Quicker Method for Long Division of Polynomials (1 Viewer)

[Undertoad]

Here's another method for long division of polynomials that I don't think a lot of people know but I found really useful.

eg. Using the fact that (x-2) is a factor of P(x)=x^3-3x^2+4x-4, factorise P(x) over the real field R.

1. Start by putting the a in (x-a) out the front of the long division signs and place each coefficient as shown (even if it is 0)

2|1 -3 4 -4

2. Take the first coefficient down two lines as shown

2|1 -3 4 -4
________
1

3. Multiply the bottom number by a (in this case 2) and place it in the middle line underneath the 2nd coefficient

2|1 -3 4 -4
2
________
1

4. Then simply add the two vertical numbers and place down on the bottom line

2|1 -3 4 -4
2
________
1 -1

5. Continue along until finished

2|1 -3 4 -4
2 -2 3
________
1 -1 2 0

6. This gives you the answer:

The most far right number is ALWAYS the remainder, in this case 0
Following that, each number to the left increases its power of x by one, in this case: x^2-x+2

7. So the answer would be P(x)=(x-2)(x^2-x+2)=0 and you can continue with that until it is fully factorised

The only problem with this method is that it does not work for dividing polynomials by terms such as x^2, where normal long division methods must be used. But, as time is a large factor in success in 4unit maths, many people that I know including myself have found this method great for solving problems quickly.

It can also be used on problems where a factor and a polynomial with an unknown is given, eg. P(x)=3x^3-12x^2-11x+k

 

[Riviet]

It's certainly alot more simple, leaving out all x terms, hehe. Hence it's probably faster. Thanks for the cool tip.

 

[buchanan]

It's called synthetic division.

http://www.purplemath.com/modules/synthdiv.htm
http://www.purplemath.com/modules/synthdiv2.htm

 

[A l]

Too bad it only works for division of a polynomial by a linear factor...

 

[buchanan]

I can be used for quadratic divisors as well:
http://www.purplemath.com/modules/synthetic-division-sgnd.pdf

 

[pLuvia]

Damn this is really useful thanks

 

[Mill]

I'm of the opinion that more capable HMX2 students should practice in being able to put any polynomial function in the form A(x).Q(x) if A(x) is linear simply by inspection.

Take the polynomial, p(x) = x3 - 4x2 + x + 6.
[Sorry no idea how to do powers on the forum but you get the idea... ]

Since we know p(2) = 0, (x-2) is a factor of p(x).

Thus,

p(x) = x3 - 4x2 + x + 6
= (x-2)( ??? )

The first term in the second brackets is obviously an x2.

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 + ???)

So what is the effect of this x2? Well it gives us the x3 as desired, and then an extra -2x2.

But we require -4x2. That is, we require a FURTHER -2x2.

So the next term in the second brackets is -2x (to mupltiply with the x from the first bracket to give that extra -2x2).

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 - 2x + ???)

So what is the effect of this -2x? Well it gives us the -2x2 as desired, and then an extra 4x.

But we require x. That is, we require a FURTHER -3x.

So the next term in the second brackets is -3 (to mupltiply with the x from the first bracket to give that extra -3x).

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 - 2x - 3)

And you are done.



This may be a little confusing at first but with some practice you will find it a useful technique. You may wish to write it down by pen though since I couldn't work out how to make superscript indices.

With further practice, you will be able to do this for A(x) of higher degree.

 

[buchanan]

I'm of the opinion that more capable HMX2 students should practice in being able to put any polynomial function in the form A(x).Q(x) if A(x) is linear simply by inspection.

You forgot the remainder. It should be A(x).Q(x)+R, for some number R.

If it is a more general divisor, A(x).Q(x)+R(x) where deg(R(x)) < deg(A(x)).

The inspection method works for these as well. Just use the last term in Q(x) to get R(x).

Also, in your example, I would suggest checking that the remainder is 0. This will help to see if a mistake had been made.

You are correct however in saying that students should be able to do it by inspection. Nevertheless, it's a bit more complicated if deg(A(x)) > 1.

I couldn't work out how to make superscript indices.

Without the spaces, x< sup >2< /sup > will give x².

 

[Mill]

I thought I'd start them off on one that didn't have a remainder since it is something that a lot of students actually do struggle with I've noticed. I think the main problem is how much they can remember in their head at any given moment which is sort of what it is. Calculations, addition and subtraction, in your head. But you are right, it can be done for non-zero R(x).



I would like to point out that I did say that P(2) = 0 before I started which implies that R(2) = 0. I assume that is what you mean by checking there is no remainder? If not, I'm curious to hear what you propose. Perhaps, you want students to 'expand' the two constant terms from the factorised form to recognise that it is the same as the required constant term of the original expression?

 

[Undertoad]

I can be used for quadratic divisors as well:
http://www.purplemath.com/modules/synthetic-division-sgnd.pdf

Wish I'd seen that! cool.

 

[buchanan]

Perhaps, you want students to 'expand' the two constant terms from the factorised form to recognise that it is the same as the required constant term of the original expression?

Yes. That's right. If it's different, a mistake has been made.

Students should check their solutions as well as just doing them.

 

[Rax]

So Synthetic division is a valid method for division in the EX2 HSC exam?

 

[Mill]

Why wouldn't it be?

You aren't required to long divide.

You are required to factorise polynomials.

How you choose to perform that factorisation is up to you!

 

[Undertoad]

As far as I've heard it is. Expect if you are asked to prove something, but in most cases the remainder theorem is used then. If it is just for your own working in a part of a bigger question there shouldn't be a problem with it.

 

[bboyelement]

hahah sick method thanks

 

[nickthebiscuit]

Synthetic division -> aka. Ruffini's rule~

For those who are interested

 

[zzdfa]

I'm of the opinion that more capable HMX2 students should practice in being able to put any polynomial function in the form A(x).Q(x) if A(x) is linear simply by inspection.

Take the polynomial, p(x) = x3 - 4x2 + x + 6.
[Sorry no idea how to do powers on the forum but you get the idea... ]

Since we know p(2) = 0, (x-2) is a factor of p(x).

Thus,

p(x) = x3 - 4x2 + x + 6
= (x-2)( ??? )

The first term in the second brackets is obviously an x2.

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 + ???)

So what is the effect of this x2? Well it gives us the x3 as desired, and then an extra -2x2.

But we require -4x2. That is, we require a FURTHER -2x2.

So the next term in the second brackets is -2x (to mupltiply with the x from the first bracket to give that extra -2x2).

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 - 2x + ???)

So what is the effect of this -2x? Well it gives us the -2x2 as desired, and then an extra 4x.

But we require x. That is, we require a FURTHER -3x.

So the next term in the second brackets is -3 (to mupltiply with the x from the first bracket to give that extra -3x).

p(x) = x3 - 4x2 + x + 6
= (x-2)( x2 - 2x - 3)

And you are done.



This may be a little confusing at first but with some practice you will find it a useful technique. You may wish to write it down by pen though since I couldn't work out how to make superscript indices.

With further practice, you will be able to do this for A(x) of higher degree.

Isn't this exactly the same as long division, but in your head?

 

[Dumbledore]

i tried to check it, i kept getting a wrong answer....2 hours later... i found out i did the long division wrong

 

[Timothy.Siu]

Isn't this exactly the same as long division, but in your head?

nope, its just finding the other terms
its a lot faster

 

[maths-bbqz]

Yes, 7 steps to get the cubic into a product of a linear factor (which also happened to be given) and a quadratic. Much faster than long division