[PDF] AQA Core 1 Polynomials Section 2: The factor and remainder

101368_6C1_factor_and_remainder_theorem_notes.pdf

AQA Core 1 Polynomials

1 of 6 13/11/13 © MEI

Section 2: The factor and remainder theorems

Notes and Examples

These notes contain subsections on

The factor theorem Dividing polynomials The remainder theorem

The factor theorem

You already know that you can solve some quadratics by factorising them. e.g. to solve the quadratic equation x² + 3x 10 = 0 you factorise: (x + 5)(x 2) = 0 and deduce the solutions x = -5 and x = 2 Clearly, for f(x) = x² + 3x 10, f(-5) = 0 and f(2) = 0. (x + 5) is a factor of f(x) f(-5) = 0 (x 2) is a factor of f(x) f(2) = 0 This idea can be extended to other polynomials such as cubics.

For example, for the cubic function

g( ) ( 1)( 2)( 3)x x x x , g(1) = 0, g(-2) = 0 and g(3) = 0. (x 1) is a factor of g(x) g(1) = 0 (x + 2) is a factor of g(x) g(-2) = 0 (x 3) is a factor of g(x) g(3) = 0

In general, the factor theorem states that:

You can use the factor theorem to solve cubic and higher order equations. These are the steps you need to take to solve a cubic equation of the form f(x) = 0, where f(x) is a cubic function: First, work out f(x) for different values of x until you find one for which f(x) = 0. Using the factor theorem, you now know one factor of the function. Divide the function by this linear factor. You can now express the function as the product of the linear factor and a quadratic factor. Factorise the quadratic factor, if possible. If the quadratic factor does factorise, you now have three linear factors, from which you can deduce the three solutions to the equation. If xa is a factor of f(x), then f(a) = 0 and x = a is a root of the equation f(x) = 0.

Conversely, if f(a) = 0, then

xa is a factor of f(x). AQA C1 Polynomials 2 Notes and Examples 2 of 6 13/11/13 © MEI If the quadratic factor does not factorise, you can use the quadratic formula to find the two further solutions, if they exist. For higher order equations, you will have to find more than one factor by trial and error, and you will have to divide more than once. The following example shows how this method works.

Example 1

(i) Solve the equation x³ + 2x² 5x 6 = 0 (ii) Sketch the graph of y = x³ + 2x² 5x 6

Solution

(i) Let f(x) = x³ + 2x² 5x 6 f(1) = 1 + 2 5 6 = -8 f(-1) = -1 + 2 + 5 6 = 0 f(-1) = 0 so by the factor theorem x + 1 is a factor of f(x). x³ + 2x² 5x 6 = (x + 1) quadratic factor. Let the quadratic factor be ax² + bx + c. x³ + 2x² 5x 6 = (x + 1)(ax² + bx + c) = ax³ + bx² + cx + ax² + bx + c = ax³ + (a + b)x² + (b + c)x + c Equating coefficients of x³ a = 1 Equating constant term c = -6 Equating coefficients of x² a + b = 2 b = 1 Check coefficient of x: b + c = 1 6 = -5 x³ + 2x² 5x 6 = (x + 1)(x² + x 6) = (x + 1)(x 2)(x + 3) The solutions of the equation are x = -1, x = 2 and x = -3. (ii) Part (i) shows that the graph of y = x³ + 2x² 5x 6 crosses the x-axis at (-3, 0), (-1, 0) and (2, 0). By putting x = 0 you can see that it crosses the y-axis at (0, -6).

This information allows you to sketch the graph.

You need to find a value of

x for which f(x) = 0.

