WHRHS
Unit 10 – Division of Polynomials & Rational Root Theorem
2011-2012
Algebra 2A
Unit 10 Day A – Long Division & Synthetic Division of Polynomials
Division of Polynomials –
1. Terminology –
a. Dividend – Polynomial to be divided
b. Divisor – Polynomial divided into the Dividend
c. Quotient – Whole Polynomial Answer when Dividend is
divided by the Divisor
d. Remainder – Part of the Divisor that remains when division is
completed
e. To Check Division Use Formula:
Dividend = (Quotient)(Divisor) + Remainder !
2. Long Division – Similar method to number division.
*Remember to distribute the minus sign when you are dividing!
Example 1:
Divisor
Quotient Dividend
Remainder
Solution =
3. Synthetic Division – Shorter method of dividing a polynomial by a binomial.
* The binomial must be in the form of (x – a).
* This method uses the coefficients of the variable terms.
* Remember to account for missing terms by using 0 as a place holder!
* Remember write the Polynomial with the variables in descending
exponent order!
Example 1:
3
2  x   4( x) 2  3( x)  2 divided by (x – 6)
(form is (x – a) with a = 6 which is moved to the left of the coefficients of
the variable terms.)
6
2
4
+12
3
+96
-2
+594
2
16
99
592
x
(The 1st term of the variable of the Quotient is 1 degree less than the degree of
the 1st term of the dividend.)
Solution = 2 x 2  16 x  99 
592
x6
Unit 10 – Division of Polynomials & Rational Root Theorem
Algebra 2A
WHRHS
2011-2012
4. Remainder Theorem –
a. If polynomial P(x) is divided by (x - a) then the Remainder = P(a).
b. This theorem is used to Evaluate a polynomial at a given value.
c. Use the Synthetic Substitution Method and the Remainder will be
the polynomial evaluated at the given value.
Example 1:
Use Example 1 from 3. above and you will see that the Remainder is 592.
3
Therefore the polynomial 2  x   4( x) 2  3( x)  2 evaluated at x=6 is 592!
5. Remainder Theorem & Factoring –
If the Remainder = 0;
Then the Divisor and the Quotient of the Dividend are FACTORS of the Dividend.
Examples:
1. (14 x 2  3x  3x3  7)  ( x  5)
2.
 3c
4
 2c 3  8c  48    c 2  4 
Synthetic Division
3.
14 x
2
 3x  3x3  7   ( x  5)
Synthetic Division
4.
 3x
4
 3 x 2  4 x  3  ( x  2i )
Synthetic Division
5.
15 x
3
 x 2  12 x  12   (3 x  2)
WHRHS
Unit 10 – Division of Polynomials & Rational Root Theorem
2011-2012
Algebra 2A
Unit 10 Day A HW Page:
Use synthetic division for 1-9. Give the quotient and remainder.
1.
3x3  4 x 2  x  2
x 1
2.
y3  2 y 2  2 y  3
y2
3.
2a 4  5a3  a 2  9a  6
a 3
4.
2 x3  11x 2  10 x  8
x4
5.
2z 4  z3  2z  4
z  3i
6.
6t 3  7t 2  14t  6
2t  1
7.
4v3  2v  24
2v  3
8.
3x 4  6 x3  3x  4
x  2i
9.
2 x3  3x 2  12 x  9
x 3
WHRHS
Unit 10 – Division of Polynomials & Rational Root Theorem
2011-2012
Algebra 2A
Use Long Division to find the quotient and remainder. Determine if it is a factor.
10.
 2c
2

 c  2   c  1
11. 15  14n  8n 2    4n  5 
12.
a
13.
 2n
3

 13a  12   a  3
4
 
 4n3  7n 2  12n  3  n 2  3
