22: Division and The Remainder Theorem © Christine Crisp “Teach A Level Maths” Vol. 1: AS Core Modules

Division and the Remainder Theorem Module C1 AQA Edexcel OCRMEI/OCR Module C2 "Certain images and/or photos on this presentation are the copyrighted property of JupiterImages and are being used with permission under license. These images and/or photos may not be copied or downloaded without permission from JupiterImages"

Division and the Remainder Theorem We’ll first look at what happens when we divide numbers. e.g. Quotient This can be written as Remainder 3 is called the quotient 1 is the remainder Algebraic expressions can be divided in a similar way Division

Division and the Remainder Theorem Quotient Remainder e.g. 1Find the quotient and remainder when is divided by Solution: e.g. 2Write in the form Solution:

Division and the Remainder Theorem Exercises Find the quotient and remainder when is divided by x 1. Solution: The quotient is and the remainder is Write in the form

Division and the Remainder Theorem Dividing by an expression of the form x - a can be done in 2 ways Solution: Long division. Divide the 1 st term of the numerator by the 1 st term of the denominator. Write the division as follows: e.g.1 Divide by

Division and the Remainder Theorem Write the division as follows: Write this answer above the polynomial being divided. e.g.1 Divide by Dividing by an expression of the form x - a can be done in 2 ways Solution: Long division.

Division and the Remainder Theorem Write the division as follows: e.g.1 Divide by Dividing by an expression of the form x - a can be done in 2 ways Multiply x – 1 by this number... and write the answer below Subtract: 6 – (– 2) = 8 8 The quotient is 2 and the remainder is 8. So, Solution: Long division.

Division and the Remainder Theorem –2x 3 x4x4 x3x3 – 2x 2 – 8x+ 1 –2x 3 –8x 2 x x Ex.2 Long Division – 2x 2 + 6x x + 24x – 3 – 3x 3 x into x 4 goes x 3 –5x 3 – –3X 3 = –2x 3 x into –2x 3 goes –2x 2 –2x 2 –6X 2 = –8x 2 x into –8x 2 goes –8x No remainder so x–3 is a factor

Division and the Remainder Theorem Divide by x Solution: The quotient is x + 2 and the remainder is The solution is on the next slide Exercises2. Divide by

Division and the Remainder Theorem Solution: The quotient is and the remainder is

Division and the Remainder Theorem Finding the remainder without doing long division The quotient is and the remainder is But f(–1) = (–1)  (–1) 2 – 4  (– 1) + 1 = 7 So if dividing by x+1 substitute x = –1 to find the remainder i.e find f(–1) If dividing by (x – 2) the remainder is given by f(2) If dividing by (x + 3) the remainder is given by f(–3)

Division and the Remainder Theorem e.g. Find the remainder when is divided by x - 1 The remainder theorem gives the remainder when a polynomial is divided by a linear factor It doesn’t enable us to find the quotient The remainder is 4 The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by

Division and the Remainder Theorem Proof of the Remainder theorem Let be a polynomial that is divided by x - a The quotient is another polynomial and the remainder is a constant. We can write Multiplying by x – a gives So,

Division and the Remainder Theorem e.g.1 Find the remainder when is divided by Solution: Let So, Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too!

Division and the Remainder Theorem e.g.2 Find the remainder when is divided by Solution: Let To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a. so,

Division and the Remainder TheoremExercises Find the remainder when is divided by x Solution: Let 2. If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b.

Division and the Remainder TheoremExercises 2. If x + 2 is a factor of and if the remainder on division by x – 1 is – 3, find the values of a and b. Solution: Let (2) (1) (1) + (2) Substitute in (2)

Division and the Remainder Theorem

The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied. For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

Division and the Remainder Theorem Dividing by an expression of the form x - a can be done in 2 ways: Method 1: Long Division Method 2: Inspection Method 2 is usually easier but an example of method 1 is given next.

Division and the Remainder Theorem Write the division as follows: e.g.1 Divide by Multiply x – 1 by this number... and write the answer below Subtract: 6 – (– 2) = 8 8 The quotient is 2 and the remainder is 8. So, Solution: Method 1 Long division. Divide the 1 st term of the numerator by the 1 st term of the denominator. Write this answer above the polynomial being divided.

Division and the Remainder Theorem e.g.1 Divide by Write Adjust the constant term Separate the 2 terms: The quotient is 2 and the remainder is 8 Solution: Method 2 ( Inspection ) Copy the denominator onto the top line Divide the 1 st term of the numerator Multiply by the 1 st term of the denominator so the 1st term at the top is now correct

Division and the Remainder Theorem The remainder theorem says that if we divide a polynomial by x – a, the remainder is given by Proof of the Remainder theorem Let be a polynomial that is divided by x - a The quotient is another polynomial and the remainder is a constant. We can write Multiplying by x – a gives So,

Division and the Remainder Theorem Solution: Let So, Tip: Use the remainder theorem to check the remainder when using long division. If the remainder is correct the quotient will be too! e.g.1 Find the remainder when is divided by

Division and the Remainder Theorem e.g.2 Find the remainder when is divided by Solution: Let To find the value of a, we let 2x + 1 = 0. The value of x gives the value of a. so,