Synthetic and Remainder Theorem of Polynomials.ppt
Main
1. Core 1
Core 1
Exam January 2013
H Mort
April 4, 2013
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2. Core 1
Factors, remainders and cubic graphs
Factorisation
Expanding (x − 2)(x + 5) gives x2 + 3x − 10. The reverse of this
process is called factorisation.
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6. Core 1
Factors, remainders and cubic graphs
The factor theorem
Consider (x − 1)(x − 2)(x + 4).
Expand to give
P (x) = x3 + x2 − 10x + 8
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7. Core 1
Factors, remainders and cubic graphs
The factor theorem
Consider (x − 1)(x − 2)(x + 4).
Expand to give
P (x) = x3 + x2 − 10x + 8
To factorise this polynomial, find values of x that make P (x) = 0.
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8. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
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9. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
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10. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
This shows that (x − 1) is a factor of P (x).
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11. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
This shows that (x − 1) is a factor of P (x).
Test to see if (x − 2) is a factor
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12. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
This shows that (x − 1) is a factor of P (x).
Test to see if (x − 2) is a factor
P (2) = 23 + 22 − 10 × 2 + 8 = 0
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13. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
This shows that (x − 1) is a factor of P (x).
Test to see if (x − 2) is a factor
P (2) = 23 + 22 − 10 × 2 + 8 = 0
Show that (x + 4) is a factor
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14. Core 1
Factors, remainders and cubic graphs
The factor theorem
To show that (x − 1) is a factor of P (x) = x3 + x2 − 10x + 8 test
P (1):
P (1) = 13 + 12 − 10 × 1 + 8 = 0
This shows that (x − 1) is a factor of P (x).
Test to see if (x − 2) is a factor
P (2) = 23 + 22 − 10 × 2 + 8 = 0
Show that (x + 4) is a factor
P (−4) = (−4)3 + (−4)2 − 10 × (−4) + 8 = 0
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15. Core 1
Factors, remainders and cubic graphs
The factor theorem
The factor theorem states that (x − a) is a factor of the
polynomial
P (x) ⇐⇒ P (a) = 0
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16. Core 1
Factors, remainders and cubic graphs
The factor theorem
Have a go at the following question, then check your answers in
worked example 6.1 on page 81
Use the factor theorem to show that (x − 2) is a factor of
x3 + x2 − 7x + 2
Have a go at the following question, then check your answers in
worked example 6.2 on page 82
Substitute x = 2, x = 3, x = 1 and x = −1 into the polynomial
P (x) = x3 − 4x2 + x + 6 and hence write down the three linear
factors of P (x).
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17. Core 1
Factors, remainders and cubic graphs
The factor theorem
Exercise
Complete exercise 6A page 83. Questions 1 to 3.
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18. Core 1
Factors, remainders and cubic graphs
The factor theorem
Example
The polynomial P (x) = x3 + qx2 + 11x + 6 has a factor of
(x + 3). Find q.
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19. Core 1
Factors, remainders and cubic graphs
The factor theorem
Example
The polynomial P (x) = x3 + qx2 + 11x + 6 has a factor of
(x + 3). Find q.
Solution
Given that (x + 3) is a factor means that P (−3) = 0
P (−3) = (−3)3 + q × (−3)2 + 11 × (−3) + 6 = 0
−27 + 9q − 33 + 6 = 0
9q = 54
q=9
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20. Core 1
Factors, remainders and cubic graphs
The factor theorem
Have a go at the following questions, then check your answers in
worked example 6.4 on page 82
The polynomial x3 + ax + bx − 20 has factors x + 2 and x − 5.
Find the values of a and b.
Hint: You will need to solve a pair of simultaneous equations.
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22. Core 1
Factors, remainders and cubic graphs
Further factorisation
Further factorisation
Have a go at the following questions, then check your answers in
worked example 6.5 on page 84
Use the factor theorem to show that (x + 3) is a factor of
P (x) = x3 − 6x + 9. Hence factorise P (x) such that it has one
linear and one quadratic factor.
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23. Core 1
Factors, remainders and cubic graphs
Further factorisation
Example
P (x) = x3 + 3x2 − 6x − 8
• Use the factor theorem to show that (x + 4) is a factor.
