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**UNIT OUTCOME PART SLIDE**

Higher Maths Polynomials 1 UNIT OUTCOME PART SLIDE

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**UNIT OUTCOME PART SLIDE NOTE**

Higher Maths Polynomials 2 UNIT OUTCOME PART SLIDE NOTE Introduction to Polynomials Any expression which still has multiple terms and powers after being simplified is called a Polynomial. Examples Polygon means ‘many sides’ 2 x x x x + 7 a 9 – 5 a 7 + 3 ( 2 x )( 3 x )( x – 8 ) Polynomial means ‘many numbers’ This is a polynomial because it can be multiplied out...

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**2 x 4 + 7 x 3 + 5x 2 – 4 x + 3 UNIT OUTCOME PART SLIDE NOTE !**

Higher Maths Polynomials 3 UNIT OUTCOME PART SLIDE NOTE Polynomials are normally written in decreasing order of power. Coefficients and Degree ! Degree 2 x x 3 + 5x 2 – 4 x + 3 Coefficient Term Degree of a Polynomial The ‘number part’ or multiplier in front of each term in the polynomial. The value of the highest power in the polynomial. 4 x x 6 + 9x 3 3 x x 3 – x 2 is a polynomial of degree 6. has coefficients 3, 5 and -1

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Higher Maths Polynomials 4 UNIT OUTCOME PART SLIDE NOTE Roots of Polynomials The root of a polynomial function is a value of x for which f (x) f (x) = 0 . Example Find the roots of g(x) = 3 x 2 – 12 3 x 2 – 12 = 0 3 ( x 2 – 4 ) = 0 3 x 2 – 12 = 0 or... 3 ( x + 2 )( x – 2 ) = 0 3 x 2 = 12 x 2 = 4 x = 0 or x – 2 = 0 x = ± 2 x = -2 x = 2

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Higher Maths Polynomials 5 UNIT OUTCOME PART SLIDE NOTE Polynomials and Nested Brackets Polynomials can be rewritten using brackets within brackets. This is known as nested form. Example f ( x ) = ax bx cx dx + e = ( ax bx cx + d ) x + e = (( ax bx + c ) x + d ) x + e = (((ax + b ) x + c ) x + d ) x + e = (((ax + b ) x + c ) x + d ) x + e a × x + b × x + c × x + d × x + e f (x)

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Higher Maths Polynomials 6 UNIT OUTCOME PART SLIDE NOTE Evaluating Polynomials Using Nested Form Nested form can be used as a way of evaluating functions. Example Evaluate g ( x ) = 2 x x 3 – 10 x 2 – 5 x + 7 for x = 4 = (((2 x + 3 ) x – 10 ) x – 5 ) x + 7 g (4 ) = (((2 × ) × 4 – 10 ) × 4 – 5 ) × = 531 2 × 4 + 3 × 4 – 1 × 4 – 5 × 4 + 7 531

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**UNIT OUTCOME PART SLIDE NOTE**

Higher Maths Polynomials 7 UNIT OUTCOME PART SLIDE NOTE The Loom Diagram x a b c d + + + + Evaluation of nested polynomials can be shown in a table. × x × x × x f ( x ) = ax 3 + bx 2 + cx + d = (( ax + b ) x + c) x + d f (x) a × x b × x + c × x d (i.e. the answer) + + Example 2 4 -3 5 -6 h ( x ) = 4 x 3 – 3 x x – 6 8 10 30 Evaluate h ( x ) for x = 2 . 4 5 15 24

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**UNIT OUTCOME PART SLIDE NOTE NOTICE ! f ( x ) = 8 x 7 – 6 x 4 + 5**

Higher Maths Polynomials 8 UNIT OUTCOME PART SLIDE NOTE Division and Quotients quotient In any division, the part of the answer which has been divided is called the quotient. remainder 6 r 2 5 3 2 f ( x ) = 8 x 7 – 6 x Example 4 x 6 – 3 x r 5 Calculate the quotient and remainder for f ( x ) ÷ 2 x. 2 x 8 x 7 – 6 x cannot be divided by 2 x NOTICE ! The power of each term in the quotient is one less than the power of the term in the original polynomial.

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**coefficients of quotient**

Higher Maths Polynomials 9 UNIT OUTCOME PART SLIDE Try evaluating f ( 3)… NOTE Investigating Polynomial Division Example 3 2 -1 -16 7 f ( x) = (2 x2 + 5x – 1)( x – 3) + 4 6 15 -3 2 5 -1 4 = 2 x3 – x2 – 16 x + 7 coefficients of quotient alternatively we can write remainder ! NOTICE f ( x) ÷ ( x – 3) When dividing f ( x) by ( x – n), evaluating f ( n) in a table gives: = 2 x x – 1 r 4 • the coefficients of the quotient quotient • the remainder remainder

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**coefficients of quotient**

Higher Maths Polynomials 10 UNIT OUTCOME PART SLIDE NOTE Synthetic Division 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: n a b c d e + + + + + × n × n × n × n coefficients of quotient This is called Synthetic Division. remainder

