1. Dividing Polynomials Long Division Synthetic Division Factor Theorem Remainder Theorem

2. 6.2 Dividing Polynomials Algorithm – a systematic procedure for performing a computation. Linear polynomial – largest degree is 1 Synthetic Division is an algorithm for long division.

3. Examples Dividing Polynomial by a Monomial = = 2ab 3 +4a 4 b 2- 3a 3

4. Simple division (Polynomial)/(Monomial) Try this one using the method explained in the previous slide. This method is only used for dividing when the divisor is a monomial. How about dividing a polynomial by another polynomial? Answer

5. Dividing Polynomials Dividing a Polynomial is called long division, and it is the same skills used when you are dividing the monomials You can use a process similar to long division to divide a polynomial by a polynomial with more than one term. The process is known as the division algorithm

6. Division Algorithm Does Mcdonalds Sell Burgers Dad Mom Sister Brother Divide Multiply Subtract Bring-down

7. Division Algorithm Eliminate the highest exponent by writing the quotient 2x since Subtract and bring down the next term Eliminate the highest exponent here by writing the quotient -2 since Subtract and get the remainder 7 Multiply the 2x by the divisor 2x-1 Multiply the -2 by the divisor 2x-1

8. PRACTICE! Find the remainder of using division algorithm. Answer You should remember writing down in the dividend in order not to overlook the quadratic term. It still exists even though its coefficient is 0 and can ’ t be seen. Remainder

9. Shortcut-Synthetic Division 조립제법 Synthetic division is a shortcut for polynomial division when the divisor is a binomial. In this method, we only use the coefficients of the polynomial that we divide.

10. Process of synthetic division Let the binomial divisor take the form of  Ex) Write down the coefficients of the dividend from the highest degree term to the lowest.  1 4 -3 8 Bring the first coefficient down.

11. Multiply the first coefficient by and write down the multiple just below the second coefficient.  1*3=3 Add the multiple and the second coefficient and write the sum below the multiple.  3+4=7 Next, multiply the sum by and write the multiple below the next(third) coefficient.  7*3=21 Add the multiple and the next(third) coefficient and write the sum below the multiple.  -3+21=18 Continue the process until the last column.  18*3=54, 8+54=62

12. Not clear? Then look at this. 3 | 1 4 -3 8 _______________ 1 3 7 21 18 54 62 (Dividend) (Example previously used)

13. What do the numbers in the below mean then? 3 l 1 4 -3 8 3 21 54 1 7 18 62 Coefficients of the quotient Remainder The power of the first coefficient of the quotient is one less than the degree of the dividend since the divisor is a linear expression.

14. Divide 4x 3 + 0x 2 + x + 7 by (x – 2) ________________ x – 2 ) 4x 3 + 0x 2 + x + 7 Use the additive inverse of -2 as the divisor 2| 4 0 1 7 bring down the 4 (1 st #) + 8 16 34 4 8 17 41 Multiply the number by the 2 (divisor) and add to the next column. Your answer starts with one degree lower than the problem 4x 2 + 8x + 17 + 41 With the last term as remainder x-2

15. What if the binomial divisor isn’t in the form of (x-a)? In this situation, we find the value that makes the binomial divisor 0. We use that value and later multiply the coefficients of the quotient (without the remainder) by a certain number to make the answer true. Let’s take an example! Use ½ here since it makes the divisor 0 when substituted.

16. ½ l 2 -3 4 -6 1 -1 3/2 2 -2 3 -9/2  As a result, the remainder remains the same but the coefficients of the quotient are multiplied by ½ since synthetic division was done by multiplying ½ to the divisor.

17. How does it work? Then, why is it all right to use this method? For further development let’s look at the proof of synthetic division. When polynomial P(x) is divided by (x-k) and gets the quotient Q(x) and remainder R, it can be expressed this way.

18. If we let P(x) a cubic expression, we can generalize it as all polynomials, so we can let P(x) be a cubic expression. Since Q(x) is a quadratic expression, it can be expressed this way. Then, Synthetic division

19. k l a b c d kp kq kr a b+kp c+kq d+kr p q r R

20. Practice - Divide f(x) by g(x)

22. How do we do synthetic division when the divisor is a quadratic expression? Here is one method although there are others.

23. Divide x 4 + 8x 3 + 15x 2 + 4x + 1 by x 2 + 3x + 2, using synthetic division: Coefficients of dividend Opposite of b term and opposite of c term Add the next column -10 add down and repeat steps

24. Remainder Theorem For a polynomial P(x), the function value P( r) is the remainder when P(x) is divided by x-r. For P(x) = x 3 +8x 2 +8x-32 Find P(1) P(1) = 1 + 8 + 8 -32 = -15 Therefore is we divide P(x) by x-1, we would get a remainder of -15.

25. Factor Theorem A polynomial f(x) has a factor x-k iff f(k)=0 if and only if (shortened iff) is a biconditional logical connective between statements; either both statements are true, or both are false

26. Factor Theorem For a polynomial P(x), if P(r) = 0, then the polynomial x-r is a factor of P(x). For example P(x) = x 3 +8x 2 +8x-32 P(-4) = 0 and if we divide by x-(-4), x+4, we will get 0 as a remainder (no remainder). Therefore, x+4 is a factor of x 3 +8x 2 +8x-32

27. Zeros x-intercepts are when the function equals 0, the y-value is 0. a(x-p)(x-q)=0 means the x-intercepts are at (p,0),(q,0). p and q are called zeros of the function.

28. Find the other zeros of f(x) given that f(-2)=0

29. Find the missing dimension x+2 x+4 2x+5