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Algebra and Functions

Algebra and Functions. 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. Teachings for Exercise 1A. Algebra and Functions.

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Algebra and Functions

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  1. Algebra and Functions

  2. 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

  3. Teachings for Exercise 1A

  4. 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  You can therefore ‘cancel’ an x from each part Cancel x’s Don’t need to divide by 1! 1A

  5. 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 1A

  6. 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! 1A

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

  8. 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 can cancel out brackets as before Two numbers that multiply to give +12 and add to give +7 Cancel the (x + 3)s 1A

  9. 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 can 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 1A

  10. 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 can 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 1A

  11. Teachings for Exercise 1B

  12. 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 1 1 7 • First, ‘How many 700s in 800’ • 1 • We now take away ‘1 x 700’ from what we started with • Second, ‘How many 70s in 119’ • 1 • We now take away ‘1 x 70’ from what we had (119) • Finally, ‘How many 7s in 49’ • 7 • We now take away ‘7 x 7’ from what we had (49) 7 8 1 9 7 0 0 1 1 9 7 0 4 9 4 9 0 1B

  13. 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 9746 by 9  So 9 divides into 9746 exactly 1082 times, with 8 remainder 1 0 8 2 9 9 7 4 6 9 0 0 0 7 4 6 • Finally, ‘How many 9s in 26’ • 2 • We now take away ‘2 x 9’ from what we had left (26) • First, ‘How many 9000s in 9000’ • 1 • We now take away ‘1 x 9000’ from what we started with • Second, ‘How many 900s in 746’ • 0 • We now take away ‘0 x 900’ from what we had left (746) • Third, ‘How many 90s in 746’ • 8 • We now take away ‘8 x 90’ from what we had left (746) 0 7 4 6 7 2 0 2 6 1 8 8 1B

  14. Algebra and Functions x2 + 5x - 2 You can divide a polynomial (an equation with a power of x in) by (x ± p) We are now going to look at some algebraic examples.. 1) Divide x3 + 2x2 – 17x + 6 by (x – 3) So the answer is x2 + 5x – 2, and there is no remainder This means that (x – 3) is a factor of the original equation x - 3 x3 + 2x2 – 17x + 6 x3 – 3x2 5x2 - 17x + 6 5x2 - 15x • Third, Divide -2x by x • -2 • We then subtract -2(x – 3) from what we have left • Second, Divide 5x2 by x • 5x • We then subtract 5x(x – 3) from what we have left • First, Divide x3 by x • x2 • We then subtract x2(x – 3) from what we started with - 2x + 6 - 2x + 6 0 Therefore: 1B

  15. Algebra and Functions 6x2 - 2x + 3 You can divide a polynomial (an equation with a power of x in) by (x ± p) We are now going to look at some algebraic examples.. 1) Divide 6x3 + 28x2 – 7x + 15 by (x + 5) So the answer is 6x2 - 2x + 3, and there is no remainder This means that (x + 5) is a factor of the original equation x + 5 6x3 + 28x2 – 7x + 15 6x3 + 30x2 -2x2 - 7x + 15 -2x2 - 10x • Second, Divide -2x2 by x • -2x • We then subtract -2x(x + 5) from what we have left 3x + 15 • First, Divide 6x3 by x • 6x2 • We then subtract 6x2(x + 5) from what we started with • Third, Divide 3x by x • 3 • We then subtract 3(x + 5) from what we have left 3x + 15 0 Therefore: 1B

  16. Teachings for Exercise 1C

  17. Algebra and Functions Some things to be aware of when dividing algebraically… Always include all different powers of x, up to the highest that you have… Divide x3 – 3x – 2 by (x – 2)  You must include ‘0x2’ in the division…  So our answer is ‘x2 + 2x + 1. This is commonly known as the quotient x2 + 2x + 1 x - 2 x3 + 0x2 – 3x - 2 x3 – 2x2 2x2 – 3x - 2 Second, divide 2x2 by x = 2x Then, work out 2x(x – 2) and subtract from what you have left First, divide x3 by x = x2 Then, work out x2(x – 2) and subtract from what you started with Third, divide x by x = 1 Then, work out 1(x – 2) and subtract from what you have left 2x2 – 4x x – 2 x – 2 0 1C

