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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 C 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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