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Higher Maths 2 1 1 Polynomials. 1. OUTCOME. PART. SLIDE. UNIT. Higher Maths 2 1 1 Polynomials. 2. OUTCOME. PART. SLIDE. UNIT. NOTE. Introduction to Polynomials.

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

  2. Higher Maths 2 1 1 Polynomials 2 OUTCOME PART SLIDE UNIT 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’ 2x4 + 6x3 + 5x2 + 4x + 7 a 9 – 5a7 + 3 (2x + 3)(3x + 1)(x – 8) Polynomial means ‘many numbers’ This is a polynomial because it can be multiplied out...

  3. ! Higher Maths 2 1 1 Polynomials 3 OUTCOME PART SLIDE UNIT NOTE Polynomials are normally written in decreasing order of power. Coefficients and Degree Degree 2x4+7x3+5x2–4x+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. 4x5 + 2x6 + 9x3 3x4 + 5x3 – x2 is a polynomial of degree 6. has coefficients 3, 5 and -1

  4. The root of a polynomial function is a value ofxfor which f(x) f(x)=0. Higher Maths 2 1 1 Polynomials 4 OUTCOME PART SLIDE UNIT NOTE Roots of Polynomials Example 3x2 – 12= 0 g(x)= 3x2 – 12 Find the roots of 3(x2 – 4)= 0 3x2 – 12= 0 or... 3(x + 2)(x – 2)= 0 3x2= 12 x2= 4 x+ 2 = 0 or x– 2 = 0 x= ±2 x = -2 x = 2

  5. Higher Maths 2 1 1 Polynomials 5 OUTCOME PART SLIDE UNIT NOTE Polynomials and Nested Brackets Polynomials can be rewritten using brackets within brackets. This is known as nested form. f(x) = ax4+ bx3+ cx2+ dx+ e Example = (ax3+ bx2+ cx+d)x+ e = ((ax2+bx+c)x+d)x+ e = (((ax+b)x+ c)x+d)x+ e = (((ax+b)x+ c)x+d)x+ e a ×x ×x ×x ×x +c +e f(x) +b +d

  6. Higher Maths 2 1 1 Polynomials 6 OUTCOME PART SLIDE UNIT NOTE Evaluating Polynomials Using Nested Form Nested form can be used as a way of evaluating functions. Example x=4 g(x) = 2x4+ 3x3 – 10x2 –5x + 7 Evaluate for = (((2x+3)x –10)x – 5)x+ 7 g(4) = (((2×4+3)×4 –10)×4 – 5)×4+ 7 531 = 531 2 0 ×4 +3 ×4 1 ×4 –5 ×4 +7 –

  7. ×x Higher Maths 2 1 1 Polynomials 7 OUTCOME PART SLIDE UNIT NOTE x a b c d The Loom Diagram + + + + Evaluation of nested polynomials can be shown in a table. ×x ×x ×x f(x) = ax3+bx2+cx+d = ((ax+b)x+c)x+d f(x) (i.e. the answer) ×x +c ×x a b d + + Example 2 4 -3 5 -6 h(x) = 4x3 – 3x2+ 5x –6 8 10 30 x=2. h(x) Evaluate for 4 5 15 24

  8. NOTICE ! Higher Maths 2 1 1 Polynomials 8 OUTCOME PART SLIDE UNIT NOTE Division and Quotients quotient remainder In any division, the part of the answer which has been divided is called the quotient. 6 r2 5 3 2 f(x) = 8x7– 6x4+ 5 Example 4x6–3x3r5 Calculate the quotient and remainder for f(x) ÷2x. 8x7– 6x4+ 5 2x cannot be divided by 2x The power of each term in the quotient is one less than the power of the term in the original polynomial.

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

  10. ×n ×n ×n ×n Higher Maths 2 1 1 Polynomials 10 OUTCOME PART SLIDE UNIT NOTE Synthetic Division For any polynomial function f(x) = ax4 + bx3 + cx2 + dx + e , f(x)divided by(x–n) can be found as follows: n a b c d e + + + + + This is called Synthetic Division. coefficients of quotient remainder

  11. ! Higher Maths 2 1 1 Polynomials 11 OUTCOME PART SLIDE UNIT NOTE Examples of Synthetic Division Missing terms have coefficient zero. Example g(x) = 3x4 – 2x2 + x +4 Find the quotient and remainder for g(x) ÷(x+2). Evaluate g(-2) : -2 3 0 -2 1 4 -6 12 -20 38 g(x) ÷(x+2) 3 -6 10 -19 42 = (3x3 – 6x2 + 10x – 19)with remainder 42 Alternatively, g(x)=(3x3 – 6x2 + 10x – 19)(x+2)+ 42

  12. Higher Maths 2 1 1 Polynomials 12 OUTCOME PART SLIDE UNIT 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)= 2x4 – 9x3 + 5x2 – 3x – 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 0 zero remainder f(x) = 2x4 – 9x3 + 5x2 – 3x – 4 = (x–4)(4x3 – x2 + x + 1) + 0

  13. If f(h)=0then (x–h) is a factor. ! Higher Maths 2 1 1 Polynomials 13 OUTCOME PART SLIDE UNIT Factors of-15: ±1 NOTE Factorising with Synthetic Division ±3 f(x)=2x3+5x2–28x –15 Example Factorise ±5 ±15 Try evaluating f(3) : Consider factors of the number term... f(3) = 0 3 2 5 -28 -15 (x–3) 6 33 15 2 11 5 0 is a factor zero! f(x)=2x3+5x2–28x –15 =(x–3)(2x2 + 11x + 5) Evaluatef(h)by syntheticdivision for every factorh. =(x–3)(2x+1)(x+5)

  14. Higher Maths 2 1 1 Polynomials 14 OUTCOME PART SLIDE UNIT NOTE Finding Unknown Coefficients Example (x+3)is a factor of f(x)= 2x4 + 6x3 + px2 + 4x – 15 Find the value of p. Evaluate f(-3) : (x+3) is a factor f(-3)= 0 p -3 2 6 4 -15 -3p 9p – 12 -6 0 9p – 27 = 0 9p= 27 9p – 27 p -3p + 4 2 0 p= 3 zero remainder

  15. d d d NOTE ! Higher Maths 2 1 1 Polynomials 15 OUTCOME PART SLIDE UNIT 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 k can be found by substituting x a c b (0,d)

  16. Higher Maths 2 1 1 Polynomials 16 OUTCOME PART SLIDE UNIT 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 f(x)=3(x+2)(x–1)(x–5) k(0+2)(0–1)(0–5)=30 =3x3–12x2–21x+30 10k=30 k=3

  17. Higher Maths 2 1 1 Polynomials 17 OUTCOME PART SLIDE UNIT 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.

  18. Higher Maths 2 1 1 Polynomials 18 OUTCOME PART SLIDE UNIT NOTE The approximate root can be calculated by an iterative process: Finding Approximate Roots Example x f(x) root between f(x)=x3– 4x2– 2x +7 Show that f(x)has a root between 1and2. 1 2 -5 1 and 2 2 -0.163 1 and 1.3 1.3 1.2 and 1.3 0.568 1.2 f(1)= 2 (above x-axis) 1.25 and 1.3 0.203 1. 25 f(2)=-5 (below x-axis) -0.016 1.25 and 1.28 1. 28 1.27 and 1.28 0.057 1. 27 f(x) crosses the x-axis between 1 and 2. 1.275 and 1.28 0.020 1. 275 The root is at approximatelyx=1.28

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