
Algebraic
Expressions 


Algebraic expressions  preliminaries 
Evaluating algebraic
expressions 
Simplifying algebraic
expressions 
Expanding
algebraic expression by removing parentheses (i.e. brackets) 
The square of a
binomial  the perfect square trinomial 
The square of a trinomial 
The cube of binomial 
The
binomial expansion algorithm 
Using a variety of methods including combinations of the above to factorize
algebraic expressions 





Algebraic expressions  preliminaries

An
algebraic expression is one or more algebraic terms
containing variables and constants connected by mathematical
operations. Terms
are the elements separated by the plus or minus signs.

In
algebraic expressions, variables are letters, such as
a,
b, c,
or x,
y,
z,
that can have different values.

Constants
are the terms or elements represented only by numbers. Coefficients
are the number part of the terms that multiply a variable or
powers of a variable.

An
algebraic expression consisting of a single term is called a monomial, expression consisting of two terms is
binomial, three
terms trinomial and an expression with more than three terms is
called polynomial.


Evaluating algebraic expressions 
To
evaluate an algebraic expression means to replace (substitute)
the variables in the expression with numeric values that are
assigned to them and perform the operations in the expression. 

Example: 

Evaluate
the expression: 1
+
a^{2}b
+ 1/4a^{4}b^{2 }
for a
= 1 and
b = ^{ }2. 
Solution: 

1 + (1)^{2
}·^{
}(2)
+ 1/4 ·^{ }(1)^{4
}·^{
}(2)^{2}
= 1  2
+ 1/4
·^{ }1
·
4 = 0. 


Or,
since 1+
a^{2}b
+ 1/4a^{4}b^{2 }
= (1+ 1/2a^{2}b)^{2} 


^{ }then,
by substituting a
= 1 and
b
= ^{ }2,
[1 + 1/2·(1)^{2
}·^{ }(2)]^{2
}= 0. 


Simplifying algebraic
expressions 
By
simplifying an algebraic expression, we mean reducing it in the
simplest possible form which mainly involves: multiplication
and division, removing (expanding) brackets and collecting (adding
and subtracting) like
terms. 
Like
terms are those terms which contain the same powers of same
variables and which can only differ in coefficients. 
Examples: 

a)
 4a^{3
}+
3a^{2
}+
5a^{3
} 7a^{2
} = ( 4
+
5) · a^{3
}+
(3  7)
· a^{2 } = a^{3 } 4a^{2}, 


b)
(x^{2
} x^{
}+
1) ·
(x +
1) =
x^{3 } x^{2}
+
x
+
x^{2} ^{
} x^{
}+
1 = x^{3}
+
1. 


Expanding
algebraic expression by removing parentheses (i.e. brackets) 
The
operation of multiplying out algebraic expressions that involve
parentheses using the distributive property is often
described as expanding the brackets. 
Some
important binomial products like perfect squares, and difference
of two squares are used to help with factoring algebraic
expressions. 

Examples: 

a)
(a
 b)^{2}
= (a
 b)
· (a
 b) =
a^{2
} ab^{
} ab^{
}+
b^{2}
= a^{2
} 2ab^{
}+
b^{2}, 


b)
(a
 b)
·
(a
+
b) =
a^{2 } ab^{
}+
ab^{
} b^{2} =
a^{2 }
b^{2}, 


c)
(x^{
}+
y)
· (x^{2}
 xy
+
y^{2}) =
x^{3 } x^{2}y
+
xy^{2}
+
x^{2}y ^{
} xy^{2
}+
y^{3 }
= x^{3}
+
y^{3}. 


The
square of a binomial (or binomial square ) 
To the
square of the first term add twice the product of the two terms
and the square of the last term. 
Examples: 

a)
(a +
b)^{2}
= (a
+ b)
·
(a
+ b) =
a^{2 }+
ab^{
}+ ab^{
}+
b^{2}
= a^{2
}+ 2ab^{
}+
b^{2}, 


b)
(2x +
3)^{2} =
(2x)^{2}
+ 2
· (2x) ·
3
+
3^{2}
= 4x^{2} +
12x +
9, 


c)
(x^{
} 2y)^{2} =
x^{2} +
2 · x
·
(2y)
+
(2y)^{2} =
x^{2} 
4xy +
4y^{2}. 


Squaring
trinomial (or trinomial square) 
To the sum of squares of the 1st, the 2nd and the 3rd term add, twice the product of the 1st and the 2nd term, twice the product of the 1st and the 3rd term, and twice the product of the 2nd and the 3rd term. 