• Given that P (x) has two other linear factors with integer
coefficients, factorise P (x) completely
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24. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
If (x + 4) is a factor then
P (x) = (x + 4)(x − a)(x − b)
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25. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
If (x + 4) is a factor then
P (x) = (x + 4)(x − a)(x − b)
The constant term is given by 4 × (−a) × (−b), so
4ab = −8 =⇒ ab = −2
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26. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
If (x + 4) is a factor then
P (x) = (x + 4)(x − a)(x − b)
The constant term is given by 4 × (−a) × (−b), so
4ab = −8 =⇒ ab = −2
The options of a and b are limited to the factors of -2, i.e. ±1 and
±2.
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27. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
If (x + 4) is a factor then
P (x) = (x + 4)(x − a)(x − b)
The constant term is given by 4 × (−a) × (−b), so
4ab = −8 =⇒ ab = −2
The options of a and b are limited to the factors of -2, i.e. ±1 and
±2.
P (1) = −10 =⇒ (x − 1) is not a factor
P (−1) = 0 =⇒ (x + 1) is a factor
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28. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
If (x + 4) is a factor then
P (x) = (x + 4)(x − a)(x − b)
The constant term is given by 4 × (−a) × (−b), so
4ab = −8 =⇒ ab = −2
The options of a and b are limited to the factors of -2, i.e. ±1 and
±2.
P (1) = −10 =⇒ (x − 1) is not a factor
P (−1) = 0 =⇒ (x + 1) is a factor
Therefore a = −1. Using the fact ab = −2 =⇒ b = 2. So
P (x) = (x − 4)(x + 1)(x − 2).
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29. Core 1
Factors, remainders and cubic graphs
Further factorisation
When factorising a cubic function, once a linear factor is found
then the remaining quadratic factor can be found by comparing
coefficients (like Chapter 5).
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30. Core 1
Factors, remainders and cubic graphs
Further factorisation
When factorising a cubic function, once a linear factor is found
then the remaining quadratic factor can be found by comparing
coefficients (like Chapter 5).
Example
A polynomial is given by P (x) = x3 − 4x2 − 5x + 24.
• Show that (x − 3) is a factor
• Find the exact solutions of the equation P (x) = 0
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31. Core 1
Factors, remainders and cubic graphs
Further factorisation
When factorising a cubic function, once a linear factor is found
then the remaining quadratic factor can be found by comparing
coefficients (like Chapter 5).
Example
A polynomial is given by P (x) = x3 − 4x2 − 5x + 24.
• Show that (x − 3) is a factor
• Find the exact solutions of the equation P (x) = 0
Recall the quadratic formula
√
−b ± b2 − 4ac
x=
2a
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32. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
See 6.7 page 85. Factorise to P (x) = √ − 3)(x2 − x − 8), then
(x
1 ± 33
use the quadratic formula to give for the other 2 roots.
2
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33. Core 1
Factors, remainders and cubic graphs
Further factorisation
Have a go at the following question, then check your answers in
worked example 6.8 on page 85
Factorise the polynomial P (x) = x3 + 5x2 − 2x − 24 completely.
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34. Core 1
Factors, remainders and cubic graphs
Further factorisation
Exercise
Complete exercise 6B page 86. Questions 1 to 6.
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35. Core 1
Factors, remainders and cubic graphs
Further factorisation
Example
1 Factorise the polynomial P (x) = x3 + 6x2 + 12x + 8
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36. Core 1
Factors, remainders and cubic graphs
Further factorisation
Example
1 Factorise the polynomial P (x) = x3 + 6x2 + 12x + 8
2 Hence solve the equation y 6 + 6y 4 + 12y 2 + 8
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37. Core 1
Factors, remainders and cubic graphs
Further factorisation
Solution
1 P (x) = (x + 2)3 .
2 Either look the x term and notice that x = y 2 or solve
x3 = y 6 =⇒ x = y 2 . Then substitute to give
(y 2 + 2)3 = 27 =⇒ y 2 + 2 = 3 =⇒ y = ±1.
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39. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Graphs of cubic functions
Using values of x between −4 and 4:
• Plot the function f (x) = x3
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40. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Graphs of cubic functions
Using values of x between −4 and 4:
• Plot the function f (x) = x3
• Plot the function f (x) = −x3
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41. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Graphs of cubic functions
Using values of x between −4 and 4:
• Plot the function f (x) = x3
• Plot the function f (x) = −x3
• Describe what you notice about the 2 graphs
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42. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
x3
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43. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
x3 x3
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44. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
x3 x3
Complete transformations on these graphs in a similar way to how
quadratics are transformed.
