# Polynomials. Jackie Nicholas Jacquie Hargreaves Janet Hunter

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1 Mathematics Learning Centre Polnomials Jackie Nicholas Jacquie Hargreaves Janet Hunter c 26 Universit of Sdne

2 Mathematics Learning Centre, Universit of Sdne 1 1 Polnomials Man of the functions we will eamine will be polnomials. In this Chapter we will stud them in more detail. Definition A real polnomial, P (), of degree n is an epression of the form P () =p n n + p n 1 n 1 + p n 2 n p p 1 + p where p n,p, p 1,, p n are real and n is an integer. All polnomials are defined for all real and are continuous functions. We are familiar with the quadratic polnomial, Q() =a 2 + b + c where a. This polnomial has degree 2. The function f() = + is not a polnomial as it has a power which is not an integer and so does not satisf the definition. 1.1 Polnomial equations and their roots If, for a polnomial P (), P (k) = then we can sa 1. = k is a root of the equation P () =. 2. = k is a zero of P (). 3. k is an -intercept of the graph of P () Zeros of the quadratic polnomial The quadratic polnomial equation Q() =a 2 + b + c = has two roots that ma be: 1. real (rational or irrational) and distinct, 2. real (rational or irrational) and equal, 3. comple (not real). We can determine which one of these we have if we use the quadratic formula = b ± b 2 4ac. 2a If b 2 4ac >, there will be two real distinct roots. If b 2 4ac =, there will be two real equal roots. If b 2 4ac <, there will be comple roots. We will illustrate all of these cases with eamples, and will show the relationship between the nature and number of zeros of Q() and the -intercepts (if an) on the graph.

3 Mathematics Learning Centre, Universit of Sdne 2 1. Let Q() = We find the zeros of Q() b solving the equation Q() = = ( 1)( 3) = Therefore = 1 or 3. The roots are rational (hence real) and distinct Let Q() = Solving the equation Q() = we get, = = 4 ± Therefore = 2± 7. 2 The roots are irrational (hence real) and distinct Let Q() = Solving the equation Q() = we get, = ( 2) 2 = Therefore = 2. 3 The roots are rational (hence real) and equal. Q() = has a repeated or double root at =2. 1 Notice that the graph turns at the double root =2.

4 Mathematics Learning Centre, Universit of Sdne 3 4. Let Q() = Solving the equation Q() = we get, = = 4 ± Therefore = 2± 4. There are no real roots. In this case the roots are comple. 3 1 Notice that the graph does not intersect the -ais. That is Q() > for all real. Therefore Q is positive definite. We have given above four eamples of quadratic polnomials to illustrate the relationship between the zeros of the polnomials and their graphs. In particular we saw that: i. if the quadratic polnomial has two real distinct zeros, then the graph of the polnomial cuts the -ais at two distinct points; ii. if the quadratic polnomial has a real double (or repeated) zero, then the graph sits on the -ais; iii. if the quadratic polnomial has no real zeros, then the graph does not intersect the -ais at all. So far, we have onl considered quadratic polnomials where the coefficient of the 2 term is positive which gives us a graph which is concave up. If we consider polnomials Q() =a 2 + b + c where a< then we will have a graph which is concave down. For eample, the graph of Q() = ( ) is the reflection in the -ais of the graph of Q() = The graph of Q() = The graph of Q() = ( ).

5 Mathematics Learning Centre, Universit of Sdne Zeros of cubic polnomials A real cubic polnomial has an equation of the form P () =a 3 + b 2 + c + d where a,a, b, c and d are real. It has 3 zeros which ma be: i. 3 real distinct zeros; ii. 3 real zeros, all of which are equal (3 equal zeros); iii. 3 real zeros, 2 of which are equal; iv. 1 real zero and 2 comple zeros. We will not discuss these here. 1.2 Factorising polnomials So far for the most part, we have looked at polnomials which were alread factorised. In this section we will look at methods which will help us factorise polnomials with degree > Dividing polnomials Suppose we have two polnomials P () and A(), with the degree of P () the degree of A(), and P () is divided b A(). Then P () A() = Q()+R() A(), where Q() is a polnomial called the quotient and R() is a polnomial called the remainder, with the degree of R() < degree of A(). We can rewrite this as P () =A() Q()+R(). For eample: If P () = and A() = 2, then P () can be divided b A() as follows: The quotient is and the remainder is 21. We have = This can be written as =( 2)( )+21. Note that the degree of the polnomial 21 is.

6 Mathematics Learning Centre, Universit of Sdne The Remainder Theorem If the polnomial f() is divided b ( a) then the remainder is f(a). Proof: Following the above, we can write f() =A() Q()+R(), where A() =( a). Since the degree of A() is 1, the degree of R() is zero. That is, R() =r where r is a constant. f() = ( a)q()+r where r is a constant. f(a) = Q(a)+r = r So, if f() is divided b ( a) then the remainder is f(a). Eample Find the remainder when P () = is divided b a. +1,b Solution a. Using the Remainder Theorem: Remainder = P ( 1) = 3 ( 1) 3 1 = 27 b. Remainder = P ( 1 2 ) = 3( 1 2 )4 ( 1 2 )3 + 3( 1 2 ) 1 = = Eample When the polnomial f() is divided b 2 4, the remainder is What is the remainder when f() is divided b ( 2)? Solution Write f() =( 2 4) q()+(5 + 6). Then Remainder = f(2) = q(2)+16 = 16 A consequence of the Remainder Theorem is the Factor Theorem which we state below.