Multiply out the

brackets

Factorise the

quadratic factor The first step is to find one solution by trial and error. If there is an integer solution x = a, then by the factor theorem (x a) must be a factor of x³ + 2x² 5x 6. So a must be a factor of 6. a could therefore be 1, -1, 2, -2, 3, -3, 6 or 6. The next step is to factorise f(x) into the linear factor x + 1 and a quadratic factor. AQA C1 Polynomials 2 Notes and Examples 3 of 6 13/11/13 © MEI In the example above, the quadratic factor was found by equating coefficients. There are a number of other methods of finding this quadratic factor. You can do it by polynomial division, or by inspection (which means doing it in your head). Which one you use is really down to personal preference. You can see some different approaches using the Flash resources Polynomial division by inspection, Polynomial division box method (no remainder) and Factorising a cubic, and the PowerPoint presentation Factorising polynomials. The Geogebra resource Polynomial division shows the box method.

You can also look at the Solving cubics video.

The Flash resource Sketching factorised cubics and the Mathcentre video

Polynomial functions may also be useful.

For some additional practice, try the interactive questions Sketching polynomial curves and Finding a polynomial from its roots.

Example 2

f(x) = 2x³ + px² + 5x 6 has a factor x 2. Find the value of p and hence factorise f(x) as far as possible.

Solution

x 2 is a factor of f(x) f(2) = 0 f(2) = 16 + 4p + 10 6 = 20 + 4p

20 + 4p = 0 p = -5

f(x) = 2x³ - 5x² + 5x 6 2x³ - 5x² + 5x 6 = (x 2)(ax² + bx + c) = ax³ + bx² + cx 2ax² - 2bx 2c AQA C1 Polynomials 2 Notes and Examples 4 of 6 13/11/13 © MEI = ax³ + (b 2a)x² + (c 2b)x 2c

Equating coefficients of x³ a = 2

Equating constant terms -2c = -6 c = 3

Equating coefficients of x² b 2a = -5 b 4 = -5 b = -1

Check coefficient of x: c 2b = 3 + 2 = 5

2x³ - 5x² + 5x 6 = (x 2)(2x² x + 3)

The discriminant of the quadratic factor is (-1)² - 4 2 3 = 1 24 = -23 As this is negative, the quadratic factor cannot be factorised further.

Dividing polynomials

When you divide one polynomial by another, it may divide exactly, or there may be a remainder, just as in arithmetic.

For example: 26 6 = 4 remainder 2

You could rewrite this statement as:

26 = 6 4 + 2
This rearrangement helps to give one method of dividing polynomials. When working with polynomials, remember the following points: If you are dividing by a linear expression, the quotient is of order one less than the dividend (e.g. for a quartic, the quotient is cubic) and the remainder, if any, is a constant term. If you are dividing by a quadratic term, the quotient is of order two less than the dividend (e.g. for a quartic, the quotient is quadratic) and the remainder, if any, could be linear or a constant term. This idea can be extended to a polynomial divisor of any order. Most, if not all, of the examples you meet will only involve dividing by a linear expression.

Example 3

Divide 2x³ + 3x² - x + 1 by x + 2

Solution

Let the quotient be ax² + bx + c and the remainder be d. 2x³ + 3x² x + 1 = (x + 2)(ax² + bx + c) + d = x(ax² + bx + c) + 2(ax² + bx + c) + d = ax³ + bx² + cx + 2ax² + 2bx + 2c + d = ax³ + (b + 2a)x² + (c + 2b)x + 2c + d

26 is called the dividend,

6 is called the divisor,

4 is the quotient

and 2 is the remainder.

When you divide a cubic expression

by a linear expression, as in this example, the quotient is a quadratic expression and the remainder, if any, is a constant term. AQA C1 Polynomials 2 Notes and Examples 5 of 6 13/11/13 © MEI

Equating coefficients of x³ a = 2

Equating coefficients of x² b + 2a = 3 b + 4 = 3 b = -1 Equating coefficients of x c + 2b = -1 c 2 = -1 c = 1 Equating constant terms 2c + d = 1 2 + d = 1 d = -1 2x³ + 3x² x + 1 = (x + 2)(2x² x + 1) 1 The quotient is 2x² x + 1 and the remainder is -1.