solving polynomial equations...solving polynomial equations using the rational root theorem note:...
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SOLVING POLYNOMIAL EQUATIONS
Unit Overview
This unit is about solving polynomial equations by using variable substitution and by using the
rational root theorem. The unit concludes with verifying and proving polynomial identities.
Solving Polynomial Equations by Using Variable Substitution and
Factoring
One of the most important theorems in mathematics is the Fundamental Theorem of Algebra.
The polynomial, 4 29 14 0x x , is a polynomial of degree 4 so it should have 4 roots. The roots
may be real numbers, complex numbers, and/or the same number. These roots will be found in the
following example.
If a polynomial has a degree of more than 2, variable substitution can be used find the roots.
Example #1: Solve for x: 4 29 14 0x x
Step #1: Let u represent 2x and substitute u for
2x .
2 2 2 4 2 2
2
( ) 9( ) 14 0 express as ( )
9 14 0
x x x x
u u
Fundamental Theorem of Algebra
If P(x) is a polynomial of degree n, where n > 0, then P(x)
has exactly n roots including multiple and complex roots.
Step #2: Factor and solve.
( 7)( 2) 0
7 0 2 0
7 2
u u
u u
u u
Step #3: Remember that u represents 2x and replace u with 2x .
2 27 2
7 2
x x
x x
Example #2: Solve for z: 5 328 27 0z z z
5 328 27 0z z z factor out a GCF of z
4 2( 28 27) 0z z z substitute u for 2z
2( 28 27) 0z u u factor
z(u – 27)(u – 1) = 0 solve for u
z = 0 u – 27 = 0 u – 1 = 0
u = 27 u = 1 replace u with 2z
z = 0 2 27z 2 1z solve for z
z = 0, z = 27 , z = 1 1
Therefore, the roots of the polynomial 5 328 27 0z z z are 0, 1, –1, 27 , and
27 .
Finding the Zeros (04:08)
Stop! Go to Questions #1-5 about this section, then return to continue on to the next
section.
Rational Root Theorem
Example #1: List all the possible rational roots of the function
P(x) = 2x3 – 11x2 + 12x + 9.
According to the Rational Root Theorem, p
q is a root of 3 22 911 12x x x .
If p is a factor of the constant term, 9, and q is a factor of the leading coefficient, 2.
Step #1: Make an organized list of all factors of 9 and 2.
factors of 9: ±1, ±3, ±9
factors of 2: ±1, ±2
Step #2: Form all quotients that have factors of 9 in the numerator and factors of 2 in
the denominator.
These numbers represent all the possible rational zeros of the function.
Factors of 9 over 1
1
1 ,
3
1 ,
9
1
Factors of 9 over 2
1
2 ,
3
2 ,
9
2
Rational Root Theorem
Let P be a polynomial function with integer coefficients in standard form.
If p
q (in lowest terms) is a root of P(x) = 0, then
p is a factor of the constant term of P and
q is a factor of the leading coefficient of P
Let's practice determining the possible rational zeros of the polynomial, P(x) = 2x3 – x2 + 2x + 5.
Name the constant term. P(x) = 2x3 – x2 + 2x + 5
“Click here” to check the answer.
5
List the factors of the constant term. P(x) = 2x3 – x2 + 2x + 5
“Click here” to check the answer.
±1, ±5
Name the leading coefficient. P(x) = 2x3 – x2 + 2x + 5
“Click here” to check the answer.
2
List the factors of the leading coefficient. P(x) = 2x3 – x2 + 2x + 5
“Click here” to check the answer.
±1, ±2
List the possible rational roots. P(x) = 2x3 – x2 + 2x + 5
“Click here” to check the answer.
1 51, 5, ,
2 2
The Zero Property (06:50)
Stop! Go to Questions #6-7 about this section, then return to continue on to the next
section.
Solving Polynomial Equations Using the Rational Root Theorem
Note: The following examples are shown using a graphing calculator. The activities can be also
be done using an online graphing program. Click here to navigate to the online grapher.
