

Polynomial and/or Polynomial
Functions and Equations 
Definition
of a polynomial or polynomial
function 
Division
of polynomials 
Division of polynomials
examples

Factoring
polynomials and solving
polynomial equations by factoring 
Solving quadratic and
cubic equations by factoring, examples 





Definition
of a polynomial or polynomial function 
A
polynomial and/or polynomial function in the variable x
is an expression of the general (or standard) form 
f (x)
= a_{n}x^{n}
+ a_{n}_{}_{1}x^{n}^{}^{1}
+
.
. . +
a_{1}x
+ a_{0} 
consisting
of n
+ 1 terms each of which is a product of a real coefficient a_{i}
and the variable x
raised to a nonnegative
integral power. 
If
the leading coefficient of a polynomial a_{n }
is not
0
then, the
degree of the polynomial is n. 
The constant term a_{0}
is the yintercept
of the polynomial. 

Division of polynomials

Dividing two polynomials,
p (x) and
q (x)
obtained is quotient Q
(x)
and remainder R (x),
that we write as 
p (x)/ q(x)
= Q (x) + R (x)/q (x)
or
p (x)
= Q
(x)
· q
(x)
+ R (x). 
Two polynomials are divisible if
R (x)
= 0. 
Divide
the highest degree term of dividend by the highest degree term
of the divisor to get the first term of the quotient. 
Take the first term of the quotient and multiply it by every term of divisor.
Write this result below the dividend, making sure you line up all the terms with the terms of the dividend that has the same degree. 
Subtract the result from the dividend, i.e., reverse all the signs of the terms of the result and add like terms. 
Repeat the process of long division until the degree of the new obtained dividend is less than the degree of the divisor. 

Division of polynomials examples


Note,
since each second line should be subtracted, the sign of each term is reversed. 

b)
(
3x^{4} +
x^{2} +
5) ¸
( x^{2} 
x 
1) = 
2x^{2} +
x 
1 






like 
17 ¸
5 = 3 +
2/5 

15 

2 



Factoring
polynomials and solving
polynomial equations by factoring 
A
polynomial and/or polynomial
function with
real coefficients can be expressed as a product of its leading
coefficient a_{n
}and
n
linear factors of the form (x
 x_{i}),
where x_{i}_{
}denotes its real root and/or complex root, 
f (x)
= a_{n}x^{n}
+ a_{n}_{}_{1}x^{n}^{}^{1}
+
.
. . +
a_{1}x
+ a_{0}
= a_{n}(x
 x_{1})(x
 x_{2})
.
. . (x
 x_{n}).

By
multiplying the parentheses on the right side and collecting
like terms and then comparing the resulting coefficients with
the coefficients of the given polynomial obtained are Vieta's
formulas that show relations between coefficients and roots of a
polynomial. 
Thus,
for a quadratic function or a second
degree polynomial 

a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{2}(x
 x_{1})(x
 x_{2})
= a_{2}[x^{2}

(x_{1} +
x_{2})x
+
x_{1}x_{2}], 



and
similarly, for a cubic function or a third
degree polynomial 

a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{3}(x
 x_{1})(x
 x_{2})(x
 x_{3})
= 



=
a_{3}[x^{3}

(x_{1} +
x_{2}
+
x_{3})x^{2}
+
(x_{1}x_{2} +
x_{1}x_{3}
+
x_{2}x_{3})x
 x_{1}x_{2}x_{3}]. 


Solving quadratic and
cubic equations by factoring, examples 
Example:

Factorize (2/3)x^{2}
 (2/3)x
 4
using
the above theorem. 
Solution: 
(2/3)x^{2}
 (2/3)x
 4
=
(2/3)(x^{2}
 x
 6)
=
(2/3)[x^{2
}
 (3
+ (
2))x
+ 3(
2)] =


= (2/3)(x^{2
}
 3x
+ 2x
 6)
=
(2/3)[x(x^{
}
 3)
+ 2(x^{
}
 3)]
=
(2/3)(x^{
}
 3)(x
+ 2) 


Example:
Given
are leading
coefficient a_{2}_{
} =
1and
the pair
of conjugate complex
roots,
x_{1 } =
1 +
i
and 
x_{2
} =
1  i,
of
a second
degree polynomial, find
the polynomial using
the above theorem. 
Solution: 
By
plugging the given values into a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{2}(x
 x_{1})
(x
 x_{2}) 

a_{2}x^{2}
+ a_{1}x
+ a_{0} = 1[x

(1
+ i)]
· [x

(1  i)]
= [(x
 1)
 i]
· [(x  1)
+ i]
=


= [(x  1)^{2}
 i^{2}]
=
(x^{2}
 2x
+ 1 +
1)
=
 x^{2}
+ 2x
 2 


Example:

The
real root of the polynomial 
x^{3}

x^{2}
+ 4x
 6
is
x_{1 } =
 3,
factorize the polynomial. 
Solution: 
We
divide given polynomial by one of its known
factors, 

a_{3}x^{3}
+ a_{2}x^{2}
+ a_{1}x
+ a_{0} = a_{3}(x
 x_{1})(x
 x_{2})(x
 x_{3}) 



then
we calculate another two roots of given cubic by solving
obtained quadratic trinomial, 



Finally
we use the theorem to factorize given polynomial (see
the previous example), 

a_{3}(x
 x_{1})(x
 x_{2})(x
 x_{3})
= 1(x
+ 3)[x

(1
+ i)][x

(1  i)]
= 1(x
+ 3)(x^{2}
 2x
+ 2). 

Notice that given cubic has one real root and the pair of the conjugate complex roots. 

Odd
degree polynomials must have at least one real root. 


Example:
Solve polynomial
equation x^{3}_{
}+_{ } 2x^{2 }
x 
2 = 0 by factoring. 
Solution:
x^{2}(x
+_{ } 2) ^{ }
(x +_{ } 2) = 0 
(x +_{ }
2)·(x^{2} _{ }
1) = 0 
(x +_{ }
2)·(x +_{ }1)·(x _{
}1) = 0 
x +_{ }
2 = 0 =>
x_{1} = 2 
x +_{ }
1 = 0 =>
x_{2} = 1 
x _{
}1 = 0 =>
x_{3} = 1 
The
roots are: x_{1}
= 2,
x_{2} = 1
and x_{3}
= 1. 








Functions
contents C 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 