Long Division

mathxyz · 2024-05-18 05:50:47

The last long division for today. This is tedious work.

Problem 103 on 49 goes like this:

Divide x^3 - a^3 by x - a.

I posted this in a FB math group, and several members responded.

Here is one reply:

x^2 + ax + a^2
_______________
x - a | x^3 - a^3
x^3 - ax^2
—————
ax^2 - a^3
ax^2 - (a^2)x
———————
(a^2)x - a^3
(a^2)x - a^3
———————
0

Look at this line:

ax^2 - a^3
ax^2 - (a^2)x

This is where I get stuck. I think it is easier to use the difference of cubes to expand x^3 - a^2.

You say?

Bob · 2024-05-18 06:06:38

'Difference of cubes' ? I only know difference of two squares.

There's a theorem known as the remainder theorem ... that leads to the factor theorem.

If F(x) = Q(x).(x-a)+ R

If we put x = a we get F(a) = R

This can be useful in getting a remainder quickly.

example. An earlier question was "Find the remainder if (4x^3 - 3x^2 + x + 1) ÷ (x + 2)

a in this case is -2 so put that value into the dividend:

4 times (-2)^3 - 3 times (-2)^2 + (-2) + 1 = -32 -12 -2+ 1 = -46 + 1 = -45

If that is no remainder then R = 0 so F(x) = Q(x) . (x-a) Putting x=a leads to F(a) = 0.

Example. x^3 - a^3. Put x = a and we get a^3 - a^3 = 0. So (x-a) is a factor.

Bob

mathxyz · 2024-05-18 06:11:16

Bob wrote:

'Difference of cubes' ? I only know difference of two squares.
There's a theorem known as the remainder theorem ... that leads to the factor theorem.
If F(x) = Q(x).(x-a)+ R
If we put x = a we get F(a) = R
This can be useful in getting a remainder quickly.
example. An earlier question was "Find the remainder if (4x^3 - 3x^2 + x + 1) ÷ (x + 2)
a in this case is -2 so put that value into the dividend:
4 times (-2)^3 - 3 times (-2)^2 + (-2) + 1 = -32 -12 -2+ 1 = -46 + 1 = -45
If that is no remainder then R = 0 so F(x) = Q(x) . (x-a) Putting x=a leads to F(a) = 0.

Example. x^3 - a^3. Put x = a and we get a^3 - a^3 = 0. So (x-a) is a factor.
Bob

Very informative. Thanks.

Can we expand x^3 - a^3 using the difference of cubes?

Look on page 44 for the difference of cubes. This allows us to cancel out the binomial x - a.

Yes?

Bob · 2024-05-18 19:20:24

p44 example 10 has (x-1). But the quesion here shows this can be generalised.

Bob

mathxyz · 2024-05-19 01:48:35

Bob wrote:

p44 example 10 has (x-1). But the quesion here shows this can be generalised.
Bob

On page 44 you can see the difference of cubes formula that can be used to expand x^3 - a^3 in our long division problem.

mathxyz · 2024-05-19 04:23:33

Bob wrote:

p44 example 10 has (x-1). But the quesion here shows this can be generalised.
Bob

Bob,

Using formula 5 on page 44, I can easily divide x^3 - a^3 by x - a.

Look:

x^3 - a^3 = (x - a)(x^2 + ax + a^2).

So, [(x - a)(x^2 + ax + a^2)]/(x - a) = (x^2 + ax + a^2).

Much easier than using long division. However, I do understand why Sullivan instructs students to use long division.

Bob · 2024-05-19 04:37:16

Easier .... like using the quadratic formula rather than completing the square. But where did the formula come from? Mathematicians like to know the answer to that question ... understanding rather than rote learning.

Bob

mathxyz · 2024-05-19 07:03:48

Bob wrote:

Easier .... like using the quadratic formula rather than completing the square. But where did the formula come from? Mathematicians like to know the answer to that question ... understanding rather than rote learning.
Bob

I totally get it. Sullivan wants students to practice long division on that particular page.

Math Is Fun Forum

#1 2024-05-18 05:50:47

Long Division

#2 2024-05-18 06:06:38

Re: Long Division

#3 2024-05-18 06:11:16

Re: Long Division

#4 2024-05-18 19:20:24

Re: Long Division

#5 2024-05-19 01:48:35

Re: Long Division

#6 2024-05-19 04:23:33

Re: Long Division

#7 2024-05-19 04:37:16

Re: Long Division

#8 2024-05-19 07:03:48

Re: Long Division

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