7 Mathematics Learning Centre, Universit of Sdne The Factor Theorem If = a is a zero of f(), that is f(a) =, then ( a) is a factor of f() and f() ma be written as f() =( a)q() for some polnomial q(). Also, if ( a) and ( b) are factors of f() then ( a)( b) is a factor of f() and for some polnomial Q(). f() =( a)( b) Q() Another useful fact about zeros of polnomials is given below for a polnomial of degree The cubic polnomial revisited If a (real) polnomial P () =a 3 + b 2 + c + d, where a,a, b, c and d are real, has eactl 3 real zeros α, β and γ, then P () =a( α)( β)( γ) (1) Furthermore, b epanding the right hand side of (1) and equating coefficients we get: i. ii. iii. α + β + γ = b a ; αβ + αγ + βγ = c a ; αβγ = d a. This result can be etended for polnomials of degree n. Eample Let f() = a. Factorise f(). b. Sketch the graph of = f(). c. Solve f().

8 Mathematics Learning Centre, Universit of Sdne 7 Solution a. Consider the factors of the constant term, 2. We check to see if ±1 and ±2 are solutions of the equation f() = b substitution. Since f(2) =, we know that ( 2) is a factor of f(). We use long division to determine the quotient So, f() = ( 2)(4 2 1) = ( 2)(2 1)(2 +1) b The graph of f() = c. f() when or 2. Eample Show that ( 2) and ( 3) are factors of P () = , and hence solve =. Solution P (2)= = and P (3) = = so ( 2) and ( 3) are both factors of P () and ( 2)( 3) = is also a factor of P (). Long division of P () b gives a quotient of ( + 5). So, P () = =( 2)( 3)( +5).

9 Mathematics Learning Centre, Universit of Sdne 8 Solving P () = we get ( 2)( 3)( +5)=. That is, =2or =3or = 5. Instead of using long division we could have used the facts that i. the polnomial cannot have more than three real zeros; ii. the product of the zeros must be equal to 3. Let α be the unknown root. Then 2 3 α = 3, so that α = 5. Therefore the solution of P () = = is =2or =3or = Eercises 1. When the polnomial P () is divided b ( a)( b) the quotient is Q() and the remainder is R(). a. Eplain wh R() is of the form m + c where m and c are constants. b. When a polnomial is divided b ( 2) and ( 3), the remainders are 4 and 9 respectivel. Find the remainder when the polnomial is divided b c. When P () is divided b ( a) the remainder is a 2. Also, P (b) =b 2. Find R() when P () is divided b ( a)( b). 2. a. Divide the polnomial f() = bg() = Hence write f() = g()q() + r() where q() and r() are polnomials. b. Show that f() and g() have no common zeros. (Hint: Assume that α is a common zero and show b contradiction that α does not eist.) 3. For the following polnomials, i. factorise ii. solve P () =. a. P () = b. P () = c. P () = d. P () = e. P () = Solutions 1. a. Since A() =( a)( b) is a polnomial of degree 2, the remainder R() must be a polnomial of degree < 2. So, R() is a polnomial of degree 1. That is, R() =m + c where m and c are constants. Note that if m = the remainder is a constant.

10 Mathematics Learning Centre, Universit of Sdne 9 b. Let P () =( )Q()+(m + c) =( 2)( 3)Q()+(m + c). Then and P (2) = ()( 1)Q(2) + (2m + c) = 2m + c = 4 P (3) = (1)()Q(3) + (3m + c) = 3m + c = 9 Solving simultaneousl we get that m = 5 and c = 6. R() =5 6. c. Let P () =( a)( b)q()+(m + c). Then So, the remainder is P (a) = ()(a b)q(a)+(ma + c) = am + c = a 2 and P (b) = (b a)()q(b)+(mb + c) = bm + c = b 2 Solving simultaneousl we get that m = a + b and c = ab provided a b. So, R() =(a + b) ab. 2. a =( )( ) 2 b. Let α be a common zero of f() and g(). That is, f(α) =andg(α) =. Then since f() = g()q()+ r() we have f(α) = g(α)q(α)+r(α) = ()q(α)+ r(α) since g(α) = = r(α) = since f(α) = But, from part b. r() = 2 for all values of, so we have a contradiction. Therefore, f() and g() do not have a common zero. This is an eample of a proof b contradiction. 3. a. i. P () = =( + 1)( + 2)( 4)

11 Mathematics Learning Centre, Universit of Sdne 1 ii. = 1, = 2 and = 4 are solutions of P () =. b. i. P () = = ( +2) 2 ( 5). ii. = 2 and = 5 are solutions of P () =. = 2 is a double root. c. i. P () = =(+2)( ) = (+2)( ( 1+ 5))( ( 1 5)) ii. = 2, = 1+ 5 and = 1 5 are solutions of P () =. d. i. P () = =( 2)( ) = has no real solutions. ii. = 2 is the onl real solution of P () =. e. i. P () = =( + 2)( 3)(2 1). ii. = 2, = 1 and = 3 are solutions of P () =. 2

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