WHRHS
Unit 10 – Division of Polynomials & Rational Root Theorem
2011-2012
Algebra 2A
Unit 10 Day B – Factor, Remainder & Integer Zero Theorems
1. 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 Unit 10J Day A #3. Synthetic Division 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!
2. Remainder Theorem & Factoring –
If the Remainder = 0;
Then the Divisor and the Quotient of the Dividend are FACTORS of the Dividend.
3. Factor Theorem -  x  a  is a factor of a polynomial if and only if a is a root/zero of the
polynomial.  x  a  is a factor if the Remainder = 0.
4. Integer Zero Theorem – To use try all integral factors of the Constant term. Use these factors to find
the roots/zeros of the polynomial (Synthetic Division Remainder=0 then factor using any method).
Examples:
1. Evaluate P  x   2 x3  7 x2  5x  1 when x = 3 using Direct Substitution.
2. Using the same polynomial from 1. above find the Remainder using Synthetic Division and dividing the
polynomial by  x  3 .
3. Evaluate P  x   x4  14 x2  5x  3 for x = - 4.
4. a. Is  x 1 a factor of 3x3  4 x2  x  2?
b. Is 3x3  4 x2  x  2 divisible by  x 1 ?
c. Is x  1 a root(zero/solution) of 3x3  4 x 2  x  2?
d. Is P 1  2 ?
5. Find k so that x  3 is a factor of 3x3  2kx2   k  2 x  3.
WHRHS
2011-2012
6. Is x  1 a factor of x 200  1?
Unit 10 – Division of Polynomials & Rational Root Theorem
Algebra 2A
7. Find all Roots/Zeros of x3  6 x 2  11x  6 .
8. Solve by finding all Roots/Zeros of 2 x 4  7 x3  4 x 2  7 x  6  0 .
9. Solve by finding all Roots/Zeros of x3  8 x  3  0 .
Unit 10 – Division of Polynomials & Rational Root Theorem
Algebra 2A
WHRHS
2011-2012
Unit 10 Day B HW Page:
I.
Evaluate for the given value of x. Use Synthetic Division/Substitution.
1. x 4  3 x 2  7 x  2
for x  3.
2. 2 x 4  9 x 3  x  1
for x  5.
3.  2 x 3  15 x 2  7
for x  6.
II.
III.
Is the first polynomial a FACTOR of the second? Use the most appropriate method. Explain your
answer.
4.
 t  1 ;
5.
 x  5 ;
6.
 x  1 ;
x100  5 x 99  6.
7.
b  2 ;
b 4  3b3  3b 2  2b  16.
t 4  3t 2  2t  4.
x3  5 x  3 x 2  15.
Is “c” a root/zero of the given equation?
8. x3  4 x 2  8 x  5  0; c  1.
9. 3 y 3  2 y 2  8 y  3  0; c  3.
10. 6 s 4  29 s 3  40 s 2  7 s  12  0; c  
4
3.
Unit 10 – Division of Polynomials & Rational Root Theorem
Algebra 2A
WHRHS
2011-2012
Unit 10 Day B HW Page Continued:
Find “k” so that the first polynomial is a FACTOR of the second.
IV.
V.
11.
t  2 ;
3t 3  2kt 2   k  1 t  10.
12.
 v  1 ;
2v 3   k  1 v 2  6kv  11.
Use the Integer Zero Theorem to find the Roots/Zeros of the polynomial. Then Factor the
polynomial.
13. 3 x3  10 x 2  9 x  4
14. 2 x3  3 x 2  32 x  15
15. 2 x3  3 x 2  17 x  30  0
16. 2 x 4  x3  8 x 2  x  6  0
17. x 4  6 x 2  27  0