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Higher Maths 2.1.1 - Polynomials

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Higher Maths 2.1.1 - Polynomials

  1. 1. SLIDE Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  2. 2. Any expression which still has multiple terms and powers after being simplified is called a Polynomial . Introduction to Polynomials NOTE SLIDE Polynomial means ‘many numbers’ 2 x 4 + 6 x 3 + 5 x 2 + 4 x + 7 Examples 9 – 5 a 7 + a 3 ( 2 x + 3 )( 3 x + 1 )( x – 8 ) Polygon means ‘many sides’ This is a polynomial because it can be multiplied out... Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  3. 3. 2 x 4 + 7 x 3 + 5 x 2 – 4 x + 3 SLIDE Coefficients and Degree NOTE The value of the highest power in the polynomial. 4 x 5 + 2 x 6 + 9 x 3 is a polynomial of degree 6 . Coefficient 3 x 4 + 5 x 3 – x 2 has coefficients 3 , 5 and -1 Degree Term The ‘number part’ or multiplier in front of each term in the polynomial. Degree of a Polynomial Polynomials are normally written in decreasing order of power. Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  4. 4. Roots of Polynomials NOTE SLIDE Find the roots of g ( x ) = 3 x 2 – 12 3 x 2 – 12 = 0 3 x 2 = 12 x 2 = 4 x = ± 2 3 ( x 2 – 4 ) = 0 3 ( x + 2 )( x – 2 ) = 0 x + 2 = 0 x – 2 = 0 x = 2 x = -2 3 x 2 – 12 = 0 or... or Example Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART The root of a polynomial function is a value of x for which f ( x ) f ( x ) = 0 .
  5. 5. Polynomials and Nested Brackets NOTE Polynomials can be rewritten using brackets within brackets. This is known as nested form . Example SLIDE f ( x ) = ax 4 + bx 3 + cx 2 + dx + e = ( ax 3 + bx 2 + cx + d ) x + e = ( ( ax 2 + bx + c ) x + d ) x + e = ( ( ( ax + b ) x + c ) x + d ) x + e × x a + b + c + e × x × x + d f ( x ) = ( ( ( ax + b ) x + c ) x + d ) x + e × x Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  6. 6. Evaluating Polynomials Using Nested Form NOTE Example SLIDE g ( x ) = 2 x 4 + 3 x 3 – 10 x 2 – 5 x + 7 = ( ( ( 2 x + 3 ) x – 10 ) x – 5 ) x + 7 2 Evaluate for x = 4 g ( 4 ) = ( ( ( 2 × 4 + 3 ) × 4 – 10 ) × 4 – 5 ) × 4 + 7 = × 4 531 + 3 1 – 0 × 4 – 5 × 4 × 4 + 7 531 Nested form can be used as a way of evaluating functions. Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  7. 7. The Loom Diagram NOTE SLIDE Evaluation of nested polynomials can be shown in a table. f ( x ) = ax 3 + bx 2 + cx + d = ( ( ax + b ) x + c ) x + d b × x + c × x + d + a a b c d x × x + × x + × x + + Example h ( x ) = 4 x 3 – 3 x 2 + 5 x – 6 Evaluate h ( x ) for x = 2 . 2 4 -3 5 -6 4 5 15 24 8 10 30 f ( x ) (i.e. the answer) Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART × x
  8. 8. Division and Quotients NOTE SLIDE Example f ( x ) = 8 x 7 – 6 x 4 + 5 Calculate the quotient and remainder for f ( x ) ÷ 2 x . 4 x 6 – 3 x 3 r 5 2 x 8 x 7 – 6 x 4 + 5 5 3 2 6 r 2 quotient remainder In any division, the part of the answer which has been divided is called the quotient . cannot be divided by 2 x The power of each term in the quotient is one less than the power of the term in the original polynomial. Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART NOTICE
  9. 9. Investigating Polynomial Division NOTE SLIDE Example f ( x ) = ( 2 x 2 + 5 x – 1 )( x – 3 ) + 4 = 2 x 3 – x 2 – 16 x + 7 f ( x ) ÷ ( x – 3 ) = 2 x 2 + 5 x – 1 r 4 alternatively we can write quotient Try evaluating f ( 3 ) … 3 2 -1 -16 7 2 5 -1 4 6 15 -3 When dividing f ( x ) by ( x – n ) , evaluating f ( n ) in a table gives: • the coefficients of the quotient • the remainder remainder coefficients of quotient remainder Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART NOTICE
  10. 10. Synthetic Division NOTE SLIDE a b c d n + + + + e + coefficients of quotient remainder For any polynomial function f ( x ) = ax 4 + bx 3 + cx 2 + dx + e , f ( x ) divided by ( x – n ) can be found as follows: This is called Synthetic Division. Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART × n × n × n × n
