Algebra and Functions This chapter focuses on Algebra

Algebra and Functions • This chapter focuses on Algebra and Algebraic manipulation • We will look at Algebraic Division • We will also learn the Remainder and Factor theorems

Algebra and Functions You can simplify algebraic fractions by division Sometimes you need to look for common factors to each term In this case, every term, top and bottom, contains an x Cancel x’s You can therefore ‘cancel’ an x from each part Don’t need to divide by 1! 1 A

Algebra and Functions You can simplify algebraic fractions by division Sometimes you will have to break up the fraction You may find that on one part, the letters can simplify, and on another part, it is the numbers that simplify… Split the fraction apart Simplify whatever you can 1 A

Algebra and Functions You can simplify algebraic fractions by division Sometimes you will have to break up the fraction You can always simplify in stages to make it easier to follow! Split the fraction apart Simplify algebra (3/-3 = -1) (4/-3 = -4/3) The negatives cancel out! 1 A

Algebra and Functions You may also need to use factorisation in the simplifying process This equation has already been put into brackets You cancel out brackets which are on the top and bottom Cancel the (2 x – 1)s Don’t need to divide by 1! 1 A

Algebra and Functions You may also need to use factorisation in the simplifying process Sometimes you will have to factorise one of the equations first Once this is done, you cancel out brackets as before Two numbers that multiply to give +12 and add to give +7 Cancel the (x + 3)s 1 A

Algebra and Functions You may also need to use factorisation in the simplifying process Sometimes you will have to factorise both of the equations first Once this is done, you cancel out brackets as before Two numbers that multiply to give +5 and add to give +6 Two numbers that multiply to give -10 and add to give +3 Cancel the (x + 5)s 1 A

Algebra and Functions You may also need to use factorisation in the simplifying process Sometimes you will have to factorise one of the equations first Once this is done, you cancel out brackets as before Two numbers that multiply to give +12 and add to give +11, when one is doubled Cancel the (x + 4)s 1 A

Algebra and Functions You can divide a polynomial (an equation with a power of x in) by (x ± p) First though, we will look at numerical long division, and what the process actually means… 1) Divide 819 by 7 So 7 divides into 819 exactly 117 times, with no remainder Finally, First, ‘How Second, ‘How many 7 s 700 s 70 s inin 49’ in 119’ 800’ 7 1 We now take away ‘ 7 ‘ 1 xx 70’ 700’ 7’ from we what had we(49) (119) started with 7 1 8 7 1 1 7 1 9 00 1 9 70 49 49 0 1 B

Algebra and Functions You can divide a polynomial (an equation with a power of x in) by (x ± p) 9 First though, we will look at numerical long division, and what the process actually means… Finally, First, ‘How many Second, ‘How Third, ‘How many 9000 s 9 s inin 9000’ 900 s in 746’ 90 s in 26’ 1) Divide 9746 by 9 2 18 0 So 9 divides into 9746 exactly 1082 times, with 8 remainder We now take away ‘ 1 x 9000’ ‘ 2 from what ‘ 0 900’ ‘ 8 x 9’ 90’ from what we started we had with left from what we had (26) left (746) 1 0 82 97 46 90 00 746 720 26 1 8 8 1 B

Algebra and Functions x 2 + 5 x - 2 You can divide a polynomial (an equation with a power of x in) by (x ± p) x-3 x 3 + 2 x 2 – 17 x + 6 x 3 – 3 x 2 We are now going to look at some algebraic examples. . 1) Divide x 3 + 2 x 2 – 17 x + 6 by (x – 3) So the answer is x 2 + 5 x – 2, and there is no remainder This means that (x – 3) is a factor of the original equation Therefore: 5 x 2 - 17 x + 6 3 5 x 2 xby Third, Divide First, Divide Second, Divide x-2 x byby xx x 2 -2 5 x 5 x 2 - 15 x - 2 x + 6 0 We then subtract x 2 We We (x –then 3) from subtract what 2(x – – 3)3)from what we started 5 x(x from with whatwe have left we have 1 B

Algebra and Functions 6 x 2 - 2 x You can divide a polynomial (an equation with a power of x in) by (x ± p) x+5 6 x 3 + 28 x 2 – 7 x + 15 6 x 3 + 30 x 2 We are now going to look at some algebraic examples. . 1) Divide by (x + 5) 6 x 3 + 28 x 2 – 7 x + 15 So the answer is 6 x 2 - 2 x + 3, and there is no remainder This means that (x + 5) is a factor of the original equation Therefore: + 3 -2 x 2 - 7 x + 15 Second, Divide -2 x 2 First, Divide 6 x byxx Third, 3 x 3 by by x -2 x 36 x 2 -2 x 2 - 10 x 3 x + 15 0 We then subtract Wethensubtract. We 6 x 2(x + 5) from what 3(x ++ 5) from what 2 x(x 5) from what wehave started we have leftwith we left 1 B

Algebra and Functions Some things to be aware of when dividing algebraically… x 2 + 2 x + 1 x-2 Always include all different powers of x, up to the highest that you have… Divide x 3 – 3 x – 2 by (x – 2) x 3 + 0 x 2 – 3 x - 2 x 3 – 2 x 2 – 3 x - 2 First, divide Second, Third, divide xx 3 2 x by by 2 xxby x 2 x 2 – 4 x x – 2 You must include ‘ 0 x 2’ in the division… = 12 x x 2 So our answer is ‘x 2 + 2 x + 1. This is commonly known as the quotient Then, work out 1(x x 2(x– – 2) 2 x(x – 2) 2) and subtract from whathave you started you left have with left x – 2 0 1 C

