Maths Extension 2 - Polynomials. Polynomials

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1 Maths Extension - Polynomials Polynomials! Definitions and properties of polynomials! Factors & Roots! Fields ~ Q Rational ~ R Real ~ C Complex! Finding zeros over the complex field! Factorization & Division of polynomials! Remainder & Factor Theorem! Rational Roots! Multiplicity Theorem/ Repeated Roots! Relationship between the roots and coefficients of a polynomial equation

2 Maths Extension - Polynomials Definitions and properties of polynomials Polynomial Expression P(x) = p x n + p x n- + p x n- + + p n- x + p n where p Coefficients Leading term p, p, p, p, p n x n Constant p If p n = If p = p = p = It is a monic Then P(x) is a zero polynomial P(x) = x x + 7x x + Coefficient of.x Is.x Is Leading term is x Constant is More points: _a i _a _a n x n _a n _P(x) _a n = _P(x) _P(x) _P(x) = Are the coefficients of the polynomial Is the constant term Is the leading term Is the leading coefficient Polynomial of degree n The polynomial is a monic Null polynomial Expression of polynomial Polynomial equation

3 Maths Extension - Polynomials Factors and Roots Factor A polynomial that divides into another and has remainder (x ) is a factor of x Root P() is a root of x _x = x = Fields Q Rational R Real Integer numbers ±,,, Irrational Numbers Surdic roots occur in conjugate pairs If a + b is a root, so too will a b C Complex Numbers over the complex field a ± ib Complex roots occur in conjugate pairs If a + ib is a root, so too will a ib Factorize x x 5 over Q, R, C Y Y - 5 = (Y 5)(Y + ) Q = (x 5)(x + ) R = (x + 5 )(x 5 )(x + ) C = (x + 5 )(x 5 )(x + i )(x ) i

4 Maths Extension - Polynomials Finding zeros over the complex field. If one root is complex, then one of the other roots is it s conjugate. If ( + i ) is a root, [x ( + i )] is the factor So [x ( + i )][x ( i )] = x x + How do we get this??? (x z)(x - z ) Where: _z = + i = x (z + z )x + z z z = - i _z + z = = x x + z z = Find all the zeros of P(x) = x x x + 6x over C. ( + i ) is a zero. If ( + i ) is a root, then ( i ) is also a root. By multiplying out the factors: x x + _x + x _x x + _x x x + 6x _x x +x _x x + 6x _x x + x x + x x + x The factorized equation is [x ( + i )][x ( i )](x + )(x ) The zeros are ( + i ), ( i ), -,

5 Maths Extension - Polynomials Factorization and Division of Polynomials Factorizing polynomials. Simple factorizing. Trinomials, Grouping, Difference of squares, etc. Quadratic Formula. Completing the Square Simple factorizing _x + x = (x + )(x ) _x = (x )(x + ) = (x )(x + )(x + ) = (x )(x + )(x + i )(x - i ) Quadratic formula _x + x + _x = ± () ()() ± 8 = ± = = ± i (x a) _x + x + _x = (x + + i )(x + ) i ± () ()() ± = 8 ± i = 8 i i ( x + + )( x + ) We can only find the factors of the polynomial, not the constant outside Completing the Square _x + x + = x + x + + = (x + ) + = (x + ) i = [(x + ) i ][(x + ) + ] i = (x + i )(x + + ) i x + + _x + x + = [ ] 9 = [( x + + ) + ] 6 6 i [ x + ] 8 6 = ( ) i i = [ x + + ][ x + ] ! Complete the square, then add end term to satisfy the equation.! i = -! Difference of two squares 5

6 Maths Extension - Polynomials Division of polynomials P(x) = A(x) Q(x) + R(x) Dividend = Divisor Quotient + Remainder x x + 7x x + = x x + 5x +7x LONG DIVISION!!! x + 5x + 7x + x x x + 7x x + x 6x 5x + 7x 5x x 7x 7x Divide and find a such that R(x) = x x x + x 6 67 x + (a )x + ( a) x + x + ax + ax + 6 x + ax (a )x + ax (a )x + (z )x ( a)x + 6 ( a)x + ( a) a For R(x) =, a a = = a = 6

