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Algebra 2 - PreAP/GT
Solving and Graphing Polynomial Equations #2
Name ____________________________________
Solve the following polynomial equations by factoring.
2 x3  5x2  4 x  10  0
x 4  3x 2  4  0
Notice that solutions can be rational, irrational, real, and/or complex/imaginary.
IRRATIONAL ROOT THEOREM
If a polynomial P  x  has rational coefficients and a  b c is a root of the polynomial equation
P  x   0 , where a and b are rational and
c is irrational, then a  b c is also root of P  x   0 .
COMPLEX CONJUGATE ROOT THEOREM
If a  bi is a root of a polynomial equation with real number coefficients, then a  bi is also a root of
that polynomial equation.
Write a polynomial function with zeros of 2i and 1.
Write a polynomial function with zeros of 1  i ,
2 , and -3.
What happens when we try to solve x3  3x 2  10 x  24  0 by factoring.
Not all polynomial equations will factor using the factoring rules we have been using. Could we solve
the example above by using Square Root Property? Completing the Square? Quadratic Formula? Let’s
look at a theorem that might help us solve the polynomial equations that we can not initially factor.
RATIONAL ROOT THEOREM
If the polynomial P  x  has integer coefficients, then every rational root of the polynomial equation
p
, where p is a factor of the constant term of P  x  and q is a
q
factor of the leading coefficient of P  x  .
P  x   0 can be written in the form
x3  3x2 10 x  24  0
p (factors of constant): ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
q (factors of leading coefficient): ±1
POSSIBLE rational roots
p
: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
q
Find the possible rational roots of f  x   8 x 4  2 x3  31x 2  8 x  4 .
Once we have the possible rational roots, we can use the possible roots with some other theorems to help
us solve and graph a polynomial function.
Going back to the example from above,
x3  3x2 10 x  24  0
POSSIBLE rational roots
p
: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24
q
DESCARTES’ RULE OF SIGNS
The number of positive real roots of a polynomial equation P  x   0 with real number coefficients is
equal to the number of sign changes between the coefficients of the terms of P  x   0 or is less than
this number by a multiple of 2.
x3  3x2 10 x  24  0
+
+
–
1 sign change
Therefore, the number of positive real roots is 1.
–
DESCARTES’ RULE OF SIGNS (Continued)
The number of negative real roots of a polynomial equation P  x   0 with real number coefficients is
equal to the number of sign changes between the coefficients of the terms of P   x   0 or is less than
this number by a multiple of 2.
x
3
 3   x   10   x   24  0
2
 x3  3x2  10 x  24  0
–
+
–
+
1st sign change
2nd sign change
Therefore, the number of negative real roots is either 2 or 0.
Since we know that this polynomial has 3 solutions (including multiplicities and complex/imaginary
solutions), we can make Descartes’ Chart based on Descartes’ Rule of Signs.
+
–
C/i
1
1
2
0
0
2
each row must add to 3 since we
know there has to be 3 solutions
Using synthetic division, the Remainder Theorem, the Factor Theorem along with the fact that we have
one real positive root, let’s try our possible rational roots to see if we can find the positive real root.
(Note: there is no guarantee that the positive real root is a rational root)
In case you don’t remember them –
REMAINDER THEOREM
If the polynomial function P  x  is divided by x  a , then the remainder r = P  a  .
FACTOR THEOREM
For any polynomial P  x  ,  x  a  is a factor of P  x  if and only if P  a   0 .
Remember, we are guaranteed a positive root so we will start with the possible positive numbers –
1
1
3
1
-10
4
-24
-6
1
4
-6
-30
1 is not the positive real root since the
remainder is not 0.
3
1
3
3
-10
18
-24
24
1
6
8
0
2
1
3
2
-10
10
-24
0
1
5
0
-24
2 is not the positive real root since the
remainder is not 0.
3 is the positive real root since the remainder is 0.


Since 3 is a root, then x3  3x 2  10 x  24  0 will factor to  x  3 x2  6 x  8 which will then factor
to  x  3 x  2 x  4 . Therefore, the solutions to x3  3x 2  10 x  24  0 are x = 3, -2, -4. Now that
we have solved the polynomial equation, we can also graph it using the x-intercepts, the y-intercept, and
the end behavior as guidelines.
x – intercept(s) of f  x  ___________________________
y – intercept of f  x  _______________
End behavior of f  x  _______________________________
Make Descartes’ Chart for the following polynomials:
f  x   8 x 4  2 x3  31x 2  8 x  4
f  x    x3  3x2  x  2
HOMEWORK
For each problem below, list the possible rational roots (zeros) and make Descartes’ Chart.
1)
x3  2 x 2  2 x  3  0
2)
f ( x)  x3  x2  10x  8
3)
3x3  7 x 2  15x  9  0
4)
f  x   x 4  4 x3  3x 2  14 x  8
Bookwork: p 449 #20-22