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45. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Discuss the features of cubics:
• Always cross the x-axis at at least one point
• Always cross the y-axis at one point
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46. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
To sketch a graph of a cubic function
1 Shape: coefficient of x3 is positive / or negative
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47. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
To sketch a graph of a cubic function
1 Shape: coefficient of x3 is positive / or negative
2 Let x = 0 ⇒ y-intercept
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48. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
To sketch a graph of a cubic function
1 Shape: coefficient of x3 is positive / or negative
2 Let x = 0 ⇒ y-intercept
3 Let y = 0 ⇒ roots of the equation
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49. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
To sketch a graph of a cubic function
1 Shape: coefficient of x3 is positive / or negative
2 Let x = 0 ⇒ y-intercept
3 Let y = 0 ⇒ roots of the equation
4 Look for turning points near the y-axis.
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50. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Example
Sketch the graph of the cubic f (x) = (2x + 4)(x + 1)(2 − x)
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51. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
The worked examples cover 1 root, i.e. the quadratic has no real
roots and repeated roots.
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52. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Have a go at the following question, then check your answers in
worked examples 6.10 on page 89.
6.10
Sketch the graph of y = (2x + 3)(x + 4)(5 − 2x).
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53. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Have a go at the following question, then check your answers in
worked examples 6.10 on page 89.
6.10
Sketch the graph of y = (2x + 3)(x + 4)(5 − 2x).
Example
Sketch the graph of y = (x + 1)(x2 − x + 1)
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54. Core 1
Factors, remainders and cubic graphs
Graphs of cubic functions
Have a go at the following question, then check your answers in
worked examples 6.10 on page 89.
6.10
Sketch the graph of y = (2x + 3)(x + 4)(5 − 2x).
Example
Sketch the graph of y = (x + 1)(x2 − x + 1)
Example
Sketch the graph of y = (2x + 3)(x − 1)2
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56. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Dividing a polynomial by a linear expression
12
=
7
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57. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Dividing a polynomial by a linear expression
12 7+5
= =
7 7
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58. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Dividing a polynomial by a linear expression
12 7+5 7 5
= = + =
7 7 7 7
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59. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Dividing a polynomial by a linear expression
12 7+5 7 5 5
= = + =1
7 7 7 7 7
Solve algebraic fractions in a similar way.
Substituting x = 7 above gives
x+5 x 5 5
= + =1
x x x x
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60. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Dividing a polynomial by a linear expression
12 7+5 7 5 5
= = + =1
7 7 7 7 7
Solve algebraic fractions in a similar way.
Substituting x = 7 above gives
x+5 x 5 5
= + =1
x x x x
If (x + 3)(x − 2) = x2 + x − 6 then rearranging gives
x2 + x − 6
=x+3
x−2
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61. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Remember
4+3
=3
4
but
4(4 + 3)
¡
=7
4
¡
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62. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
Divide x by x − 3
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63. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
Divide x by x − 3
Solution
x
=
x−3
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64. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
Divide x by x − 3
Solution
x x−3+3 x−3 3 3
= = + =1+
x−3 x−3 x−3 x−3 x−3
Example
Divide x + 2 by x + 5
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65. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
Divide x by x − 3
Solution
x x−3+3 x−3 3 3
= = + =1+
x−3 x−3 x−3 x−3 x−3
Example
Divide x + 2 by x + 5
Solution
x+5−3 3
=1−
x+5 x+5
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66. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
To divide x2 + 3x + 1 by x + 2:
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67. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
To divide x2 + 3x + 1 by x + 2:
Write
x2 + 3x + 1 = (x + 2)(x + q) + r
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68. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
To divide x2 + 3x + 1 by x + 2:
Write
x2 + 3x + 1 = (x + 2)(x + q) + r
Expand the right-hand side and collect like terms:
(x + 2)(x + q) + r = x2 + qx + 2x + 2q + r
= x2 + (2 + q)x + 2q + r
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69. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
To divide x2 + 3x + 1 by x + 2:
Write
x2 + 3x + 1 = (x + 2)(x + q) + r
Expand the right-hand side and collect like terms:
(x + 2)(x + q) + r = x2 + qx + 2x + 2q + r
= x2 + (2 + q)x + 2q + r
Compare coefficients:
• 2+q =3⇒q =1
• 2q + r = 1 ⇒ r = −1
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70. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
To divide x2 + 3x + 1 by x + 2:
Write
x2 + 3x + 1 = (x + 2)(x + q) + r
Expand the right-hand side and collect like terms:
(x + 2)(x + q) + r = x2 + qx + 2x + 2q + r
= x2 + (2 + q)x + 2q + r
Compare coefficients:
• 2+q =3⇒q =1
• 2q + r = 1 ⇒ r = −1
so
x2 + 3x + 1 1
= (x + 1) −
x+2 x+2
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71. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
In general
polynomial = divisor × quotient + remainder
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72. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
In general
polynomial = divisor × quotient + remainder
Have a go at the following question, then check your answer in the
worked example on page 92.