Examples:


a) (x^{2
}  2x
+ 5)^{2}
= (x^{2})^{2
}+ (2x)^{2}
+ 5^{2}
+ 2
·
x^{2 }· (2x)^{
}+ 2
·
x^{2 }· 5
+
2^{ }·
(2x)^{
}·
5
= 


= x^{4
}+ 4x^{2}
+ 25
 4x^{3
}+ 10x^{2}
 20x
= x^{4 }  4x^{3
}+ 14x^{2}
 20x
+ 25, 



b) (a^{3
}
a^{2}b  3ab^{2})^{2}
= (a^{3})^{2 }+ (a^{2}b)^{2}
+ (3ab^{2})^{2}
+ 2a^{3
}(a^{2}b)^{
}+ 2a^{3
}(3ab^{2})
+ 2(a^{2}b)^{
}(3ab^{2})
= 


= a^{6
}+ a^{4}b^{2
}+ 9a^{2}b^{4
} 2a^{5}b
 6a^{4}b^{2
}+ 6a^{3}b^{3
}= a^{6}  5a^{4}b^{2
}+ 9a^{2}b^{4
} 2a^{5}b^{
}+ 6a^{3}b^{3}. 


Cube
of a binomial 
To the
cube of the first term add, three times the product of the
square of the first term and the last term, three times the
product of the first term and the square of the last term, and
the cube of the last term. 

Examples:


a) (a  b)^{3}
= (a
 b)^{2
}·
(a  b)
= (a^{2
} 2ab^{
}+
b^{2})^{
}·
(a  b)
= 


= a^{3 }  2a^{2}b^{
}+
ab^{2
} a^{2}b^{
}+
2ab^{2
} b^{3}
= a^{3 } 3a^{2}b^{
}+
3ab^{2
} b^{3}, 





b) (x  2)^{3}
= x^{3 }+
3^{ }·
x^{2
}·
(2)^{
}+ 3^{
}·
x ·
(2)^{2
}+
(2)^{3
}= x^{3 } 6x^{2
}+
12x^{
} 8, 





c) (2x
+ y)^{3}
= (2x)^{3 }+
3^{ }·
(2x)^{2
}· y^{
}+ 3^{
}·
(2x)
· y^{2 }+ y^{3
}= 8x^{3 }+
12x^{2}y
+
6xy^{2
}+ y^{3}. 


The binomial expansion
algorithm  the binomial theorem 
The
binomial expansion of any positive integral power of a binomial,
which represents a polynomial with n
+ 1 terms, 

or written in the form of the sum formula



is
called the binomial theorem. 
The
binomial coefficients can also be
obtained by using Pascal's triangle. 
The
triangular array of integers, with 1 at the
apex, in which each number is the sum of the two
numbers above it in the preceding row, as is
shown in the initial segment in the diagram, is
called Pascal's triangle. 

So,
for example the last row of the triangle
contains the sequence of the coefficients of a
binomial of the 5th power. 


n 






1 






1 





1 

1 





2 




1 

2 

1 




3 



1 

3 

3 

1 



4 


1 

4 

6 

4 

1 


5 

1 

5 

10 

10 

5 

1 

 
1 

 

 

 

 

 

1 




Using
a variety of methods including combinations of the above to
factorize algebraic expressions 
Examples:


a) x^{2
}  2xy
+ y^{2
} + 2y
 2x
=
(x
 y)^{2}
 2(x
 y)
=
(x
 y)(x
 y
 2), 





b) x^{2
} y^{2
} + xz
 yz
=
(x
 y)(x
+ y)
+ z(x
 y)
= (x
 y)(x
+ y
+ z), 





c) 4x^{2
} 4xy
+ y^{2} ^{
} z^{2
}= (2x  y)^{2} ^{
} z^{2
}
=
(2x
 y
 z)(2x
 y
+ z), 



d) a^{3
} 7a
+ 6^{
}
= a^{3 } a^{
} 6a
+ 6^{
}
= a(a^{2} 1)^{
} 6(a
1)^{
}
= (a 1)·[a(a
+ 1)^{
} 6]
= (a 1)(a^{2}
+ a^{
} 6)^{
}
= 
^{ }
= (a 1)(a^{2}
+ 3a^{
}  2a^{
} 6)^{
}
= (a 1)[a(a
+ 3)^{
}  2(a
+ 3)]
= (a 1)(a
+ 3)(a
 2). 









Precalculus
contents A 



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