Synthetic Division/Rational Roots 
 Section 5.3  
Use synthetic division to determine if (x  1) is a factor of P(x) = x^{3 } 2x^{2 } 5x^{ }+ 6 
Put 1 in the upper left hand corner (box). Put the coefficients of the polynomial in the first row.  
Solve 
Step 


Bring the down the 1. 
Use 1 (in box) if x 
1 is the factor. For x  r , use r 

Multiply
1 × 1
= 1
Add Columns 2 + 1 = 1 


Multiply
1 × 1
= 1
Add Columns 5 + (1) = 6 


Multiply
6 × 1
= 6
Add Columns 6 + (6) = 0 

If there is a remainder of zero, 0 , then 1 is a zero and (x 1) is a factor of the polynomial. 


Use synthetic division to determine if (x  1) is a factor of P(x) = 2x^{3 } 2x^{2 } 5x^{ }+ 6 
[Solution] 


Find all rational roots of P(x) = 2x^{3 }+ 3x^{2 } 14x^{ } 15 
Find all possible rational zeros  rational numbers are ratio of integers i.e. 2/1 = 2, 0, 1/2, 3/2  
Solve 
Step 

p = ± 1, ±
3, ± 5, ±
15
q ± 1, ± 2 
Find
possible rational zeros p/q where p are factors of the last coefficient 15. q are the factors of the first coefficient 2. 

p = ± 1, ± 2, ± 3, ± 6 and 
p/1 = p and all of the p/2's.  We always get the p's 

Try the easy ones first.  
Start with 1 (I tried 1 first but it failed  You can use Descartes' Rules of Signs to narrow the search.)  
Put 1 in the upper left hand corner (box). Put the coefficients of the polynomial in the first row.  

Bring the down the 2.  coefficient of 2x^{3} is 2 

Multiply
2 × 1
= 2
Add Columns 3 + (2) = 1 

Multiply
1 × 1
= 1 Add Columns 14 + (1) = 15 

Multiply
15 × 1
= 15
Add Columns 15 + 15 = 0 

If there is a remainder of zero, 0 , then 1 is a zero and (x (1)) = (x + 1) is a factor of the polynomial. 

Q(x) = 2x^{2 }+ x^{ } 15 
The quotient is the coefficients of the bottom line  Once we have a quadratic equation, factor or use the quadratic formula. 
2x^{2 }+ x^{ } 15 = (2x  1)(x + 3) = 0 
Set equal to zero and solve for x  
2x  1 = 0 or x + 3 = 0 

x = 1/2 or x = 3 
Check in original equation.  
{ 1, 1/2, 3} 
There are three solutions for P(x)  P(1/2) = P(1) = P(3) = 0 
Side note: P(x) = 2x^{3 }+ 3x^{2 } 14x^{ } 15 = (x + 1)(2x  1)(x + 3) 


Find all possible rational roots and all rational roots of P(x) = 4x^{3 } 15x^{2 } 31x^{ }+ 30 
[Solution] 


Find all roots of P(x) = 3x^{4 }+ 10x^{3 } 9x^{2 } 40x^{ } 12 
Find all possible rational zeros  rational numbers are ratio of integers i.e. 2/1 = 2, 0, 1/2, 3/2  
Solve 
Step 
Property 
p = ± 1, ±
2, ± 3, ±
4, ± 6, ±
12
q ± 1, ± 3 
Find
possible rational zeros p/q where p are factors of the last coefficient 12. q are the factors of the first coefficient 2. 

p = ± 1, ± 2, ± 3, ± 4, ± 6, ± 12 and 
p/1 = p  We always get the p's 
and all of the p/3's.  
Reduce  
Start with 2 (I tried others first  You can use Descartes' Rules of Signs to narrow the search.)  
Bring the down the 3.  coefficient of 3x^{4} is 3  
Multiply
3 × 2
= 6 Add Columns 10 + 6 = 16 

Multiply
16 × 2
= 32 Add Columns 9 + 32 = 23 

Multiply
23 × 2
= 46 Add Columns 40 + 46 = 6 

Multiply
6 × 2
= 12 Add Columns 12 + 12 = 0 

If there is a remainder of zero, 0 , then 2 is a zero and (x  2) is a factor of the polynomial.  
Q(x) = 3x^{3 }+ 16x^{2 }+ 23x^{ }+ 6 
The quotient is the coefficients of the bottom line  
Try 2 next  I tried other numbers that failed first  
I did the work for you....  
If there is a remainder of zero, 0 , then 2 is a zero and (x  (2)) = (x + 2) is a factor of the polynomial.  
Q(x) = 3x^{2 }+ 10x^{ }+ 3  The quotient is the coefficients of the bottom line  Once we have a quadratic equation, factor or use the quadratic formula. 
3x^{2 }+ 10x^{ }+ 35 = (3x + 1)(x + 3) = 0 
Set equal to zero and solve for x  
3x + 1 = 0 or x + 3 = 0 

x = 1/3 or x = 3 
Check in original equation.  
{ 2, 2, 1/3 , 3} 
There are three solutions for P(x)  P(2) = P(2) = P(1/3) = P(3) = 0 
Side note: P(x) = 3x^{4 }+ 10x^{3 } 9x^{2 } 40x^{ } 12 = (x  2)(x + 2)(3x + 1)(x + 3)  

Tutorials and Applets by
Joe McDonald
Community College of Southern Nevada