Rational Root Theorem
Rational Root Theorem:
Suppose that a polynomial equation with integral
coefficients has the root h , where h and k are
k
relatively prime integers. Then h must be a
factor of the constant term of the polynomial
and k must be a factor of the coefficient of the
highest degree term.
(useful when solving higher degree polynomial equations)
How many solutions does the equation have?
1.
x  5x  36  0
4
2
4
2.
5x  7 x  8x 16  0
3
2
3
Factors of the constant
Factors of the leading coefficient
Example 1
a) State all possible rational zeros for
g(x) = x3 + 2x2 - 3x + 5
+1
+5,+1
Possible rational zeros: +5, +1
b) State all possible rational zeros
for g(x) = 6x3 + 6x2 - 15x - 2
+2,+1
+6,+3,+2,+1
+2,+1
Possible rational zeros:
+6,+3,+2,+1
 , ,1,2, ,
1
3
2
3
1
6
1
2
Solve using the Rational Root Theorem:

4x2 + 3x – 1 = 0
(any rational root must have a numerator
that is a factor of -1 and a denominator
that is a factor of 4)
factors of -1: ±1
factors of 4: ±1,2,4
1 1
possible rational roots: 1, , (now use synthetic division
2 4
to find rational roots)
1
4 3 -1
-1
4 3 -1
4x 1  0
4 7
-4 1
4x  1
x
4 7 6 no
4 -1 0 yes !
1
x
4
(note: not all possible rational roots are roots!)
1
1,
4
Ex : Solve using the Rational Root Theorem:
x  2 x  13x  10  0
3
h :  1,2,5,10
k : 1
2
possible rational roots:
1, 2, 5,10
x  3 x  10  0
2
1 1 2 -13 10
1
3 -10
1 3
-10 0
 x  5  x  2   0
yes !
x  5, 2
x  5,1, 2
Ex : Solve using the Rational Root Theorem:
x  4x  x  4  0
3
h :  1,2,4
k : 1
possible rational roots:
1 1 -4 -1
1
2
x 2  3x  4  0
4
 x  4  x  1  0
-3 -4
1 -3 -4 0
1, 2, 4
yes !
x  1, 4
x  1,1, 4
Ex : Solve using the Rational Root Theorem:
3x  5 x  4 x  4  0
3
h :  1,2,4
k :  1, 3
-1
possible rational roots:
3 -5 -4
-3
2
3x 2  8 x  4  0
4
 3x  2  x  2   0
8 -4
3 -8 -4
0
yes !
To find other roots can use synthetic division
using other possible roots on these coefficients.
(or factor and solve the quadratic equation)
2
3 -8 4 3x  2  0 x  2
3
6 -4
3x  2
3 -2 0
1 2 4
1, 2, 4, , ,
3 3 3
2
x ,2
3
2
x  1, , 2
3
Find all zeros of the polynomial function.
3.
f ( x)  x  4x  3x  4x  4
4
3
2
How many answers:
Factors of last term:
Factors of first term:
Possible rational zeros:
4
2,
1,
4
1
1,
2,
4
Find all real zeros of the function.
3.
f ( x)  x  4x  3x  4x  4
4
3
2
1,
Possible rational zeros:
1
1
–4
1
1
(x – 1)
–3
3
–3
0
( x3 – 3x2 + 0x + 4) = 0
4
–4
0
4
4
0
2,
4
Find all real zeros of the function.
3.
f ( x)  x  4x  3x  4x  4
4
3
2
1,
Possible rational zeros:
1
1
–4
3
–3
1
–1
1
1
(x – 1)
(x + 1)
4
4
0
0
4
–1
4
–4
–4
4
0
4
–4
0
–3
2,
( x2 – 4x + 4) = 0
x
–2
x
–2
(x – 1)(x + 1)(x – 2)(x – 2) = 0
x=
1, 2
Find all zeros of the polynomial function.
4.
f ( x)  x  x  7 x  9x 18
4
3
2
How many answers:
4
Factors of last term:
1,
Factors of first term:
1
Possible rational zeros:
2,
1,
3,
2,
6,
3,
9,
6,
18
9,
18
Find all real zeros of the function.
f ( x)  x  x  7 x  9x 18
4
4.
3
2
1,
Possible rational zeros:
1
1
1
2,
–1
7
–9
–18
1
0
7
–2
0
7
–2

