How do you multiply #(9k+q)(2k-q)#?

Question

2628 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T15:43:03

#18k^2 - 7kq - q^2#

To multiply two polynomials, simply multiply each term in the first parenthesis with each term in the second, and sum everything.
So, we have:

First element of first parenthesis times first element of second parenthesis: #9k * 2k = 18k^2#
Second element of first parenthesis times first element of second parenthesis: #q * 2k = 2kq#
First element of first parenthesis times second element of second parenthesis: #9k * -q = -9kq#
Second element of first parenthesis times second element of second parenthesis: #q * (-q) = -q^2#

Summing it up:

#18k^2 + 2kq - 9kq - q^2 = 18k^2 - 7kq - q^2#

Answer 2 · 2018-03-29T15:44:46

Use FOIL methos (First,Outer,Inner,Last)

(9k+q)(2k-q)
(9k2k) + (9k-q2k) + (q-q)
=#18k^2#-9kq+2kq- #q^2#

Hope this helps!

Answer 3 · 2018-03-29T16:06:05

#18k^2-q^2-7kq#

(9k+q)(2k-q)
=)#18k^2-q^2+2kq-9kq#
=)#18k^2-q^2-7kq#