Example #1: Find all of the rational roots of 3 28 10 11 2 0x x x .
According to the Rational Root Theorem, p
q is a root of 3 28 10 11 2 0x x x , if p is
a factor of the constant term, 2, and q is a factor of the leading coefficient, 8.
Step #1: Make an organized list of all factors of 2 and 8.
factors of 2: 1 , 2
factors of 8: 1 , 2 , 4 , 8
Step #2: Form all quotients that have factors of 2 in the numerator and factors of 8 in
the denominator.
*Some rational numbers are repeated, such as 1
4 and
2
8 .
Step #3: Graph the function on your graphing calculator. Use the button that is
marked XTθn for the variable x. Also, to raise a number to the third power, use the
button that is located above the key. The standard viewing screen is the
figure on the left. ZOOMIN ENTER is the figure on the right.
One root appears to be –2. Test whether P(–2) = 0 by using synthetic division.
1
1 ,
1
2 ,
1
4 ,
1
8
2
1 ,
2
2 ,
2
4 ,
2
8
–2 8 10 –11 2
–16 12 –2 remainder
8 –6 1 0
Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the
polynomial.
Step #4: From the graph there appears to be two real zeros between 0 and 1. Zoom in
( ZOOM 2) and use the trace feature to check. Press TRACE and move the
cursor around with the arrow keys until the y-value on your screen gets very
very close to 0. ZOOM , 2: Zoom In, ENTER if necessary to locate this point.
It looks like there are two real roots, one at 1
4 and the other at
1
2. Test whether
10
4P
and 1
02
P
by using synthetic division.
1
4 8 10 –11 2
1
2 8 10 –11 2
2 3 –2 4 7 –2
8 12 –8 0 8 14 –4 0
Since 1
04
P
and 1
02
P
, 1
4 and
1
2 are also roots of the polynomial.
Therefore, there are three roots –2, 1
4, and
1
2.
*When a polynomial function P(x) touches but does not cross the x-axis at (r, 0), then P(x) = 0
has a double root.
For example: P(x) = (x – 3)(x – 3)(x+1) or 3 2( ) 5 3 9P x x x x touches the x-axis at
(3, 0) but does not cross the x-axis there, as shown in the graphs below.
Therefore, 3 is a double root of P(x) = 0.
Example #2: Find all of the rational roots of 3 23 2 12 8 0x x x .
According to the Rational Root Theorem, p
q is a root of 3 23 2 12 8 0x x x ,
if p is a factor of the constant term, 8, and q is a factor of the leading coefficient, 3.
Step #1: Make an organized list of all factors of 8 and 3.
factors of 8: 1 , 2 , 4 , 8
factors of 3: 1 , 3
Step #2: Form all quotients that have factors of 8 in the numerator and factors of 3 in
the denominator.
1 1,
1 3
2 2,
1 3
4 4,
1 3
8 8,
1 3
Step #3: Graph the function on your graphing calculator. Use the button that is
marked XTθn for the variable x. Also, to raise a number to the third power, use the
button that is located above the key. The standard viewing screen is the
figure below on the left. ZOOMIN ENTER is the figure on the right.
There appears to be two real roots at –2 and +2. Test either P(–2) =0 or P(2)=0 by
using synthetic division. We’ll try P(–2)=0.
–2 3 –2 –12 8
–6 16 –8 remainder
3 –8 4 0
Since the remainder is 0, this means that P(–2) = 0 is true and that –2 is a root of the
polynomial.
Now we’ll try P(2) = 0.
2 3 –2 –12 8
6 8 –8 remainder
3 4 – 4 0
Since the remainder is 0, this means that P(2) = 0 is true and that 2 is a root of the
polynomial.
Step #4: From the graph there appears to be a real zero between 0 and 1. Zoom in
( ZOOM 2) and use the trace feature to check. Press TRACE and move the
cursor around with the arrow keys until the y-value on your screen gets very
very close to 0. ZOOM , 2: Zoom In, ENTER if necessary to locate this point.