7 Maths Extension - Polynomials Factor and Remainder Theorems Remainder Theorem! If a polynomial P(x) is divided by (x a), then the remainder is P(a) x ; a = P() = () () + 7() () + = = 67 x + 5x + 7x + x x x + 7x x + x 6x 5x + 7x 5x x 7x 7x x x x + x 6 67 Factor Theorem! For any polynomial P(x), if P(a) =, then (x a) is a factor of P(x) OR! For any polynomial P(x), if (x a) is a factor of P(x), then P(a) = 7

8 Maths Extension - Polynomials Rational Roots Let P(x) have degree n with integer coefficients. Suppose P(x) has a rational root of p q. _p, q, are prime integers. Then p a and q a n is divides into Given that P(x) = x x x + 6 has a rational root. Find all the zeros of P(x). Let the root be p q. _p 6_;_q The possible rational roots are: _p ± 6, ±, ±, ± _q ±, ± P(-) = P() = P( ) = The zeros of P(x) are,, The factorized equation is (x + )(x )(x - ) OR (x + )(x )(x ) Given that P(x) = x + x + x 6 has a rational root. Find all the zeros of P(x). Let the root be p q. _p -6_;_q The possible rational roots are: _p ± 6, ±, ±, ± _q ± P() = (x ) is a factor _x + 5x + 6 _x _x + x + x 6 _x x (x )(x + )(x + ) 5x + x 5x 5x 6x 6 6x 6 The zeros of P(x) are, -, - The factorized equation is (x + )(x + )(x ) 8

9 Maths Extension - Polynomials Multiplicity Repeated Roots _x = a is a repeated root of P(x) or multiplicity r. If P(x) = (x a) r.q(x) so (x a) / Q(x) If x = a is of multiplicity r of P(x), then multiplicity (r ) or P`(x) Proof Let P(x) = (x a) r.q(x) P`(x) = (x a) r.q`(x) + Q(x).r(x a) r-. Product rule = (x a) r- [(x a).q(x) + r.q(x)] = (x a) r-.r(x) P`(x) has a root at x = a with multiplicity (r ) Find roots of P(x) = x + x, given that the polynomial has a double root. P(x) = x + x P`(x) _x = - is a double root. = x + 6x P(x) has a single root, so only need to differentiate once. = x(x + ) _x = or Check both results by remainder theorem!!! P() P(-) = To find the other root, we can either use Long Division or Sum of the Roots Long Division (x + ) = x + x + _x x + x + _x + x + _x + x + x x x x x Sum of the Roots Sum of the roots one at a time = b a α + β + γ = + ( ) + γ = γ = So the roots of P(x) are: -, -, 9

10 Maths Extension - Polynomials For polynomials which have: Double roots Differentiate one time Triple roots Differentiate two times Quadruple roots Differentiate three times If number of multiple roots are not given, keep differentiating until the P`(x), P``(x), etc.. can be factorized. Factorize and find the zeros of P(x) = x + x x 5x if it has multiple roots. P(x) = x + x x 5x P`(x) = x + x 6x 5 P``(x) = x + 6x 6 = 6(x )(x + ) P(-) = α + β + γ + δ + δ δ = = = P(x) = x x 9x + C has a double root. Find C P`(x) = x 6x 9 = (x )(x + ) _x = or We can substitute both, but our C will be different. Both answers must be given P() = C C = 7 P(-) = C C = 5

11 Maths Extension - Polynomials Relationship between the roots and coefficients of a polynomial equation Quadratic : ax + bx + c α + β b = Sum of roots at a time a αβ c = a Sum of roots at a time (product of roots) Cubic : ax + bx + cx + d α + β + γ b = Sum of roots at a time a αβ + βγ + c γα = a Sum of roots at a time αβγ d = a Sum of roots at a time (product of roots) Quartic : ax + bx + cx + dx + e α + β + γ + δ b = Sum of roots at a time a αβ + βγ + γδ + δα + αγ + βδ c = a Sum of roots at a time αβγ + βγδ + γδα + δαβ d = Sum of roots at a time a αβγδ e = a Sum of roots at a time (product of roots) If n is the number of roots at a time, to find out how many combinations there are, we use: n C r. For a quartic: α + β + γ + δ = Σ α = Σα i C αβ + βγ + γδ + δα + αγ + βδ = Σ αβ = Σα i α j C αβγ + βγδ + γδα + δαβ = Σ αβγ = Σαiα jα k C αβγδ = Σ αβγδ = Σαiα jα kα l C

