§ 6.5 Synthetic Division and the Remainder Theorem

Synthetic Division of Polynomials We can use a process called synthetic division to divide polynomials if the divisor is of the form x - c . This method is just a shortcut for long division. With this process -we can save both steps and paper by writing down only what is necessary from the long division problem and by also compacting the form. Synthetic division is quick and can be extremely useful. Note that this process works only when you can express the divisor in the form x – c. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.5

To divide a polynomial by x – c: Synthetic Division of Polynomials Synthetic Division To divide a polynomial by x – c: STEPS EXAMPLES 1) Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. 2) Write c for the divisor, x – c. To the right, write the coefficients of the dividend. 3) Write the leading coefficient of the dividend on the bottom row. 4) Multiply c (in this case, 3) times the value just written on the bottom row. Write the product in the next column in the second row. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.5

To divide a polynomial by x – c: Synthetic Division of Polynomials CONTINUED Synthetic Division To divide a polynomial by x – c: STEPS EXAMPLES 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, plus the remainder above the divisor. 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 this row is the remainder. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.5

Synthetic Division of Polynomials EXAMPLE Use synthetic division to divide SOLUTION The divisor must be in the form x – c. Thus, we write 6 + x as x – (-6). This means that c = -6. Rearranging the terms of the dividend in descending order and writing a 0 for the missing constant-term we can express the division as follows: Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.5

Synthetic Division of Polynomials EXAMPLE Use synthetic division to divide SOLUTION The divisor must be in the form x – c. Thus, we write 6 + x as x – (-6). This means that c = -6. Rearranging the terms of the dividend in descending order and writing a 0 for the missing constant-term we can express the division as follows: Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.5

Synthetic Division of Polynomials EXAMPLE Use synthetic division to divide SOLUTION The divisor must be in the form x – c. Thus, we write 6 + x as x – (-6). This means that c = -6. Rearranging the terms of the dividend in descending order and writing a 0 for the missing constant-term we can express the division as follows: Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.5

Synthetic Division of Polynomials CONTINUED 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 c in x – (-6). We begin the synthetic division process by bringing down 1. This is followed by a series of multiplications and additions. 1) Bring down 1. 2) Multiply: -6(1) = -6. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.5

Synthetic Division of Polynomials CONTINUED 3) Add: -6 + -6 = -12. 4) Multiply: -6(-12) = 72. 5) Add: 1 + 72 = 73. 6) Multiply: -6(73) = -438. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.5

Synthetic Division of Polynomials CONTINUED 7) Add: -6 + -438 = -444. 8) Multiply: -6(-444) = 2664. 9) Add: 0 + 2664 = 2664. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.5

Synthetic Division of Polynomials CONTINUED 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 4, the degree of the quotient is 3. This means that the 1 in the last row represents Thus, Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.5

The Remainder Theorem The Remainder Theorem If the polynomial f (x) is divided by x –c, then the remainder is f (c). This theorem is extremely useful because it can be used to evaluate a polynomial function at c. Rather than having to substitute c for x in the function, you can just divide the function by x – c using the shortcut of synthetic division. The remainder will be the function value at c. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.5

The Remainder Theorem EXAMPLE Given , use synthetic division and the Remainder Theorem to find f (-2). SOLUTION By the Remainder Theorem, if f (x) is divided by x – (-2) = x + 2, then the remainder is f (2). We’ll use synthetic division to divide. Remainder The remainder, -4, is the value of f (-2). Thus, f (-2) = -4. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.5

The Remainder Theorem EXAMPLE Show that -1 is a solution of the equation Then solve the polynomial equation. SOLUTION One way to show that -1 is a solution is to substitute -1 for x in the equation and obtain 0. Another way is to use synthetic division and the Remainder Theorem. Proposed Solution Remainder Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.5

The Remainder Theorem Equivalently, CONTINUED Equivalently, Because the remainder is 0, the polynomial has a value of 0 when x = -1. Thus, -1 is a solution of the given equation. The synthetic division also shows that x + 1 is a factor of the polynomial, as shown to the right of the synthetic division. The other factor is the quotient found in the last row of the synthetic division. Now we can solve the polynomial equation. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.5

The Remainder Theorem This is the given equation. CONTINUED This is the given equation. Factor using the result from the synthetic division. Factor the trinomial. Set each factor equal to zero. Solve for x. The solutions are -1, 1, and 2, and the solution set is {-1,1,2}. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.5