MATHEMATICS 2 FOR SENIOR HIGH SCHOOL CHAPTER 5 POLYNOMIAL Algebraic forms , such as a linear equation and a quadratic equation are kinds of polynomial Equation of degree 1 and 2. Consider the following algebraic forms: a. b. c. 10 7 42 3+ + x x x5 3 2 52 3 4 + x x x x3 10 2 3 62 3 4 5 + + + x x x x xSeveral polynominal equations can be used in a statistical calculation to estimate The relationship between two variables. A Definition The general form of polynomial in variabel xof degree n is : The polynomials are arranged based on descending order of the exponent of x,where : 0 12211.... ) ( a x a x a x a x a x Pnnnnnn+ + + + + == 1 1,...., , a a an nReal coefficient of the polynominal and Real constant Degree of the polynominal, n s whole number 0 =na=0a= nGeneral Aspects of Polynomial Polynominal Equation Definition Suppose given polynominal off(x) and g(x), where 0 12221... ) ( a x a x a x a x a x fnnnn+ + + + + =0 12221... ) ( b x b x b x b x b x gnnnn+ + + + + =) (x f is equal to) (x g written) ( ) ( x g x f if satisfies : 0 0 1 1 2 2 2 2 1 1, , ,..., , , b a b a b a b a b a b an n n n n n= = = = = = is determined by subtitusing the value ofinto the variable ofon the polynominal. ExampleDetermine the value of polynomial for and Answer : For For Value of a polynominal Determining the Value of a polynominal by subtitution method Using this method, the value of the polynominal offor, written) (x pk x =) (k pkx3 10 4 3 2 ) (2 3 4 5+ + + = x x x x x x P1 = x 2 = x3 10 4 3 2 ) (2 3 4 5+ + + = x x x x x x P3 ) 1 ( 10 ) 1 ( 4 ) 1 ( 3 ) 1 ( 2 ) 1 ( ) 1 ( 12 3 4 5+ + + = = P x1 3 10 4 3 2 1 = + + + =3 ) 2 ( 10 ) 2 ( 4 ) 2 ( 3 ) 2 ( 2 ) 2 ( ) 2 ( 22 3 4 5+ + + = = P x49 3 20 16 24 32 32 = + + + =B 2. Determiningthe value of a polynomial by schene (chart) 1-2 3 4-103 1 -1 2 6-4 +

1-1 2 6-4-1 Thus, the value of P(1)=-1. Description Symbolmean multiply by the input number (in this case input number = 1 since we will find the value of the polynominal, P(x), with). 3 10 4 3 2 ) (2 3 4 5+ + + = x x x x x x P1 = x1 = xPolynomial Dividing Definition Consider the following expression. The expression above is a division of polynomial, where the dividend,the divisor,the quotient, and the remainder. ) ( ) ( ). ( ) ( x S x H x Q x P + == ) (x P= ) (x Q = ) (x H= ) (x SC 1. Long division Example: Determine the quotient and the remainder of the division ofby Answer : has degree of 3,has degree of 1, and the quotient of has degree of 3-1=2, the remainder of the divisionhas degree of 1-1=0. the quotient The divisor x+3 the dividend

the remainder Using this method, we get the quontient of , and the remainder 20 4 5 ) (2 3 + = x x x x P 3 + x) (x p ) (x Q ) (x H) (x S2 32 32320 4 510 2x xx x xx x+ + +x xx x6 24 222+30 1020 10 xx1010 2 ) (2 + = x x x H10 ) ( = x S2. Synthetic Division( horner Method ) This method can solve several form of division of polynomial. a. Dividing polynomial byTo understaind it, consedier the division of polynomial ofbybelow. ) ( k x c bx ax x P + + =2) () ( k x akx axc bx axak b ax+ + + +22) () ( k x 2) () (ak bk x ak bc x ak b ++ +c bk ak + +2Next, inspect the determining of the value of polynomial for a b c + a= the remainder the coefficient quotient c bx ax x P + + =2) ( k x =c bx ax + +2ak2ak bk +) ( ak b +c bk ak + +2k x =The general from of diving a polynomial by with the quontientand theremainder is: The general from of diving a polynomial by with the quontient of and the remainderof is Supposethen get Based on the fact above we have the following things. 1) The polynomial ofdivided by is resulting, with is the quotient of dividing ofby 2)The remainder of dividing ofby is the same as the remainder of dividing it by

b. Dividing a Polynomial by (ax-b) ) (x H) (x S) ( ) ( ). ( ) ( x S x H b ax x P + =) ( k x ) ( k x ) ( ) ( ). ( ) ( x s x H k x x P + =abk =) () () ( ) ( ) () () ( ) ( . ) ( x Sax Hb ax x s x Hab axx S x Habx x P + = += +|.|

\| =) (x Pax H ) () (x H|.|

\|abx) ( b ax |.|

\|abx) (x H) (x S) (x P) (x P) ( b ax The general form : With and are consecutively the quotient and the remainder of the division. Supposecan be factorized to betherefore,then: The steps to determine the quotient and the remainder of the division of a polynomial are: 1) Divided the polynomial ofby to getandas the quotient and the remainder. 2) Divide by to getandas the quotient and remainder. 3) The quotient of dividing by is, and the remainder is.c. Dividing Polynomial by with) (2c bx ax + +0 = a) ( ) ( ) ( ) (2x S x H c bx ax x P + + + =) (x H ) (x S) (2c bx ax + +2 1P P 2 12P P c bx ax = + +) ( ) ( ) ( ) ( ). ( ) (2 12x S x H P P x S x H c bx ax x P + = + + + =) (x P1P ) (0x H1S2P ) (x H2S) ) ( (2 1P P x Q =1 2 1) ( S S P x S + =) (0x H) (x P ) (x HTheorem If polynominal ofwith degree of n divided by, then the remainder is Proof: For Then, the remainder of. It is proved.