It looks like there is one real root, 2
3. Test whether
20
3P
and by using synthetic
division.
2
3 3 –2 –12 8
2 0 –8 remainder
3 0 –12 0
Since 2
03
P
, 2
3 is a root of the polynomial.
Therefore there are 3 roots –2, 2, and 2
3.
Some polynomials do not have all rational roots. We can still find all the zeros of a function
using some of the strategies already learned.
Example #3: Find all of the roots of 3 24 6 4 0x x x .
According to the Rational Root Theorem, p
q is a root of 3 24 6 4 0x x x ,
if p is a factor of the constant term, –4, and q is a factor of the leading coefficient, 1.
Step #1: Make an organized list of all factors of 4 and 1.
factors of –4: 1 , 2 , 4
factors of 1: 1
Step #2: Form all quotients that have factors of 4 in the numerator and factors of 1 in
the denominator.
Step #3: Graph the function on a graphing calculator. Use the button that is marked
XTθn for the variable x. Also, to raise a number to the third power, use the
button that is located above the key. The standard viewing screen is the figure
below on the left. ZOOMIN ENTER is the figure on the right.
1
1
2
1
4
1
The graph crosses the x-axis once. Therefore there is only one real root. This root
appears to be 2. Test whether P(2) = 0 by using synthetic division.
2 1 –4 +6 –4
2 –4 4 remainder
1 -2 2 0
Since the remainder is 0, this means that P(2) = 0 is true and that 2 is a root of the
polynomial.
The resulting numbers 1, –2 and 2 are the coefficients of the quotient. The quotient
will start with an exponent that is one less than the dividend. 21 2 2x x
This quotient is called a depressed polynomial. Since the depressed polynomial is
quadratic, use the Quadratic Formula to find the other two roots.
2 2 2 0x x a = 1 b = –2 c = 2
2
2
4Substitute 1, 2, 2
2
( 2) ( 2) 4(1)(2)
2(1)
2 4 8
2
2 4
2
2 ( 1)4
2
2 2 ( 1)
2
2 2
2
1
1
1
1 or 1
b b acx a b c
a
x
x
x
x
x
ix
ix
x i
x i x i
Therefore the roots of the equation are 2, 1 + i and 1 – i.
Stop! Go to Questions #8-12 about this section, then return to continue on to the next
section.
Proving Polynomial Identities
A polynomial identity is a statement of equality between two polynomials that is true for all
values of the variable(s) for which the expression is defined.
Example #1: Check the polynomial identity 2 2 2( ) 2a b a ab b by substituting a = 4
and b = 2.
2 2 2
2
(4 2) (4 +2(4)(2) + 2 ) Substitute 4 and 2 on both sides of the equation.
6 (16 16 4) Simplify
36 = 36
a b
Example #2: Prove the polynomial identity 2 2 2( ) 2a b a ab b by using algebraic
operations.
2
2 2
2 2
( )
( )( ) Definition of Squared
( ) ( ) Distributive Property
Distributive Property
2 Combine like terms
a b
a b a b
a a b b a b
a ab ab b
a ab b
Example #3: Check the polynomial identity 3 3 2 2( )( )a b a b a ab b by
substituting a = 4 and b = 2.
3 3 2 24 2 (4 2)(4 +4(2) + 2 ) Substitute 4 and 2 on both sides of the equation.
64 8 2(16 8 4) Simplify
56 = 2(28)
56 = 56
a b
Example #4: Prove the polynomial identity 3 3 2 2( )( )a b a b a ab b using algebraic
expressions.
2 2
2 2 2 2
3 2 2 2 2 3
3 3
( )( )
( ) ( ) Distributive Property
Distributive Property
Collect like terms
x y x xy y
x x xy y y x xy y
x x y xy x y xy y
x y
Stop! Go to Questions #13-33 to complete this unit.