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The area of a rectangle whose length and breadth are given by 6p2−5p+26p^{2}-5p+26p2−5p+2​ and p−2p-2p−2​ is
Question

The area of a rectangle whose length and breadth are given by 6p25p+26p^{2}-5p+2​ and p2p-2​ is

A.

6p312p2+17p46p^{3}-12p^{2}+17p-4​​

B.

6p24p6p^{2}-4p​​

C.

6p317p2+12p46p^{3}-17p^{2}+12p-4​​

D.

6p35p2+2p46p^{3}-5p^{2}+2p-4​​

Correct option is C

Given:
Length of the rectangle = 6p25p+26p^{2}-5p+2​​
Breadth of the rectangle = p-2

Formula Used:
Area of Rectangle =Length×Breadth= Length\times Breadth​​

Solution:
Multiply the given polynomials to find the area:
Area=(6p25p+2)(p2)Area=(6p^{2}-5p+2)(p-2)​​
Distribute each term of the first polynomial into the second polynomial:
Area=6p2(p2)5p(p2)+2(p2)Area=6p^{2}(p-2)-5p(p-2)+2(p-2)​​
Expand the terms:
Area=(6p312p2)(5p210p)+(2p4)Area=(6p^{3}-12p^{2})-(5p^{2}-10p)+(2p-4)​​
Combine the like terms:
Area=6p312p25p2+10p+2p4Area=6p317p2+12p4Area=6p^{3}-12p^{2}-5p^{2}+10p+2p-4\\Area=6p^{3}-17p^{2}+12p-4​​
This represents the area of the rectangle.

Final Answer
So the correct answer is (c)

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