MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §6.4 Divide PolyNomials

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §6.3 → Complex Rational Expressions  Any QUESTIONS About HomeWork §6.3 → HW MTH 55

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 3 Bruce Mayer, PE Chabot College Mathematics §6.4 Polynomial Division  Dividing by a Monomial  Dividing by a BiNomial  Long Division

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 4 Bruce Mayer, PE Chabot College Mathematics Dividing by a Monomial  To divide a polynomial by a monomial, divide each term by the monomial.  EXAMPLE – Divide: x x 4 − 12x 3 by 6x  Solution

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 5 Bruce Mayer, PE Chabot College Mathematics Example  Monomial Division  Divide:  Solution:

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 6 Bruce Mayer, PE Chabot College Mathematics Dividing by a Binomial  For divisors with more than one term, we use long division, much as we do in arithmetic.  Polynomials are written in descending order and any missing terms in the dividend are written in, using 0 (zero) for the coefficients.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 7 Bruce Mayer, PE Chabot College Mathematics Recall Arithmetic Long Division  Recall Whole-No. Long Division Divide: Divisor Quotient Remainder Quotient 13 Divisor 12 Remainder 1 = Dividend =

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 8 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor The quotient will be here. 2x³ + 3x² - x + 1 is the dividend  Use an IDENTICAL Long Division process when dividing by BiNomials or Larger PolyNomials; e.g.;

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 9 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step First divide the first term of the dividend, 2x³, by x (the first term of the divisor). This gives 2x². This will be the first term of the quotient.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 10 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Now multiply (x+2) by 2x² and subtract

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 11 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Bring down the next term, -x.-x.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 12 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Now divide –x², the first term of –x² - x, by x, the first term of the divisor which gives –x.–x.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 13 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Multiply (x +2) by -x and subtract

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 14 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Bring down the next term, 1

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 15 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Divide x, the first term of x + 1, by x,x, the first term of the divisor which gives 1

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 16 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step Multiply x + 2 by 1 and subtract

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 17 Bruce Mayer, PE Chabot College Mathematics Binomial Div.  Step by Step The remainder is –1. The quotient is 2x² - x + 1

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  BiNomial Division  Divide x 2 + 7x + 12 by x + 3.  Solution Subtract by changing signs and adding Multiply (x + 3) by x, using the distributive law

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  BiNomial Division  Solution – Cont. Subtract Multiply 4 by the divisor, x + 3, using the distributive law Bring Down the +12

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  BiNomial Division  Divide 15x 2 − 22x + 14 by (3x − 2)  Solution  The answer is 5x − 4 with R6. We can also write the answer as:

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  BiNomial Division  Divide x 5 − 3x 4 − 4x x by (x − 3)  Solution  The Result

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 22 Bruce Mayer, PE Chabot College Mathematics Divide 3x 2 − 4x − 15 by x − 3  SOLUTION: Place the TriNomial under the Long Division Sign and start the Reduction Process Divide 3x 2 by x: 3x 2 /x = 3x. Multiply x – 3 by 3x. Subtract by mentally changing signs and adding − 4x + 9x = 5x.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 23 Bruce Mayer, PE Chabot College Mathematics Divide 3x 2 − 4x − 15 by x − 3  SOLUTION: next divide the leading term of this remainder, 5x, by the leading term of the divisor, x. Divide 5x by x: 5x/x = 5. Multiply x – 3 by 5. Subtract. Our remainder is now 0.  CHECK: (x − 3)(3x + 5) = 3x 2 − 4x − 15  The quotient is 3x + 5.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 24 Bruce Mayer, PE Chabot College Mathematics Formal Division Algorithm  If a polynomial F(x) is divided by a polynomial D(x), with D(x) ≠ 0, there are unique polynomials Q(x) and R(x) such that F(x) = D(x) Q(x) + R(x) Dividend Divisor Quotient Remainder  Either R(x) is the zero polynomial, or the degree of R(x) is LESS than the degree of D(x).

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 25 Bruce Mayer, PE Chabot College Mathematics PolyNomial Long Division 1.Write the terms in the dividend and the divisor in descending powers of the variable. 2.Insert terms with zero coefficients in the dividend for any missing powers of the variable 3.Divide the first terms in the dividend by the first terms in the divisor to obtain the first term in the quotient.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 26 Bruce Mayer, PE Chabot College Mathematics PolyNomial Long Division 4.Multiply the divisor by the first term in the quotient, and subtract the product from the dividend. 5.Treat the remainder obtained in Step 4 as a new dividend, and repeat Steps 3 and 4. Continue this process until a remainder is obtained that is of lower degree than the divisor.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  TriNomial Division  Divide  SOLN

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 28 Bruce Mayer, PE Chabot College Mathematics Example  TriNomial Division  SOLN cont.

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 29 Bruce Mayer, PE Chabot College Mathematics Example  TriNomial Division  Divide  The Quotient =  The Remainder =  Write the Result in Concise form:

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 30 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §6.4 Exercise Set 30, 32, 40  BiNomial Division

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 31 Bruce Mayer, PE Chabot College Mathematics All Done for Today Polynomial Division in base2 From UC Berkeley Electrical-Engineering 122 Course

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 32 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 33 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-31_sec_6-3_Complex_Rationals.ppt 34 Bruce Mayer, PE Chabot College Mathematics