Applying the distributive law....

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SUMMARY

The discussion focuses on applying the distributive law in algebra, particularly with expressions involving multiple sets of parentheses. The user seeks clarification on how to expand expressions like y = 3(x + 5)(x - 2) and 3(12 - 7r²)(10r - 5). The distributive law is confirmed to be applicable in these scenarios, allowing for the multiplication of each term in one factor by each term in another, resulting in expressions such as ac + ad + bc + bd. The method involves systematically multiplying terms and combining like terms at the end.

PREREQUISITES
  • Understanding of basic algebraic expressions
  • Familiarity with the distributive law (a(b+c) = ab + ac)
  • Knowledge of combining like terms in algebra
  • Ability to manipulate polynomial expressions
NEXT STEPS
  • Practice expanding polynomial expressions using the distributive law
  • Learn about the FOIL method for multiplying binomials
  • Explore factoring techniques for polynomial expressions
  • Study the concept of like terms and how to combine them effectively
USEFUL FOR

This discussion is beneficial for students learning algebra, educators teaching algebraic concepts, and anyone looking to strengthen their skills in expanding and simplifying polynomial expressions.

tmy942
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I'm a beginner and I'm trying to wrap my head around some of the basics of algebra. So I have something like this...

y = 3 ( x + 5 ) ( x - 2 )

or this...

3(12-7r^2)(10r-5)

How would I apply the distributive law to these kinds of equations? I've been trying to research this but the only thing I can find is the distributive law in it's simplest form, i.e. a(b+c) = ab + ac. I already understand that. How would I apply it in situations with multiple sets of parentheses being multiplied?
 
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Suppose you have:

$$(a+b)(c+d)$$

You want to take each term in the first factor, and multiply it by each term in the second factor:

$$ac+ad+bc+bd$$

What I did was to begin with the first term in the first factor and multiply it by each of the terms in the second factor, and then do the same for the second term in the first factor.

You could also do it in two steps:

$$(a+b)(c+d)=a(c+d)+b(c+d)=ac+ad+bc+bd$$

When you are finished, you then want to combine like terms, if there are any. Can you use this method on the two examples you posted?
 

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