A2-Level Maths: Core 4 for Edexcel C4.1 Algebra and functions This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 22 © Boardworks Ltd 2006

Contents Partial fractions Partial fractions Denominators with distinct linear factors Denominators with a repeated linear factor Improper fractions Contents 2 of 22 © Boardworks Ltd 2006

Partial fractions We know that two or more algebraic fractions can be added or subtracted to give a single fraction. For example: Now, suppose we want to reverse the process. In other words, suppose we are given and asked to express it as a sum of two separate fractions.

Two distinct linear factors This process is called expressing in partial fractions. For example: Express in partial fractions. The first step is to set up an identity. The denominator of this fraction has two distinct linear factors so let Tell students that “distinct linear factors” means different linear factors. where A and B are constants to be found.

Two distinct linear factors 1 There are now two ways to continue: using suitable substitutions, Remind students that identities are true for any chosen value of x. So if we choose x to be 3, A will be eliminated to give B, and if we choose x to be –1, B will be eliminated to give A. by equating coefficients. To solve by substitution, we choose values of x that make the brackets zero.

Two distinct linear factors Here substitute x = –1 into : 1 Now substitute x = 3 into : 1 Therefore

Denominators with distinct linear factors Partial fractions Denominators with distinct linear factors Denominators with a repeated linear factor Improper fractions Contents 7 of 22 © Boardworks Ltd 2006

Two distinct linear factors Express in partial fractions. Let This can be simplified by multiplying through by (3x – 2)(2x –1): This time if we use the substitution method we’ll have to substitute fractional values for x. Let’s multiply out the brackets and equate coefficients instead.

Two distinct linear factors Equate the coefficients of x: 1 Now equate the constants: 2 + (2 × ) gives: 2 1 Substitute this into to find A: 1 Therefore

Three distinct linear factors Express as a sum of partial fractions. This time we have three distinct linear factors, so: Let Multiply through by (x – 3)(x +1)(2x +1): 1 Explain that choosing x to be 3 eliminates B and C from the identity. To find A, substitute x = 3 into : 1

Three distinct linear factors We can find B by substituting x = –1 into : 1 We can now find C either by substituting x = or by reverting to the method of equating coefficients.

Three distinct linear factors To avoid awkward arithmetic involving fractions, let’s form an equation in C by equating the constant terms in . 1 1 We could also equate the coefficients of x2 or x. Therefore

Denominators with a repeated linear factor Partial fractions Denominators with distinct linear factors Denominators with a repeated linear factor Improper fractions Contents 13 of 22 © Boardworks Ltd 2006

Denominators with a repeated linear factor Suppose we wish to express in partial fractions. This is an example of a fraction whose denominator contains a repeated linear factor. In this case, the partial fractions will be of the form: We can now find A, B and C using a combination of substitution and equating the coefficients.

Denominators with a repeated linear factor 1 Substitute x = –4 into : 1 Substitute x = 3 into : 1 To find B we can switch to the method of comparing coefficients.

Denominators with a repeated linear factor Equate the coefficients of x2 in : 1 But A = 2 so: Therefore Express as a sum of partial fractions. An alternative to comparing coefficients to find B would be to substitute x = 0 to give an equation in terms of A, B and C. Substituting the values we have already found for A and C would then leave an equation in B only. Let

Denominators with a repeated linear factor Multiply through by x2(4 – x): 1 Substitute x = 0 into : 1 Substitute x = 4 into : 1

Denominators with a repeated linear factor We can find A by comparing the coefficients of x2. But C = 4 so: Therefore

Contents Improper fractions Partial fractions Denominators with distinct linear factors Denominators with a repeated linear factor Improper fractions Contents 19 of 22 © Boardworks Ltd 2006

Improper fractions Remember, an algebraic fraction is called an improper fraction if the degree of the polynomial is equal to, or greater than, the degree of the denominator. To express an improper fraction in partial fractions we start by expressing it in the algebraic equivalent of mixed number form. Any proper fractions contained in this form can then be expressed in partial fractions. Express in partial fractions. Note that dividing 2x2 –3x + 13 by x2 – 2x – 15 gives 2 remainder x + 43. x + 43 over (x + 3)(x – 5) would then have to be split into partial fractions. Setting up an identity as shown on the next slide removes the need for this step. We can either use long division to divide 2x2 – 3x + 13 by x2 – 2x – 15 or we can set up an identity as follows:

Improper fractions Start by factorizing the denominator. The numerator and the denominator are both of degree 2 and so they will divide to give a constant, A. The part that is a proper fraction will have two distinct linear factors. So we can let Multiply through by (x + 3)(x – 5): 1

Improper fractions Substitute x = –3 into : Substitute x = 5 into : 1 Substitute x = 5 into : 1 We can find A by equating the coefficients of x2 in . 1 Therefore