  11. 11. = ( 3 x 3 – 6 x 2 + 10 x – 19 ) with remainder 42 Examples of Synthetic Division NOTE SLIDE Example g ( x ) = 3 x 4 – 2 x 2 + x + 4 Find the quotient and remainder for g ( x ) ÷ ( x + 2 ) . -2 3 0 -2 1 3 -6 10 -19 -6 12 -20 4 42 38 Evaluate g ( -2 ) : Missing terms have coefficient zero. g ( x ) ÷ ( x + 2 ) Alternatively, g ( x ) = ( 3 x 3 – 6 x 2 + 10 x – 19 )( x + 2 ) + 42 Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  12. 12. The Factor Theorem NOTE SLIDE If a polynomial f ( x ) can be divided exactly by a factor ( x – h ) , then the remainder, given by f ( h ) , is zero. Example Show that ( x – 4 ) is a factor of f ( x ) = 2 x 4 – 9 x 3 + 5 x 2 – 3 x – 4 4 2 -9 5 -3 4 -1 1 1 8 -4 4 -4 0 4 Evaluate f ( 4 ) : ( x – 4 ) is a factor of f ( x ) zero remainder f ( 4 ) = 0 f ( x ) = 2 x 4 – 9 x 3 + 5 x 2 – 3 x – 4 = ( x – 4 )( 4 x 3 – x 2 + x + 1 ) + 0 Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  13. 13. Factorising with Synthetic Division NOTE SLIDE Factorise Try evaluating f ( 3 ) : ± 1 ± 3 ± 5 ± 15 Example Evaluate f ( h ) by synthetic division for every factor h . 3 2 5 -28 -15 2 11 5 0 6 33 15 f ( x ) = 2 x 3 + 5 x 2 – 28 x – 15 ( x – 3 ) f ( 3 ) = 0 is a factor = ( x – 3 )( 2 x 2 + 11 x + 5 ) f ( x ) = 2 x 3 + 5 x 2 – 28 x – 15 = ( x – 3 )( 2 x + 1 )( x + 5 ) Consider factors of the number term... Factors of - 15 : zero! Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART If f ( h ) = 0 then ( x – h ) is a factor.
  14. 14. 9 p – 27 Finding Unknown Coefficients NOTE SLIDE ( x + 3 ) is a factor of f ( x ) = 2 x 4 + 6 x 3 + p x 2 + 4 x – 15 Example Find the value of p . Evaluate f ( - 3 ) : - 3 2 6 p 4 2 0 p - 3 p + 4 - 6 0 - 3 p -15 9 p – 12 ( x + 3 ) is a factor f ( - 3 ) = 0 9 p – 27 = 0 9 p = 27 p = 3 zero remainder Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  15. 15. Finding Polynomial Functions from Graphs NOTE SLIDE The equation of a polynomial can be found from its graph by considering the intercepts. x b a c f ( x ) = k ( x – a )( x – b )( x – c ) Equation of a Polynomial From a Graph k can be found by substituting ( 0 , d ) with x -intercepts a , b and c f ( x ) Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART d NOTE
  16. 16. Finding Polynomial Functions from Graphs (continued) NOTE SLIDE Example x f ( x ) - 2 1 5 30 Find the function shown in the graph opposite. f ( x ) = k ( x + 2 )( x – 1 )( x – 5 ) f ( 0 ) = 30 k ( 0 + 2 )( 0 – 1 )( 0 – 5 ) = 30 10 k = 30 k = 3 f ( x ) = 3 ( x + 2 )( x – 1 )( x – 5 ) = 3 x 3 – 12 x 2 – 21 x + 30 Substitute k back into original function and multiply out... Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  17. 17. Location of a Root NOTE SLIDE x f ( x ) b a f ( a ) > 0 f ( b ) < 0 A root of a polynomial function f ( x ) lies between a and b if : and or... x f ( x ) b a f ( a ) < 0 f ( b ) > 0 and If the roots are not rational, it is still possible to find an approximate value by using an iterative process similar to trial and error. root root Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART
  18. 18. Finding Approximate Roots NOTE SLIDE Example Show that f ( x ) has a root between 1 and 2. f ( x ) = x 3 – 4 x 2 – 2 x + 7 f ( 1 ) = 2 f ( 2 ) = - 5 ( above x - axis ) ( below x - axis ) f ( x ) crosses the x - axis between 1 and 2. f ( x ) x root between 1 2 1 and 2 2 - 5 1 and 1.3 1.3 - 0.163 1.25 and 1.3 1. 25 0.203 1.25 and 1.28 1. 28 - 0.016 1.27 and 1.28 1. 27 0.057 1.275 and 1.28 1. 275 0.020 The approximate root can be calculated by an iterative process: 1.2 and 1.3 1.2 0.568 The root is at approximately x = 1.28 Higher Maths 2 1 1 Polynomials UNIT OUTCOME PART

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