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Revision:Algebraic Long Division

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TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Algebraic Long Division

Algebraic long division is very similar to traditional long division (which you may have come across earlier in your education).


Contents

Example

We want to simplify \dfrac{2x^3-3x^2-3x+2}{x-2} . The procedure is as follows:

Step 1:

image:algdiv1.gif

Set up the problem like a normal long division problem as above.

Step 2:

image:algdiv2.gif

Divide the first term of the numerator by the first term of the denominator. In this case, 2x^3 / x = 2x^2.

Step 3:

image:algdiv3.gif

Multiply this result by the whole of the denominator. Write this underneath. So here: 2x^2(x-2) = 2x^3 - 4x^2.

Step 4:

image:algdiv4.gif

Subtract: (2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2. Bring the next term "down".

Step 5:

image:algdiv5.gif

Divide the first term of the new expression by the first term of the denominator. (Note that we have, at this stage, deduced that \dfrac{2x^3-3x^2-3x+2}{x-2} = 2x^2 + \dfrac{x^2-3x+2}{x-2}, and are now looking to divide out the smaller fraction; then step 5 is equivalent to step 2.)

Step 6:

image:algdiv6.gif

Multiply as in step 3.

Step 7:

image:algdiv7.gif

Continue. Note that, when we have finished, we end up with a 0 - that is, no remainder. This won't be the case for all polynomial division, we were just lucky!

NB: If the numerator has a term in x missing, add the missing term by placing a zero in front of it. For example, if you are dividing \ x^3 + x - 4 by something, rewrite it as \ x^3 + 0x^2 + x - 4 .

For algebraic long division practice makes perfect- the best way to learn how to do them properly is to do loads of examples until you get them right every time!


The Remainder Theorem

When dividing one algebraic expression by another, more often than not there will be a remainder. It is often useful to know what this remainder is and it can be calculated without going through the process of dividing as above. The rule is:

  • If a polynomial \ f(x) is divided by ax - b , the remainder is \ f(b/a)


Example

In the above example, \ 2x^3 - 3x^2 - 3x + 2 was divided by \ x - 2 .

Let \ f(x) = 2x^3 - 3x^2 - 3x + 2 , a = 1, b = 2 . The remainder is therefore \ f(2) = 2\times 2^3 - 3\times 2^2 - 3\times 2 + 2 = 0, as we saw when we divided the whole thing out. However, dividing by x-3 would have given us a remainder of f(3) = 20.


The Factor Theorem

This states:

  • If \ x - a is a factor of the polynomial \ f(x) , then \ f(a) = 0 . Conversely, if f(a) = 0 and f is a polynomial function, then x-a is a factor of f(x).


Example

In the above worked example, \ f(2) = 0 . This means that (x - 2) is a factor of the equation.

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