  18. Algebra and Functions Some things to be aware of when dividing algebraically… Always include all different powers of x, up to the highest that you have… Divide 3x3 – 3x2 – 4x + 4 by (x – 1)  So our answer is ‘3x2 – 4’ 3x2 - 4 x - 1 3x3 - 3x2 – 4x + 4 3x3 – 3x2 – 4x + 4 - 4x + 4 Second, divide -4x by x = -4 Then, work out -4(x – 1) and subtract from what you have left First, divide 3x3 by x = 3x2 Then, work out 3x2(x – 1) and subtract from what you started with 0 1C

  19. 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; 2x3 – 5x2 – 16x + 10 is divided by (x – 4)  So the remainder is -6. 2x2 + 3x - 4 x - 4 2x3 - 5x2 – 16x + 10 2x3 – 8x2 3x2 – 16x + 10 Second, divide 3x2 by x = 3x Then, work out 3x(x – 4) and subtract from what you have left First, divide 2x3 by x = 2x2 Then, work out 2x2(x – 4) and subtract from what you started with Third, divide -4x by x = -4 Then, work out -4(x – 4) and subtract from what you have left 3x2 – 12x -4x + 10 -4x + 16 -6 1C

  20. Teachings for Exercise 1D

  21. Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) f(x)  A function of x, any equation f(p)  The function of x with a value p substituted in For example; Show that (x – 2) is a factor of x3 + x2 – 4x - 4 x3 + x2 – 4x - 4 Substitute in x = 2 23 + 22 – (4x2) - 4 Work out each term 8 + 4 – 8 - 4 = 0 So because f(2) = 0, (x – 2) is a factor of the original equation 1D

  22. Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) f(x)  A function of x, any equation f(p)  The function of x with a value p substituted in For example; Factorise 2x3 + x2 – 18x - 9 2x3 + x2 – 18x - 9 Substitute in values of x to find a factor x = 1 2 + 1 – 18 - 9 = -24 x = 2 16 + 4 – 36 - 9 = -25 x = 3 54 + 9 – 54 - 9 So (x – 3) is a factor = 0 1D

  23. Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) f(x)  A function of x, any equation f(p)  The function of x with a value p substituted in For example; Factorise 2x3 + x2 – 18x – 9  Now we know (x – 3) is a factor, divide by it to find the quotient The quotient is 2x2 + 7x + 3 2x2 + 7x + 3 x - 3 2x3 + x2 – 18x - 9 2x3 – 6x2 7x2 – 18x - 9 Second, divide 7x2 by x = 7x Then, work out 7x(x – 3) and subtract from what you have left First, divide 2x3 by x = 2x2 Then, work out 2x2(x – 3) and subtract from what you started with Third, divide 3x by x = 3 Then, work out 3(x – 3) and subtract from what you have left 7x2 – 21x 3x - 9 3x - 9 0 1D

  24. Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) f(x)  A function of x, any equation f(p)  The function of x with a value p substituted in For example; Factorise 2x3 + x2 – 18x – 9  (x – 3) is a factor  (2x2 + 7x + 3) is the quotient (x – 3)(2x2 + 7x + 3) You can also factorise the quotient 2 numbers that multiply to give +3, and add to give +7 when one has doubled… (x – 3)(2x + 1)(x + 3) 1D

  25. Algebra and Functions If f(p) = 0, then (x – p) is a factor of f(x) Given that (x + 1) is a factor of 4x4 – 3x2 + a, find the value of a. 4x4 – 3x2 + a If (x + 1) is a factor, then using -1 will make the equation = 0 0 = 4(-14) – 3(-12) + a Work out each term 0 = 4 – 3 + a Solve the equation to find the value of a 0 = 1 + a -1 = a 1D

  26. Teachings for Exercise 1E

  27. 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) (2x – 1) (3x + 2) (5x + 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 1E

  28. Algebra and Functions The remainder when f(x) is divided by (ax - b) will be given by f(b/a) Find the remainder when x3 – 20x + 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) x3 – 20x + 3 Substitute in x = 4 43 – 20(4) + 3 Work out each term 64 – 80 + 3 = -13 1E

  29. Algebra and Functions The remainder when f(x) is divided by (ax - b) will be given by f(b/a) When 8x4 – 4x3 + ax2 – 1 is divided by (2x + 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 8x4 – 4x3 + ax2 - 1 Substitute in x = -1/2 8(-1/2)4 – 4(-1/2)3 + a(-1/2)2 - 1 = 3 Work out each term 8(1/16) – 4(-1/8) + a(1/4) - 1 = 3 1/2 – -1/2 + 1/4a - 1 = 3 Group like terms 1/4a = 3 Multiply by 4 a = 12 1E

  30. 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…

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