Download - The Remainder and Factor Theorems
You write the division problem in the same format you would use for numbers. If a term is missing in standard form …fill it in with a 0 coefficient.
Example: 2x4 + 3x3 + 5x – 1 = x2 – 2x + 2
1503222 2342 xxxxxx
2x4 = 2x2
x2
2x2
1503222 2342 xxxxxx
2x2
+4x2-4x32x4-( )
- 4x27x3 +5x
7x3 = 7x x2
+7x
7x3 - 14x2 +14x-( )
10x2 - 9x -1
+10
10x2 - 20x +20-( )
11x - 21
remainder
2x2 + 7x + 10 + 11x – 21 x2 – 2x + 2
Quotient + Remainder over divisor
y4 + 2y2 – y + 5 =y2 – y + 1
Answer: y2 + y + 2 + 3 y2 – y + 1
If a polynomial f(x) is divisible by (x – k), then the remainder is r = f(k).
Now you will use synthetic division (like synthetic substitution)
f(x)= 3x3 – 2x2 + 2x – 5 Divide by x - 2
Long division results in ?...... 3x2 + 4x + 10 + 15
x – 2 Synthetic Division: f(2) = 3 -2 2 -5
2
3
6
4
8
10
20
15
Which gives you: 3x2 + 4x + 10 + 15 x-2
Divide x3 + 2x2 – 6x -9 by (a) x-2 (b) x+3 (a) x-2 1 2 -6 -9
2
1
2
4
8
2
4
-5
Which is x2 + 4x + 2 + -5
x-2
(b) x+3 12 -6 -9 -3
1
-3
-1
3
-3
9
0
x2 – x - 3
A polynomial f(x) has factor x-k iff f(k)=0
note that k is a ZERO of the function because f(k)=0
Factor f(x) = 2x3 + 11x2 + 18x + 9 Given f(-3)=0
Since f(-3)=0 x-(-3) or x+3 is a factor So use synthetic division to find the others!!
211 18 9
-3
2
-6
5
-15
3
-9
0
(x + 3)(2x2 + 5x + 3)
So…. 2x3 + 11x2 + 18x + 9 factors to:
Now keep factoring (bustin ‘da ‘b’) gives you:
(x+3)(2x+3)(x+1)
Factor f(x)= 3x3 + 13x2 + 2x -8 given f(-4)=0
(x + 1)(3x – 2)(x + 4)
f(x) = x3 – 2x2 – 9x +18. One zero of f(x) is x=2 Find the others! Use synthetic div. to reduce the degree of
the polynomial function and factor completely.
(x-2)(x2-9) = (x-2)(x+3)(x-3) Therefore, the zeros are x=2,3,-3!!!
f(x) = x3 + 6x2 + 3x -10 X=-5 is one zero, find the others!
The zeros are x=2,-1,-5 Because the factors are (x-2)(x+1)(x+5)