What is the product of (2p + 7)(3p2 + 4p – 3)?
a=6p3 + 29p2 – 34p + 21
b=6p3 + 29p2 – 22p + 21
c=6p3 + 29p2 + 22p – 21
d=6p3 + 29p2 + 34p – 21

The answer is C=6p3 + 29p2 + 22p – 21

To calculate the product, we need to multiply each member of each multiplier:
(2p + 7)(3p2 + 4p – 3) = 2p · 3p² + 2p · 4p + 2p · -3 + 7 ·3p² + 7 · 4p + 7 · -3
                                    =    6p³     +     8p²    -     6p    + 21p²   +  28p   -     21
                                    = 6p³ + 8p² + 21p² + 28p - 6p -21
                                    = 6p³ + 29p² + 22p - 21

Therefore, the product of (2p + 7)(3p2 + 4p – 3) is 6p³ + 29p² + 22p -21

That’d be 9 years or something