Worked examples
The question
Question: find the remainder when x² + 3x − 4 is divided by (x − 2).
x2 + 3x − 4x − 2
The remainder theorem: when we divide by x minus a number, the remainder is what f gives for that number.
We don't need to divide.
divide by x − c → remainder = f(c)
1 Call it f(x)
Let's call the polynomial f(x).
f(x) = x2 + 3x − 4
2 Make x − 2 zero
Which x makes x − 2 zero?
x − 2 = 0
2 take away 2 is 0.
So x = 2.
x = 2
Careful!
Many people put in −2, the number they see.
But x − 2 is 0 when x is 2.
Careful: not f(−2) ✗
3 Work out f(2)
Put 2 in place of every x.
We write it in brackets.
f(2) = 22 + 3(2) − 4
Work out each part: the powers first, then the times.
= 4 + 6 − 4
Then add and take away from left to right.
f(2) = 6.
f(2) = 6
4 Read the remainder
So the remainder is 6.
remainder = 6
5 Check
Let's check with synthetic division, from the division lesson.
We use 2 again.
The last number in the table is the remainder: 6.
The same as f(2)!
Last number: 6 = f(2)
Let's look back.
First: call it f(x).
Then: make x − 2 zero.
Then: work out f(2).
Then: read the remainder.
Last: check.
▶ Watch it solved
The question
Question: find the remainder when x³ + 2x² − x + 3 is divided by (x + 1).
x3 + 2x2 − x + 3x + 1
The remainder theorem: when we divide by x minus a number, the remainder is what f gives for that number.
We don't need to divide.
divide by x − c → remainder = f(c)
1 Call it f(x)
Let's call the polynomial f(x).
f(x) = x3 + 2x2 − x + 3
2 Make x + 1 zero
Which x makes x + 1 zero?
x + 1 = 0
x + 1 is the same as x minus −1.
So c is −1.
x + 1 = x − (−1)
−1 plus 1 is 0.
So x = −1.
x = −1
Careful!
Many people put in 1, the number they see.
But x + 1 is 0 when x is −1.
Careful: not f(1) ✗
3 Work out f(−1)
Put −1 in place of every x.
We write it in brackets.
f(−1) = (−1)3 + 2(−1)2 − (−1) + 3
Work out each part: the powers first, then the times.
= −1 + 2 + 1 + 3
Then add and take away from left to right.
f(−1) = 5.
f(−1) = 5
4 Read the remainder
So the remainder is 5.
remainder = 5
5 Check
Let's check with synthetic division, from the division lesson.
We use −1 again.
The last number in the table is the remainder: 5.
The same as f(−1)!
Last number: 5 = f(−1)
Let's look back.
First: call it f(x).
Then: make x + 1 zero.
Then: work out f(−1).
Then: read the remainder.
Last: check.
▶ Watch it solved
The question
Question: find the remainder when 2x⁴ − x³ + 3x − 5 is divided by (x + 1).
2x4 − x3 + 3x − 5x + 1
The remainder theorem: when we divide by x minus a number, the remainder is what f gives for that number.
We don't need to divide.
divide by x − c → remainder = f(c)
1 Call it f(x)
Let's call the polynomial f(x).
f(x) = 2x4 − x3 + 3x − 5
2 Make x + 1 zero
Which x makes x + 1 zero?
x + 1 = 0
x + 1 is the same as x minus −1.
So c is −1.
x + 1 = x − (−1)
−1 plus 1 is 0.
So x = −1.
x = −1
Careful!
Many people put in 1, the number they see.
But x + 1 is 0 when x is −1.
Careful: not f(1) ✗
3 Work out f(−1)
Put −1 in place of every x.
We write it in brackets.
f(−1) = 2(−1)4 − (−1)3 + 3(−1) − 5
Work out each part: the powers first, then the times.
= 2 + 1 − 3 − 5
Then add and take away from left to right.
f(−1) = −5.
f(−1) = −5
4 Read the remainder
So the remainder is −5.
remainder = −5
5 Check
Let's check with synthetic division, from the division lesson.
We use −1 again.
The last number in the table is the remainder: −5.
The same as f(−1)!
Last number: −5 = f(−1)
Let's look back.
First: call it f(x).
Then: make x + 1 zero.
Then: work out f(−1).
Then: read the remainder.
Last: check.
▶ Watch it solved