Algebra and Functions Some things to be aware of when dividing algebraically… 3 x 2 x-1 Divide (x – 1) – 3 x 2 – 4 x + 4 by So our answer is ‘ 3 x 2 – 4’ 3 x 3 - 3 x 2 – 4 x + 4 3 x 3 – 3 x 2 Always include all different powers of x, up to the highest that you have… 3 x 3 - 4 – 4 x + 4 3 by Second, First, divide 3 x-4 x byx x - 4 x + 4 0 ==-4 3 x 2 Then, workout out-4(x 3 x 2(x – 1) – and 1) and subtract from what you what have youleft started with 1 C

Algebra and Functions Some things to be aware of when dividing algebraically… Sometimes you will have a remainder, in which case the expression you divided by is not a factor of the original equation… Find the remainder when; 2 x 3 – 5 x 2 – 16 x + 10 is divided by (x – 4) So the remainder is -6. 2 x 2 + 3 x - 4 x-4 2 x 3 - 5 x 2 – 16 x + 10 2 x 3 – 8 x 2 3 x 2 – 16 x + 10 3 by 2 by First, divide Second, Third, divide 2 x -4 x 3 x by xx x 3 x 2 – 12 x -4 x + 10 = -4 2 x 2 3 x -4 x + 16 -6 Then, work out -4(x 2 x 2(x––– 4) 3 x(x 4) and subtract from whathave you left started have leftwith 1 C

Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) A function of x, any equation f(p) The function of x with a value p substituted in x 3 + x 2 – 4 x - 4 Substitute in x = 2 23 + 22 – (4 x 2) - 4 Work out each term 8 + 4 – 8 - 4 For example; Show that (x – 2) is a factor of x 3 + x 2 – 4 x - 4 =0 So because f(2) = 0, (x – 2) is a factor of the original equation 1 D

Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) A function of x, any equation 2 x 3 + x 2 – 18 x - 9 Substitute in values of x to find a factor f(p) The function of x with a value p substituted in x=1 For example; x=2 = -24 16 + 4 – 36 - 9 = -25 Factorise 2 x 3 + x 2 – 18 x - 9 x=3 So (x – 3) is a factor 2 + 1 – 18 - 9 54 + 9 – 54 - 9 =0 1 D

Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) 2 x 2 + 7 x + 3 x-3 f(x) A function of x, any equation f(p) The function of x with a value p substituted in Factorise 2 x 3 – 6 x 2 7 x 2 – 18 x - 9 3 by 2 by First, divide Second, Third, divide 2 x 3 x 7 x by xx x 7 x 2 – 21 x 3 x - 9 For example; 2 x 3 + x 2 – 18 x - 9 + x 2 – 18 x – 9 Now we know (x – 3) is a factor, divide by it to find the quotient The quotient is 2 x 2 + 7 x + 3 =3 2 x 2 7 x 3 x - 9 0 Then, work out 3(x 2 x 2(x 7 x(x – – 3) and subtract from whathave you left started have leftwith 1 D

Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) (x – 3)(2 x 2 + 7 x + 3) f(x) A function of x, any equation You can also factorise the quotient f(p) The function of x with a value p substituted in 2 numbers that multiply to give +3, and add to give +7 when one has doubled… For example; Factorise 2 x 3 + x 2 – 18 x – 9 (x – 3)(2 x + 1)(x + 3) (x – 3) is a factor (2 x 2 + 7 x + 3) is the quotient 1 D

Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) If (x + 1) is a factor, Given that (x + 1) is a factor of 4 x 4 – 3 x 2 + a, find the value of a. then using -1 will make the equation = 0 4 x 4 – 3 x 2 + a 0 = 4(-14) – 3(-12) + a Work out each term Solve the equation to find the value of a 0 = 4 – 3 + a 0 = 1 + a -1 = a 1 D

Algebra and Functions The remainder when f(x) is divided by (ax - b) will be given by f(b/a) For example, to find the remainder when f(x) is divided by each of the following… (x – 4) (x + 2) (2 x – 1) (3 x + 2) (5 x + 6) Substitute x = 4 into the equation Substitute x = -2 into the equation Substitute x = 1/2 into the equation Substitute x = -2/3 into the equation Substitute x = -6/5 into the equation 1 E

Algebra and Functions The remainder when f(x) is divided by (ax - b) will be given by f(b/a) Find the remainder when x 3 – 20 x + 3 is divided by (x – 4) The remainder will be -13 (You can have a negative remainder, as we do not know what the actual numbers are) x 3 – 20 x + 3 Substitute in x = 4 Work out each term 43 – 20(4) + 3 64 – 80 + 3 = -13 1 E

Algebra and Functions The remainder when f(x) is divided by (ax - b) will be given by f(b/a) 8 x 4 4 x 3 ax 2 When – + – 1 is divided by (2 x + 1), the remainder is 3. Find the value of a. We will substitute in x = -1/ 2 We will set the equation equal to 3 Substitute in x = -1/2 8 x 4 – 4 x 3 + ax 2 - 1 8(-1/2)4 – 4(-1/2)3 + a(-1/2)2 - 1 = 3 8(1/16) – 4(-1/8) + a(1/4) - 1 = 3 Work out each term 1/ 2 – -1/ 2 Group like terms + 1 / 4 a - 1 1/ Multiply by 4 4 a a = 3 = 12 1 E

Summary • We have learnt how to divide algebraically, having seen how it is done numerically • We have looked at the factor theorem and used it in solving equations beyond quadratics…