6.14, page 92
Divide x2 + 3x + 1 by x + 2
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73. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Exercise
Complete exercise 6D page 93 question 1, right hand column.
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74. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
x3 + x2 − x + 2
Solve
x−1
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75. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Solution
Key points of solution:
• (x − 1)(x2 + px + q) + r = x3 + (p − 1)x2 + (q − p)x − q + r
• p−1=1⇒p=2
• q − p = −1 ⇒ q − 2 = −1 ⇒ q = 1
• −q + r = 2 ⇒ r = 3
x3 + x2 − x + 2 = (x − 1)(x2 + 2x + 1) + 3
so
x3 + x2 − x + 2 3
= x2 + 2x + 1 +
x−1 x−1
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76. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Example
x3 − 3x2 − 6x + 10
Solve
x+2
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77. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Solution
Key points of solution:
• (x + 2)(x2 + px + q) + r = x3 + (p + 2)x2 + (2p + q)x + 2q + r
• p + 2 = −3 ⇒ p = −5
• 2p + q = −6 ⇒ −10 + q = −6 ⇒ q = 4
• 2q + r = 10 ⇒ 8 + r = 10 ⇒ r = 2
x3 − 3x2 − 6x + 10 = (x + 2)(x2 − 5x + 4) + 2
so
x3 − 3x2 − 6x + 10 2
= x2 − 5x + 4 +
x+2 x+2
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78. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
In the exam you will only get one of the following:
x+b x2 + bx + c x3 + bx2 + cx + d
x−a x−a x−a
where a, b, c and d are integers.
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79. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Have a go at the following questions, then check your answers in
the worked examples on page 92 and 93.
6.15, page 92
Divide x3 + 2x2 − x + 3 by x − 2
6.16, page 93
Divide x3 + 2x2 − 1 by x − 1
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80. Core 1
Factors, remainders and cubic graphs
Dividing a polynomial by a linear exression
Exercise
Complete exercise 6D page 93, question 2 - alternate questions.
Then the rest of the exercise.
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82. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Example
P (x) = x2 + x − 3.
• Find the quotient and the remainder when P (x) is divided by
(x − 2).
• Find the value of P (2) and comment on the solution
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83. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Example
P (x) = x2 + x − 3.
• Find the quotient and the remainder when P (x) is divided by
(x − 2).
• Find the value of P (2) and comment on the solution
Solution
• P (x) = (x − 2)(x + 3) + 3
• P (2) = 3
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85. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Any polynomial can be written in the form
P (x) = (x − a)Q(x) + R
where
• P (x) is a polynomial, degree = n.
• (x − a) is the divisor, degree = 1.
• Q(x) is the quotient, degree = n − 1.
• R is the remainder, degree = 0.
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86. Core 1
Factors, remainders and cubic graphs
The remainder theorem
P (x) = (x − a)Q(x) + R
Letting x = a in the above formula gives
P (a) = (a − a)Q(x) + R
P (a) = 0 + R
P (a) = R
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87. Core 1
Factors, remainders and cubic graphs
The remainder theorem
P (x) = (x − a)Q(x) + R
Letting x = a in the above formula gives
P (a) = (a − a)Q(x) + R
P (a) = 0 + R
P (a) = R
The remainder theorem states that if a polynomial P (x) is
divided by (x − a) then the remainder is P (a).
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88. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Example
When x3 − 5x2 + 3x + k is divided by (x − 1), the remainder is 5.
Find k.
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89. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Example
When x3 − 5x2 + 3x + k is divided by (x − 1), the remainder is 5.
Find k.
Example
When x3 + 3x2 + x + k is divided by (x + 2), the remainder is 12.
Find k.
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90. Core 1
Factors, remainders and cubic graphs
The remainder theorem
If R = 0 in
P (x) = (x − a)Q(x) + R
then (x − a) is a factor of P (x).
We can write P (a) = 0 ⇐⇒ (x − a) is a factor of P (x). This is
the factor theorem discussed at the start of the chapter.
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91. Core 1
Factors, remainders and cubic graphs
The remainder theorem
Exercise
Complete exercise 6E page 95. Questions 1 a, c, e; 2 a, c, e; and
7, 9, 11.
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