MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §6.5 Synthetic Division

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §6.4 → PolyNomial Long Division  Any QUESTIONS About HomeWork §6.4 → HW MTH 55

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 3 Bruce Mayer, PE Chabot College Mathematics StreamLining Long Division  To divide a polynomial by a binomial of the type x − c, we can streamline the usual procedure to develop a process called synthetic division.  Compare the following. In each stage, we attempt to write a bit less than in the previous stage, while retaining enough essentials to solve the problem. At the end, we will return to the usual polynomial notation.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 4 Bruce Mayer, PE Chabot College Mathematics Stage 1: Synthetic Division  When a polynomial is written in descending order, the coefficients provide the essential information.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 5 Bruce Mayer, PE Chabot College Mathematics Leading Coefficient Importance  Because the leading coefficient in the divisor is 1, each time we multiply the divisor by a term in the answer, the leading coefficient of that product duplicates a coefficient in the answer. In the next stage, we don’t bother to duplicate these numbers. We also show where the other +1 is used and drop the first 1 from the divisor.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 6 Bruce Mayer, PE Chabot College Mathematics Stage 2: Synthetic Division Multiply Subtract Multiply Subtract  To simplify further, we now reverse the sign of the 1 in the divisor and, in exchange, add at each step in the long division.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 7 Bruce Mayer, PE Chabot College Mathematics Stage 3: Synthetic Division  The blue numbers can be eliminated if we look at the red numbers instead. Replace the 1 with -1 Multiply Add Multiply Add

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 8 Bruce Mayer, PE Chabot College Mathematics Stage 4: Synthetic Division  Don’t lose sight of how the products −2, 2, and 4 are found. Also, note that the −2 and −4 preceding the remainder 16 coincide with the −2 and −4 following the 2 on the top line. By writing a 2 to the left of the −2 on the bottom line, we can eliminate the top line in stage 4 and read our answer from the bottom line. This final stage is commonly called synthetic division.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 9 Bruce Mayer, PE Chabot College Mathematics Stage 5: Synthetic Division  For (2x 2 − 2x − 4)  (x + 1) then: quotient is 2x 2 − 2x − 4. remainder is 16. This is the remainder. This is the zero-degree coefficient. This is the first-degree coefficient. This is the second-degree coefficient.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 10 Bruce Mayer, PE Chabot College Mathematics Form of the Divisor → x − c  Remember that in order for the Synthetic Division method to work, the divisor must be of the form x – c, that is, a variable minus a constant  Both the coefficient and the exponent of the variable MUST be 1

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 11 Bruce Mayer, PE Chabot College Mathematics Synthetic Division 1.Arrange the coefficients of F(x) in order of descending powers of x, supplying zero as the coefficient of each missing power. 2.Replace the divisor x − c with c. 3.Bring the first (leftmost) coefficient down below the line. Multiply it by c, and write the resulting product one column to the right and above the line.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 12 Bruce Mayer, PE Chabot College Mathematics Synthetic Division 4.Add the product obtained in Step 3 to the coefficient directly above it, and write the resulting sum directly below it and below the line. This sum is the “newest” number below the line. 5.Multiply the newest number below the line by c, write the resulting product one column to the right and above the line, and repeat Step 4

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 13 Bruce Mayer, PE Chabot College Mathematics Example  Synthetic Division  Use synthetic division to divide  SOLUTION by Synthetic Division  Thus the Result Quotient → Remainder → −22  Or

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Synthetic Division  Use synthetic division to divide (x 4 − 9x 3 − 7x )/(x + 5)  SOLUTION: The divisor is x + 5, so write −5 at the left −5−51 99 77 010 55 70     Thus the Algebra ANS

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 15 Bruce Mayer, PE Chabot College Mathematics Long vs. Synthetic  compare the bare essentials for finding the quotient for the Two Division Methods −5−5 Long DivisionSynthetic Division Coefficients of the polynomial c in the divisor, x - c remainder coefficients of the quotient, x+2