Remainder Theorem 1. Division by) ( k x ) (x f) ( k x ) (k f S =S x H k x x f + = ) ( ) ( ) (0 = k xS k H k k k f k x + = = ) ( ) ( ) (S k fS k f=+ =) (0 ) () (k f S =D Theorem If polinomial of with degree of n divided by ,then the remainder is Example: Determine the remainder if divided by Answers : The remainder is the value of which causes the divisor value is zero, 3x + 2 = 0 a. By subtituting the remainder =b. By Horner method 6-2-17 -44-2+ 6-635 So, the remainder = 5 2. Division by ) ( b ax ) (x f ) ( b ax |.|

\|=abf S7 2 6 ) (2 3+ = x x x x f ) 2 3 ( + x) (x f32 = x5 73298916732322326322 3= + + = +|.|

\| |.|

\| |.|

\| =|.|

\| f32 = x3. Division by Dividing polynomial of bywith the quotient of and theremainder ofis: Since the degree of the divisor is 2, the degree of S is at most 1. let S = ,then the division can written as follows. ) (x f ) )( ( b x a x ) (x H) (x S) ( ) ( ) )( ( ) ( x S x H b x a x x f + =) ( q px +) ( ) ( ) )( ( ) ( q px x H b x a x x f + + =) )( ( b x a x Theorem is a factor from polynomial ofIf and only if Proof: According to the remainder theorem, polynomial of is divided bythen the remainded is such that base form of the equation is : , whereis the quotient. (i) Based on the equation above ifthen. In other wordsis a factor of (ii) If is a factor of then : for any such that From (i) and (ii) we haveif and only if is a factory of Therefore, the division of a polynomial by one of its factory resulted a remainderwhich is equal to zero. Factorization theorem ) ( k x ) (x f . 0 ) ( = k f) (x f) ( k x ) (k f) ( ) ( ) ( ) ( k f x H k x x f + =) ( ) ( ) ( x H k x x f =) ( k x ) (x f) ( ) ( ) ( x H k x x f =) (x H0 ) ( ) ( ) ( = = x H k k k f0 ) ( = k f) ( k x ) (x f) (x H0 ) ( = k f) ( k x ) (x fE Show that is a factor of Answer : Using method of subtitution, it will be : Since , thenis a factor of There are several aspects that must be considered when factorizing a polynomial. 1. If the sum of all coefficients is equal to zero, thenis a factor of the polynomial. 2. If the sum of all coefficients of variable with an even exponent is equal to the sum of all coefficients of variable with an odd exponent, then is a factor of polynomial. Example ) 4 ( x 4 3 5 9 2 ) (2 3 4 + = x x x x x f4 0 4 = = x x4 ) 4 ( 3 ) 4 ( 5 ) 4 ( 9 ) 4 ( 2 ) 4 (2 3 4 + = f4 3 5 9 2 ) (2 3 4 + = x x x x x f04 12 80 576 512= + =0 ) 4 ( = f) 4 ( x 4 3 5 9 2 ) (2 3 4 + = x x x x x f) 1 ( x) 1 (+ xRoots of a Polynomial equation 1. Definition of roots of a polynomial equation Based the teorem of factorization we have proved thatis a factor of if and only if.Similarly, we can define the root of a polynomial equation. Theorem is the root of a polynomial equation of if and only if ) ( k x ) (x f0 ) ( = k f) ( k x = 0 ) ( = x f0 ) ( = k fE Example If 2 is the root of the equation of, determine the other roots! Answer : LetX=21-7412 2-10 -12 + 1-5-60 The quotient of the division of byis, then Thus, the roots ofis , and 0 12 4 72 3= + + x x x12 4 7 ) (2 3+ + = x x x x f) (x f2 x6 52 x x) 1 )( 6 )( 2 ( ) 6 5 )( 2 ( ) (2+ = = x x x x x x x f0 ) ( = x f 6 , 2 = = x x 1 = xa. If andare the roots of, then 1) 2) 3) b. If andare the roots of four degree 1) 2) 3) 4) 2. Rational Roots of a Polynomial 2 1, x x3x02 3= + + + d cx bx axabx x x = + +3 2 1acx x x x x x = + +3 2 3 1 2 1. . .adx x x =3 2 1. .3 2 1. . x x x4x02 3 4= + + + + dx cx bx axabx x x x = + + +2 3 2 1acx x x x x x x x x x x x = + + + + +4 3 4 2 3 2 4 1 3 1 2 1. . . . . .adx x x x x x x x x x x x = + + +2 1 4 1 4 3 4 3 2 3 2 1. . . . . . . .. . . .4 3 2 1ax x x x=