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1Higher Maths 2 1 1 Polynomials. Any expression which still has multiple terms and powers after being simplified is called a Polynomial. Introduction.

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Presentation on theme: "1Higher Maths 2 1 1 Polynomials. Any expression which still has multiple terms and powers after being simplified is called a Polynomial. Introduction."— Presentation transcript:

1 1Higher Maths Polynomials

2 Any expression which still has multiple terms and powers after being simplified is called a Polynomial. Introduction to Polynomials 2 Polynomial means many numbers 2 x x x 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 Polynomials

3 2 x x x 2 – 4 x Coefficients and Degree The value of the highest power in the polynomial. 4 x x x 3 is a polynomial of degree 6. Coefficient 3 x 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 Polynomials

4 Roots of Polynomials 4 The root of a polynomial function is a value of x for which f (x)f (x) f (x) = 0.f (x) = 0. 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 Polynomials

5 Polynomials and Nested Brackets Polynomials can be rewritten using brackets within brackets. This is known as nested form. Example 5 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× x a + b+ b + c+ c + e+ e × x× x × x× x + d+ d f (x)f (x) × x× x Higher Maths Polynomials

6 Evaluating Polynomials Using Nested Form Example 6 g ( x ) = 2 x x 3 – 10 x 2 – 5 x + 7 = ( ( ( 2 x + 3 ) x – 10 ) x – 5 ) x Evaluatefor x = 4x = 4 g ( 4 ) = ( ( ( 2 × ) × 4 – 10 ) × 4 – 5 ) × = × 4× – 0 × 4× 4– 5– 5 × 4× 4 × 4× Nested form can be used as a way of evaluating functions. Higher Maths Polynomials

7 The Loom Diagram 7 Evaluation of nested polynomials can be shown in a table. f ( x ) = ax 3 + bx 2 + cx + d = ( ( ax + b ) x + c ) x + d × x× x b × x× x + c+ c × x× x + d + a abcdx × x× x + × x× x + × x× x ++ Example h ( x ) = 4 x 3 – 3 x x – 6 Evaluate h ( x )h ( x ) for x = 2.x = f (x)f (x) (i.e. the answer) Higher Maths Polynomials

8 Division and Quotients 8 Example f ( x ) = 8 x 7 – 6 x Calculate the quotient and remainder for f ( x ) ÷ 2 x. 4 x 6 – 3 x 3 r 5 2 x2 x 8 x 7 – 6 x 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 Polynomials

9 Investigating Polynomial Division 9 Example f ( x ) = ( 2 x x – 1 )( x – 3 ) + 4 = 2 x 3 – x 2 – 16 x + 7 f ( x ) ÷ ( x – 3 ) = 2 x x – 1 r 4 alternatively we can write quotient Try evaluating f ( 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 Polynomials

10 Synthetic Division 10 abcdn × n× n ++++ e + × n× n × n× n × n× n 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 Polynomials

11 = ( 3 x 3 – 6 x x – 19 ) with remainder 42 Examples of Synthetic Division 11 Example g ( x ) = 3 x 4 – 2 x 2 + x + 4 Find the quotient and remainder for g ( x ) ÷ ( x + 2 ) Evaluate g ( -2 ) : Missing terms have coefficient zero. g ( x ) ÷ ( x + 2 ) Alternatively, g ( x ) = ( 3 x 3 – 6 x x – 19 )( x + 2 ) + 42 Higher Maths Polynomials

12 The Factor Theorem 12 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 x 2 – 3 x – Evaluate f ( 4 ) : ( x – 4 ) is a factor of f ( x ) zero remainder f ( 4 ) = 0 f ( x ) = 2 x 4 – 9 x x 2 – 3 x – 4 = ( x – 4 )( 4 x 3 – x 2 + x + 1 ) + 0 Higher Maths Polynomials

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

14 9 p – 27 Finding Unknown Coefficients 14 ( x + 3 ) is a factor of f ( x ) = 2 x x 3 + px x – 15 Example Find the value of p. Evaluate f (- 3 ) : p 4 20 p - 3 p p- 3p p – 12 ( x + 3 ) is a factor f (- 3 ) = 0 9 p – 27 = 0 9 p = 27 p = 3 zero remainder Higher Maths Polynomials

15 d d d Finding Polynomial Functions from Graphs 15 The equation of a polynomial can be found from its graph by considering the intercepts. 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 )( 0, d ) with x -intercepts a, b and c f ( x)f ( x) Higher Maths Polynomials

16 Finding Polynomial Functions from Graphs (continued) 16 Example f ( x)f ( x) Find the function shown in the graph opposite. f ( x ) = k ( x + 2 )( x – 1 )( x – 5 ) f ( 0 ) = 30 k ( )( 0 – 1 )( 0 – 5 ) = k = 30 k = 3 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 Polynomials

17 Location of a Root 17 f ( 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... f ( 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 Higher Maths Polynomials

18 Finding Approximate Roots 18 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) = - 5f (2) = - 5 ( above x - axis ) ( below x - axis ) f ( x ) crosses the x - axis between 1 and 2. f ( x)f ( x) x root between 12 1 and and and and and and The approximate root can be calculated by an iterative process: 1.2 and The root is at approximately x = 1.28 Higher Maths Polynomials


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