What is the remainder when x^3 - 2x^2 + 4x - 5 is divided by x

Step by step

The question

Question: find the remainder when x³ − 2x² + 4x − 5 is divided by (x − 2).

x³ − 2x² + 4x − 5x − 2

There are four choices.

First we solve it, the normal way.

Then we pick.

1 Call it f(x)

Let's call the polynomial f(x).

We want the remainder, without the long division.

f(x) = x³ − 2x² + 4x − 5

2 Make x − 2 zero

Which x makes x − 2 zero?

We will see why that x helps.

x − 2 = 0

Is x alone now?

Not yet: − 2 is still with x.

How do we remove it?

We add 2 on both sides.

x − 2 + 2 = 0 + 2

So x = 2.

Check: 2 take away 2 is 0.

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) ✗

So after step 2: we added 2, and now we have x = 2.

← step 2: added 2

3 Why f(2) is the remainder

Think of 7 divided by 2.

It is 3, with 1 left over.

So 7 is 2 times 3, plus 1.

7 = 2 × 3 + 1

Dividing f(x) works the same way.

f(x) is (x − 2) times the answer, plus the remainder.

We stop dividing when the part left has no x.

So the remainder is just a number.

f(x) = (x − 2) × answer + remainder

Now put in x = 2.

f(2) = (2 − 2) × answer + remainder

2 take away 2 is 0.

And 0 times the answer is 0.

= 0 × answer + remainder

So f(2) is just the remainder.

We don't need to divide!

f(2) = remainder

This is called the remainder theorem.

divide by x − 2 → remainder = f(2)

4 Work out f(2)

Put 2 in place of every x.

We write it in brackets.

f(2) = 2³ − 2(2)² + 4(2) − 5

Work out each part: the powers first, then the times.

= 8 − 8 + 8 − 5

8 − 8 = 0

0 + 8 = 8

8 − 5 = 3

Then add and take away from left to right.

f(2) = 3.

f(2) = 3

5 Read the remainder

So the remainder is 3.

remainder = 3

6 Check

Let's check with synthetic division, from the division lesson.

We use 2 again.

The last number in the table is the remainder: 3.

The same as f(2)!

Last number: 3 = f(2)

7 Pick the choice

Our answer is remainder = 3.

That is choice C.

C) 3

8 Why not the others

A is −29.

Wrong: put in −2: for x − 2 we put in 2.

A) −29 ✗ put in −2: for x − 2 we put in 2

B is 1.

Wrong: 2³ is 8, not 6.

B) 1 ✗ 2³ is 8, not 6

D is 8.

Wrong: forgot the − 5.

D) 8 ✗ forgot the − 5

Let's look back.

First: call it f(x).

Then: make x − 2 zero.

Then: why f(2) is the remainder.

Then: work out f(2).

Then: read the remainder.

Then: check.

Then: pick the choice.

Last: why not the others.

Answer