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**UNIT OUTCOME PART SLIDE NOTE ! g( x ) = 3 x 4 – 2 x 2 + x + 4**

Higher Maths Polynomials 11 UNIT OUTCOME PART SLIDE NOTE Examples of Synthetic Division Missing terms have coefficient zero. ! Example g( x ) = 3 x 4 – 2 x 2 + x + 4 Find the quotient and remainder for g( x ) ÷ ( x + 2). Evaluate g ( -2) : -2 3 -2 1 4 -6 12 -20 38 g( x ) ÷ ( x + 2) 3 -6 10 -19 42 = (3 x 3 – 6 x x – 19) with remainder 42 Alternatively, g( x ) = (3 x 3 – 6 x x – 19)( x + 2) + 42

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**UNIT OUTCOME PART SLIDE NOTE**

Higher Maths Polynomials 12 UNIT OUTCOME PART SLIDE NOTE The Factor Theorem 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 – 9x x 2 – 3 x – 4 Evaluate f ( 4) : f ( 4) = 0 4 2 -9 5 -3 -4 8 -4 4 4 ( x – 4) is a factor of f ( x) 4 -1 1 1 zero remainder f ( x) = 2 x 4 – 9x x 2 – 3 x – 4 = ( x – 4)(4 x 3 – x 2 + x + 1 ) + 0

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**UNIT OUTCOME PART SLIDE NOTE ! f ( x) = 2 x 3 + 5 x 2 – 28 x – 15**

Higher Maths Polynomials 13 UNIT OUTCOME PART SLIDE Factors of -15 : ± 1 NOTE Factorising with Synthetic Division ± 3 f ( x) = 2 x x 2 – 28 x – 15 Example Factorise ± 5 ± 15 Consider factors of the number term... Try evaluating f ( 3) : f ( 3) = 0 3 2 5 -28 -15 6 33 15 ( x – 3) If f ( h) = 0 then ( x – h) is a factor. ! 2 11 5 is a factor zero! f ( x) = 2 x x 2 – 28 x – 15 = ( x – 3 )( 2 x x + 5 ) Evaluate f ( h) by synthetic division for every factor h. = ( x – 3 )( 2 x + 1 )( x + 5 )

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Higher Maths Polynomials 14 UNIT OUTCOME PART SLIDE NOTE Finding Unknown Coefficients Example ( x + 3) is a factor of f ( x) = 2 x 4 + 6x 3 + px x – 15 Find the value of p. ( x + 3) is a factor Evaluate f (- 3) : f (- 3) = 0 - 3 p 2 6 4 -15 - 6 - 3p 9p – 12 9p – 27 = 0 9p = 27 p - 3p + 4 9p – 27 2 p = 3 zero remainder

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**UNIT OUTCOME PART SLIDE NOTE NOTE ! x**

Higher Maths Polynomials 15 UNIT OUTCOME PART SLIDE NOTE Finding Polynomial Functions from Graphs The equation of a polynomial can be found from its graph by considering the intercepts. Equation of a Polynomial From a Graph f (x) = k( x – a )( x – b )( x – c ) f ( x) with x-intercepts a , b and c d d NOTE ! d k can be found by substituting x a b c ( 0 , d )

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**UNIT OUTCOME PART SLIDE NOTE x f (x) = k( x + 2)( x – 1)( x – 5)**

Higher Maths Polynomials 16 UNIT OUTCOME PART SLIDE NOTE Finding Polynomial Functions from Graphs (continued) Example f ( x) Find the function shown in the graph opposite. 30 - 2 x 1 5 f (x) = k( x + 2)( x – 1)( x – 5) Substitute k back into original function and multiply out... f (0) = 30 k(0 + 2)(0 – 1)(0 – 5) = 30 f (x) = 3 ( x + 2)( x – 1)( x – 5) 10 k = 30 = 3 x 3 – 12 x 2 – 21x + 30 k = 3

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**UNIT OUTCOME PART SLIDE NOTE x x a a b b f (a) > 0 f (b) < 0**

Higher Maths Polynomials 17 UNIT OUTCOME PART SLIDE NOTE Location of a Root A root of a polynomial function f ( x) lies between a and b if : f ( x) f ( x) root root a b or... x x a b f (a) > 0 f (b) < 0 f (a) < 0 f (b) > 0 and 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.

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Higher Maths Polynomials 18 UNIT OUTCOME PART SLIDE NOTE Finding Approximate Roots The approximate root can be calculated by an iterative process: Example x f ( x) root between f ( x) = x 3 – 4 x 2 – 2 x + 7 Show that f ( x) has a root between 1 and 2. 1 2 - 5 2 1 and 2 1.3 1 and 1.3 f (1) = 2 1.2 0.568 1.2 and 1.3 (above x-axis) 1. 25 0.203 1.25 and 1.3 f (2) = - 5 (below x-axis) 1. 28 1.25 and 1.28 1. 27 0.057 1.27 and 1.28 f ( x) crosses the x-axis between 1 and 2. 1. 275 0.020 1.275 and 1.28 The root is at approximately x = 1.28

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Aim: How do we solve polynomial equations using factoring?

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