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Polynomial and/or Polynomial
Functions and Equations |
Zeros of
a polynomial function |
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Addition and subtraction of
polynomials
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Multiplication
of polynomials |
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Division
of polynomials |
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Factoring
polynomials and solving
polynomial equations by factoring |
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| Zeros
of a polynomial function |
| The
zeros of a polynomial function are the values of x
for which the function equals zero. |
| That
is, the solutions of the equation f
(x)
= 0,
that are called roots of the polynomial, are the zeros of the
polynomial function or the x-intercepts of its
graph in a coordinate plane. |
| At
these points the graph of the polynomial function cuts or
touches the x-axis. |
| If
the graph of a polynomial intersects with the x-axis
at (a,
0), or x
= a
is a root or zero of a polynomial, then
(x
-
a)
is a factor of that polynomial. |
| Every
polynomial of degree n
has exactly n
real or complex zeros. |
| An
nth
degree polynomial has at most n
real zeros. |
| Some
of the roots may be repeated. The number of times a root is
repeated is called multiplicity or
order of the root. |
| The
number
xi
is a root of the
polynomial f
(x)
if and only if
f (x)
is divisible by
(x
-
xi).
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| Thus, finding the
roots of a
polynomial f
(x)
s equivalent to finding its
linear divisors or is equivalent to polynomial
factorization into
linear factors. |
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| Addition and subtraction of
polynomials
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| We
add or subtract polynomials by combining their like
terms. The like terms are terms that have the same
variables raised to the same exponents. |
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| Example:
a)
( -
5x3 +
2x2
- x +
4) + ( -
4x2 +
3x - 7)
= - 5x3 +
(
2 -
4) · x2 +
(
-1
+
3) · x +
4 - 7
= |
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= - 5x3 - 2x2 +
2x - 3 |
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b) (
x4 -
3x3
+
5x - 1)
-
( - 2x4 +
x3
- 3x2 +
4)
= x4 -
3x3
+
5x - 1
+ 2x4 -
x3
+
3x2 -
4
= |
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= (
1 +
2) · x4
+ (
-3
- 1)
· x3
+
3x2
+
5x - 5
= 3x4 -
4x3
+
3x2
+
5x - 5 |
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| Multiplication of polynomials
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| When
multiplying two polynomials together, multiply every term of one
polynomial by every term of the other polynomial using
distributive property. |
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| Example:
( -
2x3 +
5x2
- x +
1) ·
(
3x - 2)
= |
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= 3x · (
- 2x3)
+ 3x · 5x2
+ 3x ·
(-x) + 3x · 1
+ (- 2) · ( -
2x3)
+ (- 2) · 5x2
+ (- 2) · ( -x)
+ (-
2) · 1
= |
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= - 6x4
+ 15x3
- 3x2
+ 3x +
4x3 -
10x2
+ 2x -
2 = -
6x4
+ 19x3
-
13x2
+ 5x -
2 |
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| Division of polynomials
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| 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. |
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| Note,
since each second line should be subtracted, the sign of each term is reversed. |
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b)
(
3x4 +
x2 +
5) ¸
( x2 -
x -
1) |
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| like |
17 ¸
5 = 3 +
2/5 |
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-15 |
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2 |
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| 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 an
and
n
linear factors of the form x
-
xi,
where xi
denotes its real roots and/or complex
roots, |
| f
(x)
= anxn
+ an-1xn-1
+
. . .
+
a1x
+ a0
= an(x
- x1)(x
- x2)
. . . (x
- xn).
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| 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. |
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| Thus,
for a quadratic or a second
degree polynomial |
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a2x2
+ a1x
+ a0 = a2(x
- x1)(x
- x2)
= a2[x2
-
(x1 +
x2)x
+
x1x2], |
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| and
similarly, for a cubic or a third
degree polynomial |
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a3x3
+ a2x2
+ a1x
+ a0 = a3(x
- x1)(x
- x2)(x
- x3)
= |
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=
a3[x3
-
(x1 +
x2
+
x3)x2
+
(x1x2 +
x1x3
+
x2x3)x
- x1x2x3]. |
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Example:
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Factorize (2/3)x2
- (2/3)x
- 4
using
the above theorem. |
| Solution: |
(2/3)x2
- (2/3)x
- 4
=
(2/3)(x2
- x
- 6)
=
(2/3)[x2
- (3
+ (-
2))x
+ 3(-
2)] =
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= (2/3)(x2
- 3x
+ 2x
- 6)
=
(2/3)[x(x
- 3)
+ 2(x
- 3)]
=
(2/3)(x
- 3)(x
+ 2) |
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| Example:
Given
are leading
coefficient a2
=
-1and
the pair
of conjugate complex
roots,
x1 =
1 +
i
and |
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x2
=
1 - i,
of
a second
degree polynomial, find
the polynomial using
the above theorem. |
| Solution: |
By
plugging the given values into a2x2
+ a1x
+ a0 = a2(x
- x1)
(x
- x2) |
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a2x2
+ a1x
+ a0 = -1[x
-
(1
+ i)]
· [x
-
(1 - i)]
= -[(x
- 1)
- i]
· [(x - 1)
+ i]
=
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= -[(x - 1)2
- i2]
=
-
(x2
- 2x
+ 1 +
1)
=
- x2
+ 2x
- 2 |
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Example:
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The
real root of the polynomial -
x3
-
x2
+ 4x
- 6
is
x1 =
- 3,
factorize the polynomial. |
| Solution: |
We
divide given polynomial by one of its known
factors, |
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a3x3
+ a2x2
+ a1x
+ a0 = a3(x
- x1)
(x
- x2)
(x
- x3) |
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then
we calculate another two roots of given cubic by solving
obtained quadratic trinomial, |
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Finally
we use the theorem to factorize given polynomial (see
the previous example), |
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a3(x
- x1)(x
- x2)(x
- x3)
= -1(x
+ 3)[x
-
(1
+ i)][x
-
(1 - i)]
= -1(x
+ 3)(x2
- 2x
+ 2). |
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Notice that given cubic has one real root and the pair of the conjugate complex roots. |
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Odd
degree polynomials must have at least one real root. |
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| Example:
Solve polynomial
equation x3
+ 2x2 -
x -
2 = 0 by factoring. |
| Solution:
x2(x
+ 2) -
(x + 2) = 0 |
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(x +
2)·(x2 -
1) = 0 |
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(x +
2)·(x + 1)·(x -
1) = 0 |
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x +
2 = 0 => x1 =
-2 |
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x +
1 = 0 => x2 =
-1 |
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x -
1 = 0 => x3 = 1 |
| The
roots are: x1
= -2,
x2 = -1
and x3
= 1. |
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| Intermediate
algebra contents |
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| Copyright
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