+4)=11 and (7)(4)=28 Hence, p2+11p+28 =p2+7p+4p+28 =p(p+7)+4(p+7) =(p+4)(p+7) Therefore from (i), p(p2+11p+28)(p+4)=p(p+4)(p+7)(p+4)=p(p+7)

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