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1Higher Maths Polynomials

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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

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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

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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

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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

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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

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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

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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

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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

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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

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= ( 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

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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

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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

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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

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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

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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

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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

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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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