01Cpolydivision

toc = Division of Polynomials =

Long Division
math \text{In } P(x) \div D(x)=Q(x)+\dfrac{R(x)}{D(x)} math


 * ** D(x) ** is called the ** divisor  **
 * ** Q(x) ** is called the ** quotient  **
 * ** R(x) ** is called the ** remainder  **

For another site that explains Long Division, go here: MathIsFun

NOTE:
 * You will not be asked specifically to use long division in the final exams.
 * You will be expected to find the result of a division using some appropriate technique.
 * Long division is appropriate if the divisor is not a factor.
 * There are easier methods if the divisor is a factor (see below).

NOTE:
 * There are other methods like, for example. synthetic division.
 * I do not use synthetic division because it is most effective if the divisor is something in the form (x + a).
 * If x has a coefficient other than 1, synthetic division still works but it is easy to make mistakes.
 * If the divisor has a degree greater than one, then synthetic division can not be used.

** Example 1 **
Simplify the following using long division math . \qquad \dfrac{4x^3+8x^2+4x+1}{2x+1} math

Solution

** Example 2 **
Simplify the following using long division math . \qquad \dfrac{4x^3+6x-2}{x^2+2x+3} math

Solution

Remainder Theorem
math \text{When } P(x) \text{ is divided by } (x - a) \text{ the remainder is } P(a) math

math \text{When }P(x) \text{ is divided by } (ax - b) \text{ the remainder is } P\Big( \dfrac{b}{a} \Big) math

**Example 3**[[image:bhs-methods34/01Ceg1.gif width="208" height="212" align="right"]]
... ... Let P(x) = x 2 + 5x + 2. ... ... Find the remainder when P(x) is divided by (x + 1).

__**Solution:**__

... ... a = –1

... ... Remainder = P(–1) ... ... ... ... ... .... = –2


 * See long division at right to check that remainder is –2

Factor Theorem
A __factor__ of P(x) is a term that divides evenly into P(x) with no remainder. Apply this to the remainder theorem and we get:

math \text{If } P(a) = 0,\text{ then } (x - a) \text{ is a factor of }P(x) math

math \text{If } (ax - b) \text{ is a factor of } P(x) \text{ then } P \left( \dfrac{b}{a} \right) = 0 math


 * NOTE: ** If (x – a) is a factor of P(x) then //**a**// must be a factor of the constant term.
 * The constant term is the term independent of x, ie the coefficient of x 0

** Example 4 **
... ... Find a factor of : P(x) = x 3 – 2x 2 – 3x + 6


 * Solution:**

... ... P(1) = 1 – 2 – 3 + 6 = 2

... ... P(–1) = –1 – 2 + 3 + 6 = 6

... ... P(2) = 8 – 8 – 6 + 6 = 0

... ... P(2) = 0 so (x – 2) is a factor


 * 2 is a factor of the constant term (6)

Factorising Cubics
(Shortcut for Long Division)

If we know the divisor is a factor of the polynomial, we can do the following:

For a demonstration of this process in Powerpoint, download the following file (2.4 Mb)

** Example 5 **
... ... Factorise (x 3 – 2x 2 – 3x + 6) given that (x – 2) is a factor


 * Solution:**


 * By examination of the above, we can conclude that the other factor must be a quadratic.


 * Further the first term must be x 2 and the last term must be –3 so we can write:

... ... x 3 – 2x 2 – 3x + 6 = (x – 2)(x 2 + ** a **x – 3) ... ... where ** a ** is an unknown


 * If we expand these brackets we get:

... ... x 3 __– 2__x 2 – 3x + 6 = x 3 __+ ** a **__x 2 – 3x __– 2__x 2 – 2** a **x + 6


 * Equate the coefficients of x 2 (underlined above), we get:

... ... –2 = ** a ** – 2

... ... so ** a  ** = 0

Thus

math . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x^2 - 3 \big) math


 * The quadratic in the second bracket can then be factorised
 * in this case using difference of two squares

Thus math . \qquad x^3 - 2x^2 - 3x + 6 = \big( x - 2 \big) \big( x - \sqrt{3} \big) \big( x + \sqrt{3} \big) math Go to top of page flat

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