Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Chabot Mathematics §6.5 SyntheticDivision Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. MTH 55 6.4 Review § • Any QUESTIONS About • §6.4 → PolyNomial Long Division • Any QUESTIONS About HomeWork • §6.4 → HW-26

3. 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.

4. Stage 1: Synthetic Division • When a polynomial is written in descending order, the coefficients provide the essential information.

5. 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.

6. Multiply Subtract Multiply Subtract Multiply Subtract Stage 2: Synthetic Division • 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.

7. Multiply Add Multiply Add Multiply Add Stage 3: Synthetic Division Replace the 1 with -1 • The blue numbers can be eliminated if we look at the red numbers instead.

8. 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.

9. This is the remainder. This is the zero-degree coefficient. This is the first-degree coefficient. This is the second-degree coefficient. Stage 5: Synthetic Division • For (2x2− 2x − 4)(x + 1) then: • quotient is 2x2− 2x− 4. • remainder is 16.

10. 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

11. Synthetic Division • Arrange the coefficients of F(x) in order of descending powers ofx, supplying zero as the coefficient of each missing power. • Replace the divisor x−c with c. • 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.

12. Synthetic Division • 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. • 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

13. Example  Synthetic Division • Use synthetic division to divide • SOLUTIONby SyntheticDivision • Thus the Result • Quotient → • Remainder → −22 • Or

14. Example  Synthetic Division • Use synthetic division to divide (x4− 9x3− 7x2 + 10)/(x + 5) • SOLUTION: The divisor is x + 5, so write −5 at the left • Thus theAlgebra ANS

15. Long vs. Synthetic • compare the bare essentials for finding the quotient for the Two Division Methods Long Division Synthetic Division Coefficients of the polynomial c in the divisor, x - c −5 remainder coefficients of the quotient, x+2

16. Example  Synthetic Division • Use synthetic division to divide (2x3−x2− 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.

17. Example  Synthetic Division • SOLUTION by Synthetic Division Multiply 7 by 4 to get 28. Add −43 and 28. Multiply −15 by 4 to get -60. Add 60 and −60. • The answer is 2x2 + 7x− 15 with R 0, or just 2x2 + 7x− 15

18. Remainder Theorem • Because the remainder is 0, the last example shows that x – 4 is a factor of 2x3 – x2 – 43x + 60 and that we can write 2x3 – x2 – 43x + 60 as (x – 4)(2x2 + 7x – 15). Using this result and the principle of zero products, we know that if f(x) = 2x3 – x2 – 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.

19. 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.

20. 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

21. 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

22. Example  Evaluate PolyNomial • Find for f(−3) for PolyNomial • SyntheticDivision:(x+3) divisor • Then by the Remainder Theorem f(−3) = R = 6 • Same Result as by Direct Substitution

23. 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

24. 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

25. Example  Factor Theorem • SyntheticDivision • The remainder is indeed zero suggesting that (x−2) divides “evenly” into the original function; i.e. with the Quotient from the synthetic Division;

26. Example  Factor Theorem • ReWrite Eqn using(x−2) Factor • Use Factoring and Zero-Products to find the solution associated with the TriNomial

27. Example  Factor Theorem • Thus the Solution for • In Set-Builder form

28. WhiteBoard Work • Problems From §6.5 Exercise Set • 18, 26, 42 • A different Typeof SyntheticDivision

29. All Done for Today AnotherRemainderTheorem

30. Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

31. Graph y = |x| • Make T-table