Warm up  Divide using polynomial long division:  n 2 – 9n – 22 n+2

Lesson 5-4 Synthetic Division Objective: To use synthetic division to divide polynomials.

Synthetic Division  Used as a shortcut when dividing a polynomial by a first degree binomial.  It is very similar to polynomial division, only using the coefficients and not the variables.

Synthetic Division To divide a polynomial by x – r Example 1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x 3 + 4x 2 – 5x Write r for the divisor, x – r. To the right, write the coefficients of the dividend. 3. Write the leading coefficient of the dividend on the bottom row. Bring down Multiply r (in this case, 3) times the value just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1 To divide a polynomial by x – r Example 1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. x – 3 x 3 + 4x 2 – 5x Write r for the divisor, x – r. To the right, write the coefficients of the dividend. 3. Write the leading coefficient of the dividend on the bottom row. Bring down Multiply r (in this case, 3) times the value just written on the bottom row. Write the 3 product in the next column in the 2nd row. 1 Multiply by 3.

5. Add the values in this new column, writing the sum in the bottom row. 6. Repeat this series of multiplications and additions until all columns are filled in. 7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder. 5. Add the values in this new column, writing the sum in the bottom row. 6. Repeat this series of multiplications and additions until all columns are filled in. 7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder Add Add. Multiply by Add. Multiply by 3. 1x 2 + 7x x – 3 Synthetic Division

Example Use synthetic division to divide 5x 3 + 6x + 8 by x + 2. Solution The divisor must be in the form x – r. Thus, we write x + 2 as x – (-2). This means that c = -2. Writing a 0 coefficient for the missing x 2 - term in the dividend, we can express the division as follows: x – (-2) 5x 3 + 0x 2 + 6x + 8. Now we are ready to set up the problem so that we can use synthetic division Use the coefficients of the dividend in descending powers of x. This is r in x-(-2).

Solution We begin the synthetic division process by bringing down 5. This is following by a series of multiplications and additions. 5. Add: = Add. 7. Add: 8 + (-52) = Add. 4. Multiply: -2(-10) = Multiply: -2(26) = Bring down Add: 0 + (-10) = Add. 2. Multiply: -2(5) = Example cont.

Solution The numbers in the last row represent the coefficients of the quotient and the remainder. The degree of the first term of the quotient is one less than that of the dividend. Because the degree of the dividend is 3, the degree of the quotient is 2. This means that the 5 in the last row represents 5x 2. Thus, x + 2 5x 3 + 0x 2 + 6x + 8 5x 2 – 10x + 26 – x Example cont.

Practice Answer:

The Remainder Theorem  If the polynomial f (x) is divided by x – r, then the remainder is f (r).  Example using synthetic division  3x 2 – 11x + 5 we got a remainder of 9  x – 4  f (4) = 3(4) 2 – 11(4) + 5 = =9

Practice  x+2

Factor Theorem  P(x) has a factor x –r if and only if P(r)=0.  Example:  Is x + 2 a factor of ?  Use the remainder theorem to see if P(-2) = 0  P(-2) = (-2) 3 –(-2) 2 – 2(-2) +8  = -8 –  = 0 therefore x + 2 is a factor.

Practice  Is x – 1 a factor of