How do you multiply #(11z ^ { 2} + z ) ^ { 2}#?

1 Answer
Jun 2, 2017

See a solution process below:

Explanation:

This is a special form of multiplying terms. Use this rule:

#(a + b)^2 = a^2 + 2ab + b^2#

Substituting #11z^2# for #a# and #z# for #b# gives:

#(11z^2 + z)^2 => (11z^2)^2 + 2(11z^2)z + z^2 => #

#121x^4 + 22x^3 + x^2#