12 Maths Extension - Polynomials Roots are α, β,γ for x x x. ( α )( β )( γ ) = αβγ ( αβ + βγ + γα) + ( α + β + γ ) = + =. ( β + γ α)( γ + α β )( α + β γ ) If α + β + γ = α + β γ = γ β + γ α = α γ + α β = β = ( α )( β )( γ ) = 8( α )( β )( γ ) From part = 8 =. Σα i = ( Σαi ) ( Σαiα j ) = - ( ) = + = 7. Σα i α β γ α β γ α β γ = = = α i ( Σ ) ( Σα i ) ( Σαi) = αi Σ 8 6 = αi Σ = 7 7 Σ α i =

13 Maths Extension - Polynomials 5. Σα i P(x) is a cubic x.p(x) is now a quadratic Sub it in x x x x x x x α β γ α β γ α β γ α α α 6. Σα β Σ = α i α j α i ( Σ ) ( Σα i ) ( Σα i ) αi 7 Σ ( ) αi Σαi Σαi (7) Σ 7 97 = α β + β γ + γ α = ( Σα α ) ( Σα α α )( Σα) i j = ) ( )() ( = i j k 7. Σ α α β γ = + + α i j = ( ) = i α α α α = j k 8. α β α Σ i α j = α β + α γ + β α + β γ + γ α + γ β = ( α + β + γ )( αβ + βγ + γα) αβγ = ( Σαi)( Σαiα j ) ( Σαiα jα k ) = )( ) ( ) ( 9 =

14 Maths Extension - Polynomials Relationship & Transformation methods If α, β,γ are the roots of x 5x + x + 6 =, form the equation whose roots are: α, β, γ Relationship Method: Sum of the roots one at a time α + β + γ = ( α + β + γ ) 5 = () = 5 Sum of the roots two at a time ( α )(β ) + (β )(γ ) + (γ )(α ) = ( αβ + βγ + γα) = () = 8 Sum of the roots three at a time ( α )(β )(γ ) = 8( αβγ ) = 8( ) = The new equation is x 5x + 8x + = Transformation Method: x = α, β, γ _y = x y = α,β, γ y _x = = y y y = y y y = y 5y y + 6 = y 5y 8y Change y for x = x 5x 8x The Transformation Method is preferred

15 Maths Extension - Polynomials If α, β,γ are the roots of x + x + x 5 =, form the equation whose roots are: A),, B) α, β, γ C) + α, + β, + γ α β γ D) α, β, γ E) α, β, A) x + x + x 5 = x = y B) x + x + x 5 = x = y + y y y + + y y y = () () + 5 = 5 = + y + y 5y = 5x x x y y y = () + () + 5 y y y = y + y = y = x + x + x γ F) α, C) x + x + x 5 = x = y = (y ) + (y ) + (y ) 5 = (y )(y )(y ) + (y )(y ) + (y ) 5 = (y 6y + y 8) + (y y + ) + (y ) 5 = y 9y + y = x 9x + x β, γ, D) x + x + x 5 = x = y + = (y + ) + (y + ) + (y + ) 5 = (y + )(y + )(y + ) + (y + )(y + ) + (y + ) 5 = (y + 6y + y + 8) + (y + y + ) + (y + ) 5 = y + 5y + 7y + 5 = x + 5x + 7x

16 Maths Extension - Polynomials E) x + x + x 5 = x = y = y + y + y 5 = y + y + y 5 y 5 = y ( y + ) Square both sides 9y y + 5 = y(y + y + ) = y + y + y = y 5y + y 5 = x 5x + x 5 F) x + x + x 5 = x = y = (-y) + (-y) + (-y) 5 = y + y y 5 = y y + y + 5 = x x + x

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