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Solving Polynomial Equations
To solve polynomials equations by factoring.
To solve polynomials equations by graphing.
Use graphing technology to find approximate
solutions for polynomial equations.
Use polynomials equations to solve real world
problems.

3
3
2
2


a

b

a

b
a

ab

b
Sum of the cubes=

Difference of the cubes= a 3  b3  a  ba 2  ab  b 2 
Remember: If  x  a  is a factor of polynomial,
then the polynomial has a value 0 when x=a. If a is
a real number, then the graph of the polynomial
has (a,0), as an x-intercept.
2
When you have a polynomial ax  bx  c
factoring that polynomial will help you to solve the
2
equation like ax  bx  c  0
So to solve a polynomial equation by factoring
Write the equation in the form P(x)=0
Factor P(x). Use the Zero Property of the roots
Problem 1: Solving Polynomial Equation Using Factors.
What are the real or imaginary solutions of each polynomial equations?
A. 2 x 3  5 x 2  3 x
Make equation=0 and get GCF
B. 3 x 4  12 x 2  6 x 3
Make equation =0, Divide by 3 and get GCF
2 x 3  5 x 2  3x  0
3 x 4  6 x 3  12 x 2  0
x 2 x 2  5x  3  0
x 4  2 x3  4 x 2  0


2 x
2x  6x  x  3
2
2 x( x  3 )  ( x  3 )
( x  3 )( 2 x  1 )
Zero property and solve for x
x 2 x  1 x  3   0
1
x  0, x   , x  3
2


 5x  3
2 x2  5 x  3
factor ( 2  3  6 )
Factor
2

x2 x2  2x  4  0


Use quadratic to solve x 2 x 2  2 x  4  0
x2  0  x  0
x
  2 
 22  414
21
2   12 2  2i 3


2
2
 1 i 3
x  0, x  1  i 3 , x  1  i 3
Your turn
What are the real or imaginary solutions of each equation?



A. x 2  1 x 2  4  0
x 2  1 applied
B. x 5  4 x 3  5 x 4  2 x 3
Answers:
x  1
x  4
2
applied
x  2 i
x  1, x  1, x  2i , x  2i
x5  5x4  6x3  0


x3 x2  5x  6  0
x 3  x  3 x  2  0
x  0, x  2, x  3
Take a note
Problem 2: Solving Polynomial Equations by Factoring
What are the real or imaginary solutions of each
polynomial equation?
4
2
x

3
x
4
A.
Make the equation =0
x4  3x2  4  0
Let a  x 2 then
a 2  3a  4  0
Factor with -4 and +1
a  4a  1  0
x
2
 4x
2
 1  0
Replace back
x2
x20
x  -2
ax
2
2
2
2
Then x  4  x  2   x  2 x  2
 x  2 x  2x 2  1  0
x20
x 4  a 2 and 3 x 2  3a
x2  1  0
and
x 2  1 applied
x  i
Two imaginary roots
Verify your answer with the graphic calculator
The graphs shows zero at
x=2, and x=-2. it also shows
three turning points.
This means that are imaginary
roots, which do not appear
on the graph
Problem 2
What are the real or imaginary solutions of each
polynomial equation?
B. x 3  1
x 1  0
3
Make equation =0
Factor the difference of cubes
 x  1x 2  x  1  0
x 1  0
x 1
x  x 1  0
 1  12  411  1  i 3

x
21
2
2
use Q.F
The three solutions are 1, 1  i 3 , 1  i 3
2
2
2
2
Your turn
What are the real or imaginary solutions of each
3
x
 8x  8x  8
polynomial equations?
3
x
 8x  8x  8  0
a ) x 4  16
Answer: c)
b) x  8 x  2 x
3

x 3  23  0
2
 x  2x 2  2 x  2 2   0
 2  2 2  414 
x
21

c ) x x 2  8  8 x  1
Answer: a)
x
2


 22 x 2  22  0
 x  2 x  2x
x  2
x  2i
2

4 0
Answer b)
x  1  i 3
x3  2x2  8x  0


x x2  2x  8  0
x  x  4 x  2  0
x  0, x  4, x  2
Problem 3: Finding Real Roots by Graphing
What are the real solutions of the equation x 3  5  4 x 2  x
Method 1: graph y1  x 3  5 and y2  4 x 2  x
Use the intersect
feature to find the
x values of the
points of intersection.
Approximate solutions are
x=-1.09,x=1.16, and x=3.93
Method 2: Rewrite the equation as
Graph the related function.
x3  4x2  x  5  0
Use the zero feature
3
2
y  x  4x  x  5
The solutions are the same that
in method 1, so the approximate
solutions are x=-1.09,x=1.16, and
x=3.93
Verify the solutions by showing
that they satisfy the original
equation. Show values of
3
2
y1  x  5 and y2  4 x  x
in a table.
Your turn: a)What are the real solutions of the equation.
b)Which method seems to be easier and more reliable
way to find the solutions of an equations? Explain
x3  x2  x  1
Answer a) -1.84
b) the second method seems to be a more reliable
way to find the solutions because you do not risk
missing a point of intersection
Problem 4:Modeling a real world situation
Close friends Stacy, Una, and Amir were all born on July 4.
Stacy is one year younger than Una. Una is two years younger
than Amir. On July 4, 2010, the product of their ages was 2300
more than the sum of their ages. How old was each friend on
that day?
Let x=Una’s age on July 4,2010
Stacy’s age=x-1
Amir’s age=x+2
Answer
Sum of their ages
x+(x-1)+(x+2)+2300
3 x  2301  x( x  x  2)
2
3 x  2301  x  x  2 x
3
2
x  x  5 x  2301  0
3
2
=
Product of their ages
x(x-1)(x+2)
Graph it and look into the tabl
when y=0 and take only the
solution that makes sense
x=13. Una was 13,Stacy 12
and Amir 15
Your turn
A. What are three consecutive integers whose product
is 480 more than their sum.
Answer: 7,8,9
x x  1 x  2  480  x   x  1   x  2
x
2

 x  x  2  480  3x  3
x  2 x  x  2 x  483  3x
3
2
2
x3  3x 2  2 x  483  3x
x3  3x 2  x  483  0
Input the data in the
function feature and
look for the table
when y=0.
B. What are three consecutive even integers whose product
is 4 times their sum?
Answer: 2, 4, 6 or -6, -4, -2
x  x  2  x  4   4 x   x  2    x  4 
x 3  6 x 2  4 x  24  0
Classwork odd
Homework even
TB pgs 301 exercises 1-56