UNIT OUTCOME PART SLIDE Higher Maths 2 1 1 Polynomials 1 UNIT OUTCOME PART SLIDE

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 4 + 6 x 3 + 5 x 2 + 4 x + 7 a 9 – 5 a 7 + 3 ( 2 x + 3 )( 3 x + 1 )( x – 8 ) Polynomial means ‘many numbers’ This is a polynomial because it can be multiplied out...

2 x 4 + 7 x 3 + 5x 2 – 4 x + 3 UNIT OUTCOME PART SLIDE NOTE ! Higher Maths 2 1 1 Polynomials 3 UNIT OUTCOME PART SLIDE NOTE Polynomials are normally written in decreasing order of power. Coefficients and Degree ! Degree 2 x 4 + 7 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 5 + 2 x 6 + 9x 3 3 x 4 + 5 x 3 – x 2 is a polynomial of degree 6. has coefficients 3, 5 and -1

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 + 2 = 0 or x – 2 = 0 x = ± 2 x = -2 x = 2

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 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 = (((ax + b ) x + c ) x + d ) x + e a × x + b × x + c × x + d × x + e f (x)

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 4 + 3 x 3 – 10 x 2 – 5 x + 7 for x = 4 = (((2 x + 3 ) x – 10 ) x – 5 ) x + 7 g (4 ) = (((2 × 4 + 3 ) × 4 – 10 ) × 4 – 5 ) × 4 + 7 = 531 2 × 4 + 3 × 4 – 1 × 4 – 5 × 4 + 7 531

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 2 + 5 x – 6 8 10 30 Evaluate h ( x ) for x = 2 . 4 5 15 24

UNIT OUTCOME PART SLIDE NOTE NOTICE ! f ( x ) = 8 x 7 – 6 x 4 + 5 Higher Maths 2 1 1 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 4 + 5 Example 4 x 6 – 3 x 3 r 5 Calculate the quotient and remainder for f ( x ) ÷ 2 x. 2 x 8 x 7 – 6 x 4 + 5 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.

coefficients of quotient Higher Maths 2 1 1 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 x2 + 5x – 1 r 4 • the coefficients of the quotient quotient • the remainder remainder

coefficients of quotient Higher Maths 2 1 1 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

UNIT OUTCOME PART SLIDE NOTE ! g( x ) = 3 x 4 – 2 x 2 + x + 4 Higher Maths 2 1 1 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 2 + 10x – 19) with remainder 42 Alternatively, g( x ) = (3 x 3 – 6 x 2 + 10x – 19)( x + 2) + 42

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 3 + 5 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 3 + 5 x 2 – 3 x – 4 = ( x – 4)(4 x 3 – x 2 + x + 1 ) + 0

UNIT OUTCOME PART SLIDE NOTE ! f ( x) = 2 x 3 + 5 x 2 – 28 x – 15 Higher Maths 2 1 1 Polynomials 13 UNIT OUTCOME PART SLIDE Factors of -15 : ± 1 NOTE Factorising with Synthetic Division ± 3 f ( x) = 2 x 3 + 5 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 3 + 5 x 2 – 28 x – 15 = ( x – 3 )( 2 x 2 + 11 x + 5 ) Evaluate f ( h) by synthetic division for every factor h. = ( x – 3 )( 2 x + 1 )( x + 5 )

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 2 + 4 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

UNIT OUTCOME PART SLIDE NOTE NOTE ! x Higher Maths 2 1 1 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 )

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

UNIT OUTCOME PART SLIDE NOTE x x a a b b f (a) > 0 f (b) < 0 Higher Maths 2 1 1 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.

UNIT OUTCOME PART SLIDE NOTE Higher Maths 2 1 1 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 - 0.163 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) - 0.016 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