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How do you multiply #(p^2+5p+2)(3p^2+p)#?

Answer 1

Eva Abramson

#3p^4 + 16p^3 + 11p^2 + 2p#

Multiply each term in one bracket with each term in the other:

#(p^2 + 5p + 2)(3p^2 + p) =#

#(p^2*3p^2) + (p^2 * p)# + ... #(5p *3p^2) + (5p*p)#+ ... #(2*3p^2) + (2*p) =#

# 3p^4 + p^3 + 15p^3 + 5p^2 + 6p^2 + 2p = #

# 3p^4 + 16p^3 + 11p^2 + 2p #

Notice how I took the first term in the first bracket and multiplied it by each of the terms in the second bracket and obtained the answer by simply adding up all of these brackets together and simplifying them.

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Answer 2

Nathan Axel

To multiply (p^2+5p+2)(3p^2+p), you can use the distributive property or the FOIL method. Here's the result after multiplying each term in the first expression by each term in the second expression:

(p^2)(3p^2) + (p^2)(p) + (5p)(3p^2) + (5p)(p) + (2)(3p^2) + (2)(p)

This simplifies to:

3p^4 + p^3 + 15p^3 + 5p^2 + 6p^2 + 2p

Combining like terms, the final result is:

3p^4 + 16p^3 + 11p^2 + 2p

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.