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Notes on the Difference of 2 Squares
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Fractions, Decimals and Percents
Solving Systems of Equations by Substitution
Quotient Rule for Radicals
Prime Polynomials
Solving Nonlinear Equations by Substitution
Simplifying Radical Expressions Containing One Term
Factoring a Sum or Difference of Two Cubes
Finding the Least Common Denominator of Rational Expressions
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Expansion of a Product of Binomials
Solving Equations
Exponential Growth
Factoring by Grouping
Solving One-Step Equations Using Models
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Rationalizing the Denominator
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What is a Quadratic Equation
Laws of Exponents and Multiplying Monomials
The Slope of a Line
Factoring Trinomials by Grouping
Multiplying and Dividing Rational Expressions
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Multiplication Property of Exponents
Multiplying and Dividing Fractions 3
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Adding and Subtracting Functions
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Factoring a Polynomial by Finding the GCF
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Polar Form of a Complex Number
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Graphing Horizontal and Vertical Lines
Invariants Under Rotation
The Addition Method
Solving Linear Inequalities in One Variable
The Pythagorean Theorem
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Expansion of a Product of Binomials

Here we look at two ways to approach the expansion or removal of brackets from a product of the form

(5x + 3)(7x – 2)

in which two binomials are being multiplied together.

Method 1:

This is the “recipe method” based on the acronym FOIL. The result of the multiplication is the sum of four terms:

  F(irst) – the product of the two first terms in each binomial
  O(uter) – the product of the two outer terms when viewed as shown above
  I(nner) – the product of the two inner terms when viewed as shown above
  L(ast) – the product of the two last terms in each binomial.

Diagrammatically, using the example above, we have


(5x + 3)(7x – 2) = First + Outer + Inner + Last

= (5x)(7x) + (5x)(-2) + (3)(7x) + (3)(-2)

= 35x 2 – 10x + 21x – 6

= 35x 2 + 11x – 6

Notice that subtracts are handled as minus signs retained with the terms being multiplied. At the end of the process, we also carry out any simplification by collection of like terms that may be obvious.


Example 1:

Expand (3x – 2y) (4x + 7y) .


(3x – 2y) (4x + 7y) = First + Outer + Inner + Last

= (3x)(4x) + (3x)(7y) + (-2y)(4x) + (-2y)(7y)

= 12x 2 + 21xy – 8xy – 14y 2

= 12x 2 + 13xy – 14y 2


Method 2:

This method exploits the definition of brackets and the distributive law of multiplication. In the simpler problem of expanding and expression such as

5x(7x – 2)

the ‘5x’ multiplying onto the bracketed expression indicates that every term inside the brackets is to be multiplied by 5x:

5x(7x – 2) = (5x)(7x) + (5x)(-2)

= 35x 2 – 10x

We just extend this notion to include multiplication by a binomial instead of just a monomial. So

(5x + 3)(7x – 2)

means “multiply every term in (7x – 2) by the factor (5x + 3).” This gives

(5x + 3)(7x – 2) = (5x + 3)(7x) + (5x + 3)(-2)

= 7x(5x + 3) - 2(5x + 3)

In the second line here, we’ve just rearranged the previous expression slightly to remove some of the clutter. But now you see that the original problem of expanding a product of two binomials has turned into a problem of expanding two products of a monomial and a binomial – the sort of thing that can be handled by the procedures described in the previous note. So, we just continue the multiplication process, making sure the final result is simplified as much as possible. As usual, we handled subtractions as additions of negated quantities. The overall process in this example is:

(5x + 3)(7x – 2) = (5x + 3)(7x) + (5x + 3)(-2)

= (7x)(5x + 3) + (-2)(5x + 3)

= (7x)(5x) + (7x)(3) + (-2)(5x) + (-2)(3)

= 35x 2 + 21x – 10x – 6

= 35x 2 +11x - 6

identical to the result obtained with Method 1 earlier.


Example 2:

Expand (3x – 2y)(4x + 7y) .


We’ll demonstrate Method 2 for this problem:

(3x – 2y)(4x + 7y) = (3x – 2y)(4x) + (3x – 2y)(7y)

= (4x)(3x – 2y) + (7y)(3x – 2y)

= (4x)(3x) + (4x)(-2y) + (7y)(3x) + (7y)(-2y)

= 12x 2 – 8xy + 21xy – 14y 2

= 12x 2 + 13xy – 14y 2 .


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