powerpoint presentation

2. We can use the remainder theorem to determine the
remainder of a polynomial function divided by a binomial function
without going through the tedious process.
If the polynomial function f(x) is divided by a
binomial function g(x) – x – a, then the remainder is
f(a).

3. Example 1. Find the remainder using the remainder theorem when
f(x) = x3 – 3x2 + 7 ÷ g(x) = x + 2
Steps Solution
1. Equate to x – a . (g) = x + 2 = 0
x = 0 – 2
x = -2
2. Find f(-2). f (x) = f(x) = x3 – 3x2 + 7
f(-2) = (-2)3 – 3(-2)2 + 7
f(-2) = -8 – 3(4) + 7
f(-2) = -8 – 12 + 7
f(-2) = -13

4. Therefore the remainder when f(x) = x3 – 3x2 + 7
is divided by g(x) = x + 2 is -13.

5. Example 2. Find the remainder when f(x) = x3 – 3x2 + 5x – 1 is
divided by g(x) = x - 1
Steps Solution
1. Equate g(x) = x – a to 0. g(x) = x – 1 = 0
x = 0 + 1
x = 1
2. Find f(1). f(x) = x3 – 3x2 + 5x – 1
f(1) = (1)3 – 3(1)2 + 5(1) – 1
f(1) = 1 – 3(1) + 5 – 1
f(1) = 1 – 3 + 5 – 1
f(1) = 2

6. Therefore, the remainder when f(x) = x3 – 3x2 + 5x – 1
is divided by g(x) = x – 1 is 2.

7. ACTIVITY

8. Activity:
1. (x4 + 2x3 + 3x + 6) ÷ (x + 2)
= Write the terms in ascending
order.
= equate (x + 2) to x – a
X + 2 = 0
x = 0 – 2
x = -2
Synthetic Division:
-2 1 2 0 3 6
-2 0 0 -6
1 0 0 3 0 – (no remainder)
3 = constant term
0 = degree of 1
0 = degree of 2
0 = degree of 3
Quotient = x3 + 3

9. Using the remainder theorem,
1. (x4 + 2x3 + 3x + 6) ÷ (x + 2)
X + 2 = 0
x = 0 – 2
x = -2
= (-2)4 + 2(-2)3 + 3(-2) + 6
= 16 + 2(-8) -6 + 6
= 16 -16 – 6 + 6
= 0 – 6 + 6
= 0 (no remainder)

10. Synthetic Division
2 . (2x3 – 5x) ÷ (x + 1)
-1 2 0 -5 0
-2 2 3
2 -2 -3 3
Quotient: 2x2 – 2x – 3 +
3
𝑥+1
Remainder Theorem
2(-1)3 – 5(-1)
2(-1) + 5
-2 + 5
= 3

11. Synthetic Division:
-2 1 0 -7 -1
-2 4 6
1 -2 -3 5
= x2 – 2x – 3 +
𝟓
𝒙+𝟐