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 16 Bruce Mayer, PE Chabot College Mathematics Example  Synthetic Division  Use synthetic division to divide (2x 3 − x 2 − 43x + 60) ÷ (x − 4).  SOLUTION by Synthetic Division Write the 4 of x – 4 and the coefficients of the dividend. Bring down the first coefficient. Multiply 2 by 4 to get 8. Add −1 and 8.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 17 Bruce Mayer, PE Chabot College Mathematics Example  Synthetic Division  SOLUTION by Synthetic Division  The answer is 2x 2 + 7x − 15 with R 0, or just 2x 2 + 7x − 15 Multiply 7 by 4 to get 28. Add −43 and 28. Multiply −15 by 4 to get -60. Add 60 and −60.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 18 Bruce Mayer, PE Chabot College Mathematics Remainder Theorem  Because the remainder is 0, the last example shows that x – 4 is a factor of 2x 3 – x 2 – 43x + 60 and that we can write 2x 3 – x 2 – 43x + 60 as (x – 4)(2x 2 + 7x – 15). Using this result and the principle of zero products, we know that if f(x) = 2x 3 – x 2 – 43x + 60, then f(4) = 0.  In this example, the remainder from the division, 0, can serve as a function value. Remarkably, this pattern extends to nonzero remainders. The fact that the remainder and the function value coincide is predicted by the remainder theorem.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 19 Bruce Mayer, PE Chabot College Mathematics Remainder Theorem The remainder obtained by dividing a PolyNomial f(x) by (x − c) is f(c).  In other words, If a number c is substituted for x in a polynomial f(x), then the result, f(c), is the remainder that would be obtained by dividing f(x) by x − c. That is, if f(x) = (x − c) Q(x) + R, then f(c) = R.

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 20 Bruce Mayer, PE Chabot College Mathematics Example  Find Remainder  Find the remainder when dividing the following polynomial by (x − 1)  SOLUTION: By the Remainder Theorem, F(1) is the remainder  The Remainder is −2

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 21 Bruce Mayer, PE Chabot College Mathematics Example  Evaluate PolyNomial  Find for f(−3) for PolyNomial  Evaluate by straight Substitution  Use The Remainder Theorem If we know R for f(x)  (x − [−3]), then f(−3) is simply R → find R by Synthetic Division

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 22 Bruce Mayer, PE Chabot College Mathematics Example  Evaluate PolyNomial  Find for f(−3) for PolyNomial  Synthetic Division: (x+3) divisor  Then by the Remainder Theorem f(−3) = R = 6 Same Result as by Direct Substitution

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 23 Bruce Mayer, PE Chabot College Mathematics Factor Theorem A polynomial f(x) has (x – c) as a factor if and only if f(c) = 0  In other words, If f(x) is divided by (x − c) and the remainder is 0, then f(c) = 0. This means the c is solution of the equation f(x) = 0 Can use this to VERIFY potential Solns

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 24 Bruce Mayer, PE Chabot College Mathematics Example  Factor Theorem  Show that x = 2 is a solution to  Then find the remaining solutions to this polynomial equation  SOLUTION: If 2 is a solution, f(2) = 0 then (x – 2) is a factor of f(x). Perform synthetic division by 2 and expect Zero Remainder

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Factor Theorem  Synthetic Division  The remainder is indeed zero suggesting that (x−2) divides “evenly” into the original function; i.e. with the Quotient from the synthetic Division;

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Factor Theorem  ReWrite Eqn using (x−2) Factor  Use Factoring and Zero-Products to find the solution associated with the TriNomial

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  Factor Theorem  Thus the Solution for  In Set-Builder form

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 28 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §6.5 Exercise Set 18, 26, 42  A different Type of Synthetic Division

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 29 Bruce Mayer, PE Chabot College Mathematics All Done for Today Another Remainder Theorem

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 30 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 31 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-33_sec_6-5_Synthetic_Division.ppt 32 Bruce Mayer, PE Chabot College Mathematics