SOLUTION: Hi- how do you solve these two problems while using "FOIL"? 1. (3X+2)(X+7) 2. (X-5)(2X-6)

Algebra ->  Expressions -> SOLUTION: Hi- how do you solve these two problems while using "FOIL"? 1. (3X+2)(X+7) 2. (X-5)(2X-6)      Log On


   

Question 637096: Hi- how do you solve these two problems while using "FOIL"?
1.
(3X+2)(X+7)


2.
(X-5)(2X-6)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do the first one to get you started.


%283x%2B2%29%28x%2B7%29 Start with the given expression.


Now let's FOIL the expression.


Remember, when you FOIL an expression, you follow this procedure:


%28highlight%283x%29%2B2%29%28highlight%28x%29%2B7%29 Multiply the First terms:%283%2Ax%29%2A%28x%29=3%2Ax%5E2.


%28highlight%283x%29%2B2%29%28x%2Bhighlight%287%29%29 Multiply the Outer terms:%283%2Ax%29%2A%287%29=21%2Ax.


%283x%2Bhighlight%282%29%29%28highlight%28x%29%2B7%29 Multiply the Inner terms:%282%29%2A%28x%29=2%2Ax.


%283x%2Bhighlight%282%29%29%28x%2Bhighlight%287%29%29 Multiply the Last terms:%282%29%2A%287%29=14.


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So we have the terms: 3%2Ax%5E2, 21%2Ax, 2%2Ax, 14


3%2Ax%5E2%2B21%2Ax%2B2%2Ax%2B14 Now add every term listed above to make a single expression.


3x%5E2%2B23x%2B14 Now combine like terms.


So %283x%2B2%29%28x%2B7%29 FOILs to 3x%5E2%2B23x%2B14.


In other words, %283x%2B2%29%28x%2B7%29=3x%5E2%2B23x%2B14.