Download (5x + 3)(7x – 2)

The Mathematics 11
Competency Test
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
plus
O(uter) – the product of the two outer terms when viewed as shown above
plus
I(nner) – the product of the two inner terms when viewed as shown above
plus
L(ast) – the product of the two last terms in each binomial.
Diagrammatically, using the example above, we have
Outer = (5x)(-2)
First = (5x)(7x)
(5x + 3)(7x – 2)
Inner = (3)(7x)
Last = (3)(-2)
Then
(5x + 3)(7x – 2) = First + Outer + Inner + Last
= (5x)(7x) + (5x)(-2) + (3)(7x) + (3)(-2)
= 35x2 – 10x + 21x – 6
= 35x2 + 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.
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 1 of 6
Example 1: Expand (3x – 2y) (4x + 7y) .
solution:
(3x – 2y) (4x + 7y) = First + Outer + Inner + Last
= (3x)(4x) + (3x)(7y) + (-2y)(4x) + (-2y)(7y)
= 12x2 + 21xy – 8xy – 14y2
= 12x2 + 13xy – 14y2
Method 2: This method exploits the definition of brackets and the distributive law of multiplication
described in an earlier note. 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)
= 35x2 – 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)
= 35x2 + 21x – 10x – 6
= 35x2 +11x - 6
identical to the result obtained with Method 1 earlier.
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 2 of 6
Example 2: Expand (3x – 2y)(4x + 7y) .
solution:
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)
= 12x2 – 8xy + 21xy – 14y2
= 12x2 + 13xy – 14y2 .
Example 3: A rectangle has length equal to 3x + 7 meters and its
width is equal to 4x – 5 meters. Develop formulas for the perimeter
and area of this rectangle in terms of x. State your final results as
formulas containing no brackets.
4x -5
solution: To solve this problem requires some knowledge of the
properties of rectangles (covered in a later note on this web site).
However, you probably already know that for a rectangle,
Perimeter = distance around
= 2 x length + 2 x width
and
Area = amount of surface covered
= length x width.
Thus, in this case,
Perimeter = 2 x length + 2 x width
= 2(3x + 7) + 2(4x – 5)
= (2)(3x) +(2)(7) + (2)(4x) + (2)(-5)
= 6x + 14 + 8x – 10
= 14x + 4
as the final result. Notice that we had to introduce brackets in the second line to ensure that the
length, 3x + 7, in its entirety, was multiplied by 2, and also, that the width, 4x – 5, in its entirety,
was multiplied by 2.
Now,
Area = length x width
= (3x + 7)(4x – 5)
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 3 of 6
= (3x)(4x) + (3x)(-5) + (7)(4x) + (7)(-5)
= 12x2 – 15x + 28x – 35
= 12x2 + 13x – 35.
Here we used the FOIL method to expand the product of binomials in order to remove the
brackets. Make sure you understand why brackets had to be used in the second line here to
assure that a correct formula for the area was obtained.
Remark 1: There are some special cases of binomial products which you can now understand
and which we will exploit later when we discuss factoring. Here we’ll just list them. In what
follows, the symbols a, b, c, and d stand for specific numbers.
The product of two binomials involving just a single variable x can always be simplified to a
second degree polynomial (or a trinomial, if you like). In general
( ax + b )( cx + d ) = ( ac ) x 2 + ( ad + bc ) x + bd
Our first example in this document demonstrated this.
If the coefficients of x in the binomials are both equal to 1, then a somewhat simpler result occurs:
( x + a )( x + b ) = x 2 + ( a + b ) x + ab
Notice the pattern here:
(x + a)(x + b) = x2
coefficient is
equal to 1
+
(a + b)x
coefficient is the sum
of the two constant
terms in the binomials.
+
ab
product of the
two constant
terms in the
binomials.
Example 4: (x + 5)(x + 7) = (1)x2 + (5 + 7)x + (5)(7) = x2 + 12x + 35.
Example 5: (x + 5)(x – 7) = (1)x2 +[5 + (-7) ]x + (5)(-7) = x2 – 2x – 35.
You will get the same result in these cases using either of our earlier method 1 or method 2.
Squaring a binomial is, of course, a special case of multiplication involving the same factor twice.
The simplest instance of squaring a binomial is
( x + a)
2
= ( x + a )( x + a ) = x 2 + 2ax + a 2
Notice the pattern here:
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 4 of 6
(x + a)2 = x2
+
coefficient
is 1
2ax
+
coefficient is double
the constant term in
the binomial
a2
coefficient is the
square of the constant
term in the binomial.
Example 6: (x + 7)2 = (1)x2 + (2)(7)x + (7)2 = x2 + 14x + 49 .
Example 7: (x - 7)2 = (1)x2 + (2)(-7)x + (-7)2 = x2 - 14x + 49 .
When the coefficients of x in the two binomials are equal, and the constant terms in the two
binomials are numerically identical, but of opposite sign, the simplified expansion of the product is
itself a binomial:
( ax + b )( ax − b ) = a2 x 2 − b 2
Example 8: (2x + 3)(2x – 3) = (2x)(2x) + (2x)(-3) + (3)(2x) + (3)(-3)
= 4x2 – 6x + 6x – 9
= 4x2 – 9
or, using the pattern shown in the boxed formula just above,
(2x + 3)(2x – 3) = (2)2x2 – (3)2 = 4x2 – 9 .
You can see what happens in this case: the “inner” and “outer” contributions (when using the
FOIL method) are identical, but of opposite sign, so they cancel out, leaving just the other two
terms.
We will exploit these special patterns to some extent later when we consider the operation of
factoring.
Remark 2: Method 2 described above is more general and more powerful than the FOIL method
because it is easily extendable to products involving factors with more than two terms. Such
problems are a bit beyond the scope of the BCIT Mathematics 11 Competency Test, but it is
worth presenting one example here to help you understand the principle.
Example 8: Expand (3x + 5y)(2x + 8y – 7z) .
solution:
(3x + 5y)(2x + 8y – 7z)
= (3x + 5y)(2x) + (3x + 5y)(8y) + (3x + 5y)(-7z)
= (2x)(3x + 5y) + (8y)(3x + 5y) + (-7z)(3x + 5y)
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 5 of 6
= (2x)(3x) + (2x)(5y) + (8y)(3x) + (8y)(5y) + (-7z)(3x) + (-7z)(5y)
= 6x2 + 10xy + 24xy + 40y2 – 21xz – 35yz
= 6x2 + 34xy + 40y2 – 21xz – 35yz.
It would be difficult to come up with a generalization of the FOIL method to handle this example
(we’d need six meaningful words that are easy to remember, but distinguish between middles,
and firsts, and lasts, and inners and outers, and so forth). However, the application of Method 2
occurs quite naturally here, and you can probably even see how we would handle products
involving factors with even more terms.
David W. Sabo (2003)
Expansion of a Product of Binomials
Page 6 of 6