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Rational Root Theorem
Rational Root Theorem:
Supposethat apolynomial equation with integral
coefficientshastheroot p/q , where pand q are
relatively primeintegers. Then pmust beafactor
of theco nstant term of thepolynomial and qmust
beafactor of theco efficient of thehighest degree
term.
(usefulwhen so lving higher degree po lyno mialequatio ns)
Solveusing theRational 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
possible rational roots: (now use synthetic division
to find rational roots)
1 1
1, ,
2 4
±
1 4 3 -1
4 7
4 7 6 no
-1 4 3 -1
-4 1
4 -1 0 !yes
4 1 0
4 1
1
4
x
x
x
− =
=
=
1
1,
4
x = −
(note: not all possible rational roots are zeros!)
Ex: Solveusing theRational Root Theorem:
3 2
2 13 10 0x x x+ − + =
1 1 2 -13 10
1 3 -10
1 3 -10 0 !yes
( ) ( )
2
3 10 0
5 2 0
5, 2
x x
x x
x
+ − =
+ − =
= −
5,1, 2x = −
1, 2, 5,10±possible rational roots:
Ex: Solveusing theRational Root Theorem:
3 2
4 4 0x x x− − + =
possible rational roots: 1, 2, 4±
1 1 -4 -1 4
1 -3 -4
1 -3 -4 0 !yes
( ) ( )
2
3 4 0
4 1 0
1, 4
x x
x x
x
− − =
− + =
= −
1,1, 4x = −
Ex: Solveusing theRational Root Theorem:
3 2
3 5 4 4 0x x x− − + =
possible rational roots:
1 2 4
1, 2, 4, , ,
3 3 3
±
-1 3 -5 -4 4
-3 8
3 -8 -4
-4
0 !yes
( ) ( )
2
3 8 4 0
3 2 2 0
2
, 2
3
x x
x x
x
− − =
− − =
=
2
1, , 2
3
x = −
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 3 2 0
6 -4 3 2
3 -2 0
x
x
− =
=
2
3
x =
Ex: Solveusing theRational Root Theorem:
3 2
3 5 4 4 0x x x− − + =
possible rational roots:
1 2 4
1, 2, 4, , ,
3 3 3
±
-1 3 -5 -4 4
-3 8
3 -8 -4
-4
0 !yes
( ) ( )
2
3 8 4 0
3 2 2 0
2
, 2
3
x x
x x
x
− − =
− − =
=
2
1, , 2
3
x = −
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 3 2 0
6 -4 3 2
3 -2 0
x
x
− =
=
2
3
x =

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RRT

  • 2. Rational Root Theorem: Supposethat apolynomial equation with integral coefficientshastheroot p/q , where pand q are relatively primeintegers. Then pmust beafactor of theco nstant term of thepolynomial and qmust beafactor of theco efficient of thehighest degree term. (usefulwhen so lving higher degree po lyno mialequatio ns)
  • 3. Solveusing theRational 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 possible rational roots: (now use synthetic division to find rational roots) 1 1 1, , 2 4 ± 1 4 3 -1 4 7 4 7 6 no -1 4 3 -1 -4 1 4 -1 0 !yes 4 1 0 4 1 1 4 x x x − = = = 1 1, 4 x = − (note: not all possible rational roots are zeros!)
  • 4. Ex: Solveusing theRational Root Theorem: 3 2 2 13 10 0x x x+ − + = 1 1 2 -13 10 1 3 -10 1 3 -10 0 !yes ( ) ( ) 2 3 10 0 5 2 0 5, 2 x x x x x + − = + − = = − 5,1, 2x = − 1, 2, 5,10±possible rational roots:
  • 5. Ex: Solveusing theRational Root Theorem: 3 2 4 4 0x x x− − + = possible rational roots: 1, 2, 4± 1 1 -4 -1 4 1 -3 -4 1 -3 -4 0 !yes ( ) ( ) 2 3 4 0 4 1 0 1, 4 x x x x x − − = − + = = − 1,1, 4x = −
  • 6. Ex: Solveusing theRational Root Theorem: 3 2 3 5 4 4 0x x x− − + = possible rational roots: 1 2 4 1, 2, 4, , , 3 3 3 ± -1 3 -5 -4 4 -3 8 3 -8 -4 -4 0 !yes ( ) ( ) 2 3 8 4 0 3 2 2 0 2 , 2 3 x x x x x − − = − − = = 2 1, , 2 3 x = − 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 3 2 0 6 -4 3 2 3 -2 0 x x − = = 2 3 x =
  • 7. Ex: Solveusing theRational Root Theorem: 3 2 3 5 4 4 0x x x− − + = possible rational roots: 1 2 4 1, 2, 4, , , 3 3 3 ± -1 3 -5 -4 4 -3 8 3 -8 -4 -4 0 !yes ( ) ( ) 2 3 8 4 0 3 2 2 0 2 , 2 3 x x x x x − − = − − = = 2 1, , 2 3 x = − 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 3 2 0 6 -4 3 2 3 -2 0 x x − = = 2 3 x =