A26 MATHEMATICS SUPPORT CENTRE

1 MATHEMATICS SUPPORT CENTRE ,Coventry University, 2001 MATHEMATICS SUPPORT CENTRE Title: Remainder Theorem and Factor Theorem Target: On completion of this worksheet you should be able to use the remainder and factor theorems to find factors of polynomials. A26 Generally when a polynomial is divided by a linear expression there is a remainder. 5)123)(2()3543(52342526312335432)2()3543 ( + +=+ + + + + +++ + +xxxxxxxxxxxxxxxxxxxxxxxx Any polynomial can be written in the following form: polynomial divisor quotient + remainder.

2 In particular if the divisor is (x a) and the polynomial is f (x) then f (x) (x a) quotient + remainder. If x = a then f (a) = (a a) quotient + remainder. f (a) = remainder This gives an easy way of finding the remainder when a polynomial is divided by (x a) Examples 1. Using previous example 5)123)(2(3543)(223+ +=+ +=xxxxxxxf Now using x = -2 55)1)2(2)2(3(05)1)2(2)2(3)(22()2(22=+ =+ + = f the remainder. 2. Find the remainder when 5311512)1(0)1( require wesince 1 Substitute352)(Let )1(by divided is )352(232323 = + == = + = + fxxxxxxfxxxx The remainder is 5 3.

3 Find the remainder when 1017)3(3)3(4)3()3(0)3( require wesince 3 Substitute734)(Let )3(by divided is )734(242424= + + = =+ = ++=+ ++fxxxxxxfxxxx The remainder is 101 divisor quotient remainder Exercise Find the remainders for the following: )4()4(.5)3()562(.4)1()132(.3)1()234(.2)2 ()465(.12332342323 + + + +++ + xxxxxxxxxxxxxxxxxxx (Answers: -4, 10, 5, -29, -4) MATHEMATICS SUPPORT CENTRE ,Coventry University, 2001 Example Find the remainder when 02151413)1(1 and 2543)(Let )1(by divided is )2543(232323= + == += +fxxxxxfxxxx The remainder is 0. quotient )1(0quotient )1(254323 =+ = +xxxxx so )1( xis a factor of )2543(23 +xxx We can use the remainder theorem to check for factors of a polynomial.

4 As before remainder quotient )()(+ =axxf and remainder )(=af If )(ax is a factor then the remainder is 0 ie 0)(=af This is called the factor theorem. We can use the factor theorem to factorise polynomials, although some trial and error is involved. Examples 1. Is )3( xa factor of )3832(23 xxx? Let 3 and )3832()(23= =xxxxxf as we are checking whether )3( xis a factor. 03383332)3(23= =f so )3( xis a factor of )3832(23 xxx 2. Is )1( xa factor of )3832(23 xxx? Using f(x) as above and x = 1 0123181312)1(23 = =f so )1( xis not a factor of )3832(23 xxx Exercise ?)6103()( offactor a)1( )6103()( offactor a)3( )10134()( offactor a)2( )10134()( offactor a)2( )122()(offactor a)1( ++= ++=+++= ++=+ += xxxxfxxxxxfxxxxfxxxxfxxxxxfx (Answers: yes, yes, no, yes, no) Example Factorise )652(23 +xxx.

5 Let 652)(23 +=xxxxf. Since the constant is 6 we will consider factors of this ie. 1, 2, 3, 6. We will try )1( x 0611512)1(23= + =f so )1( xis a factor. Now we can find the quadratic factor by division or by repeating the above. )32)(2)(1()672)(1(652)(0666677722672652) 1(2232223223++ =++ = += ++ + xxxxxxxxxxfxxxxxxxxxxxxxx The quadratic factor is factorised in the normal way. Exercise Factorise the following: 617154)(.5263)(.4433)(.3122)(.2652)(.123 23232323 + =+++= = += +=xxxxfxxxxfxxxxfxxxxfxxxxf Answers: )34)(2)(1(.5)13)(2(.4)1)(4(.3)12)(1)(1(. 2)3)(2)(1(.122 ++++ + ++ +xxxxxxxxxxxxxx