Question

How do you FOIL\[(4z - u)(3z + 2u)\]?

Answer 1

Hint: FOIL is nothing but a method to multiply polynomials, FOIL (First, Outer, Inner, and Last) the sum of all these terms. Which will give the resultant expression that we want. The terms which come first on the given parameter are first, outer means either end terms. Inner means vice-versa for the Outer term. Last means second term on first parameter & last term on second parameter.

Formula Used: FOIL has a formula if the expression is in the form of above diagram \[(a + b)(c + d)\]
First: \[a \times c\]
Outer: \[a \times d\]
Inner: \[b \times c\]
Last: \[b \times d\]
The sum will be \[ac + ad + bc + bd\]

Complete step-by-step solution:
Before solving consider the above diagram in which we will use FOIL to solve the expression. In the same way we will solve this equation. We have to multiply two binomials. We can directly use FOIL,
\[(4z - u)(3z + 2u)\]
Sum of First, Outer, Inner, Last
First: \[(4z)(3z) = 12{z^2}\]
Outer (Outside): \[(4z)(2u) = 8uz\]
Inner: \[( - u)(3z) = - 3uz\]
Last:- \[( - u)(2u) = - 2{u^2}\]
Sum of all these terms will be
\[
   \Rightarrow 12{z^2} + 8uz - 3uz - 2{u^2} \\
   \Rightarrow 12{z^2} + 5uz - 2{u^2} \\
 \]
The Solution for the FOIL \[(4z - u)(3z + 2u)\] is \[12{z^2} + 5uz - 2{u^2}\]

Additional information: FOIL is a method to multiply the polynomial terms in the parameters which is a standard to multiply. You can even multiply with the first term with the other terms in parameters and second term with another two terms in another parameter. The sum will be the same either in both ways. \[(a + b)(c + d)\]
\[
  a(c + d) + b(c + d) \\
  a \times c + a \times d \\
 \]
Then
\[b \times c + b \times d\]
The sum will be same \[ac + ad + bc + bd\]

Note: When you have a trinomial multiplied with the binomial, then only consider grouping the terms and use them as one \[(a + b + c) = ((a + b) + c)\] like this to use the formula FOIL to solve the expression.