–20
3,
6,
9,
18
Find all real zeros of the function.
f ( x)  x  x  7 x  9x 18
4
4.
3
1,
Possible rational zeros:
–1
1
1
(x + 1)
–1
2,
7
–9
–18
–1
2
–9
18
–2
9
–18
0
3,
6,
9,
18
( x3 – 2x2) +( 9x – 18 )) = 0
x2 (x – 2)
(x + 1)
2
(x2 + 9)
+9 (x – 2)
(x – 2)
(x + 1)(x2 + 9)(x – 2) = 0
x=
1,  3i, 2
Find the roots of x3 + 8x2 + 16x + 5 = 0
Ex : Determine the zeros:
x  3x  2 x  6
3
2
The Fundamental
Theorem of Algebra
If f(x) is a polynomial of degree n
where n>0, then the equation f(x)
= o has at least one solution in
the set of complex numbers.
Complex Conjugates Theorem
If f (x) is a polynomial
function with real
coefficients, and a+bi is an
imaginary zero of f(x), then
a-bi is also a zero of f(x).
(imaginary numbers always come in pairs)
Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
5.
3, –2
f(x) = (x – 3)(x + 2)
x2
f(x) = x2 – x – 6
+ 2x
– 3x
–6
Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
6.
i, –2i
f(x) = (x – i)(x + i)(x + 2i)(x – 2i)
(x2 – i2)
(x2 – 4i2)
(x2 + 1)
(x2 + 4)
x4
+ 4x2
f(x) = x4 + 5x2 + 4
+ x2
+4
Some possibilities for zeros (roots)
Equation
Real Zeros
Imaginary Zeros
Linear
1
0
Quadratic
2
0
0
2
3
0
1
2
4
0
2
2
0
4
5
0
3
2
1
4
Cubic
Quartic
Quintic
Find the zeros.
f(x) = x 3 – x2 + 4x – 4 = 0
g(x) = x 4 + 2x 3 – 5x 2 – 4x + 6 = 0
Write a polynomial function of least
degree with a leading coefficient of 1.
2, 3, -4
(x – 2)(x – 3)(x + 4) = 0
(x2 – 5x + 6)(x + 4) = 0
x 3 - x2 – 14x + 24 = 0
Write a polynomial function f of least degree that has rational
coefficients, a leading coefficient of 1, and given zeros.
7.
4,
2

f ( x)   x  4  x  2
 x  4  x
x3
2
– 2x
f(x) = x3 – 4x2 – 2x + 8
2
– 4x2

 x  2 
+8
Write a polynomial function of least
degree with a leading coefficient of 1.
7, i
(x – 7)(x – i)(x + i) = 0
(x – 7)(x 2 + 1) = 0
3
2
x –7x +x–7=0
Descartes’ Rule of Signs
Let f(x) be a polynomial function with
real coefficients.


The number of positive real zeros of f(x) is
equal to the number of sign changes of the
coefficients of f(x) or is less than this by an
even number.
The number of negative real zeros of
f
(x) is equal to the number of sign changes
of the coefficients of f(-x) or is less than
this by an even number.
huh
???
Using Descartes’ Rule of Signs….
f(x) = x6-2x 5+3x 4 – 10x 3-6x 2 -8x -8
3 sign changes, so f has 3 or 1 positive real roots
f(-x) = x6+2x 5+3x 4 + 10x 3-6x 2 +8x -8
3 sign changes, so f has 3 or 1 negative real roots
g(x) =
5
-x
–
4
5x
+
3
7x
–
2
4x
– 8x + 9

Determine the possible number of
positive real zeros, negative real zeros,
and imaginary zeros of the function
given above.

g(x) = -x5 – 5x4 + 7x3 – 4x2 – 8x + 9

g(-x) = x5 – 5x4 - 7x3 – 4x2 + 8x + 9
Example

Determine the possible number of positive
real zeros, negative real zeros, and
imaginary zeros of the function given
above.
f ( x)  3x  4 x  3x  2 x  x  3
6
4
3
2
f ( x)  3( x)  4( x)  3( x)  2( x)  ( x)  3
6
4
3
f ( x )  3x  4 x  3x  2 x  x  3
6
4
3
2
2
Use Descartes’ Rule of signs to find the number of possible real
roots.
f ( x)  x  5x  4 x  12
3
2