Dividing Polynomials Objectives Use long division to divide polynomials Use synthetic division to divide polynomials Evaluate a polynomial using the Remainder Theorem Use the Factor Theorem to solve a polynomial equation

Divide using long division. 161 ÷ 7 23 12.18 ÷ 2.1 5.8 6x – 15y 3 2x + 5y 7a2 – ab a 7a – b

Long Division Vocabulary Reminders

Examples with variables 2𝑥 3 𝑦 2 4𝑤 18 𝑥 2

Polynomial long division is a method for dividing a polynomial by another polynomial of a lower degree. It is very similar to dividing numbers.

Long Division of Polynomials Arrange the terms of both the dividend and the divisor in descending order of degree (or power). Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient. Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up. Subtract the product from the dividend. Bring down the next term in the original dividend and write it next to the remainder to form a new dividend. Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. Check your answer by multiplying the quotient by the divisor to get the dividend.

Using Long Division to Divide a Polynomial Divide using long division. (–y2 + 2y3 + 25) ÷ (y – 3) Step 1: Write the dividend in standard form, including terms with a coefficient of 0. 2y3 – y2 + 0y + 25 Step 2: Write division in the same way you would when dividing numbers. y – 3 2y3 – y2 + 0y + 25

Step 3 Divide. Notice that y times 2y2 is 2y3. Write 2y2 above 2y3. 2y2 + 5y + 15 Multiply 2y2 by y – 3. Then subtract. Bring down the next term. Divide 5y2 by y. y – 3 2y3 – y2 + 0y + 25 –(2y3 – 6y2) 5y2 + 0y Multiply 5y by y – 3. Then subtract. Bring down the next term. Divide 15y by y. –(5y2 – 15y) 15y + 25 Multiply 15 by y – 3. Then subtract. –(15y – 45) 70 Find the remainder.

Step 4 Write the final answer. –y2 + 2y3 + 25 y – 3 = 2y2 + 5y + 15 + 70 y – 3

Divide using long division. (15x2 + 8x – 12) ÷ (3x + 1) Step 1: Make sure the dividend and divisor are in standard form. Include terms with a coefficient of 0. 15x2 + 8x – 12 Step 2: Write division in the same way you would when dividing numbers. 3x + 1 15x2 + 8x – 12

Step 3 Divide. Divide 15 𝑥 2 by 3x. Notice that 3x times 5x is 15x2. Write 5x above 15x2. 5x + 1 Multiply 5x by 3x + 1. Then subtract. 3x + 1 15x2 + 8x – 12 –(15x2 + 5x) 3x −12 Bring down the next term. Divide 3x by 3x. –(3x + 1) –13 Multiply 1 by 3x + 1. Then subtract. Find the remainder.

Step 4 Write the final answer. 15x2 + 8x – 12 3x + 1 = 5x + 1 – 13 3x + 1

x – 3 x2 + 5x – 28 Divide using long division. Step 1: Write the dividend in standard form, including terms with a coefficient of 0. x2 + 5x – 28 Step 2: Write division in the same way you would when dividing numbers. x – 3 x2 + 5x – 28

Step 3 Divide. Notice that x times x is x2. Write x above x2. x + 8 x – 3 x2 + 5x – 28 Multiply x by x – 3. Then subtract. Bring down the next term. Divide 8x by x. –(x2 – 3x) 8x – 28 –(8x – 24) Multiply 8 by x – 3. Then subtract. –4 Find the remainder.

Step 4 Write the final answer. x2 + 5x – 28 x – 3 = x + 8 – 4 x – 3

Using Synthetic Division Dividing Polynomials Using Synthetic Division Synthetic division is a shorthand method of dividing a polynomial by a linear binomial by using only the coefficients. For synthetic division to work, the polynomial must be written in standard form, using 0 as a coefficient for any missing terms as well as a missing constant, and the divisor must be in the form (x – a).

The degree of the first term of the quotient is one less than the degree of the first term of the dividend.

Divide using synthetic division. (3x4 – x3 + 5x – 1) ÷ (x + 2) Step 1 Find a. a = –2 For (x + 2), a = –2. Step 2 Write the coefficients and a in the synthetic division format. 3 – 1 0 5 –1 –2 Use 0 for the coefficient of x2.

Step 3 Bring down the first coefficient Step 3 Bring down the first coefficient. Then multiply and add for each column. –2 3 –1 0 5 –1 Draw a box around the remainder, 45. –6 14 –28 46 3 –7 14 –23 45 Step 4 Write the quotient. 3x3 – 7x2 + 14x – 23 + 45 x + 2 Write the remainder over the divisor.

Divide using synthetic division. (6x2 – 5x – 6) ÷ (x + 3) Step 1 Find a. a = –3 For (x + 3), a = –3. Step 2 Write the coefficients and a in the synthetic division format. –3 6 –5 –6 Write the coefficients of 6x2 – 5x – 6.

Step 3 Bring down the first coefficient Step 3 Bring down the first coefficient. Then multiply and add for each column. –3 6 –5 –6 Draw a box around the remainder, 63. –18 69 6 –23 63 Step 4 Write the quotient. 6x – 23 + 63 x + 3 Write the remainder over the divisor.

Divide using synthetic division. (x2 – 3x – 18) ÷ (x – 6) Step 1 Find a. a = 6 For (x – 6), a = 6. Step 2 Write the coefficients and a in the synthetic division format. 6 1 –3 –18 Write the coefficients of x2 – 3x – 18.

Step 4 Write the quotient. Step 3 Bring down the first coefficient. Then multiply and add for each column. 6 1 –3 –18 There is no remainder. 6 18 1 3 Step 4 Write the quotient. x + 3

You can use synthetic division to evaluate polynomials You can use synthetic division to evaluate polynomials. This process is called synthetic substitution. The process of synthetic substitution is exactly the same as the process of synthetic division, but the final answer is interpreted differently, as described by the Remainder Theorem.

Write the coefficients of the dividend. Use a = 2. Use synthetic substitution to evaluate the polynomial for the given value. P(x) = 2x3 + 5x2 – x + 7 for x = 2. 2 2 5 –1 7 Write the coefficients of the dividend. Use a = 2. 4 18 34 2 9 17 41 P(2) = 41 Check Substitute 2 for x in P(x) = 2x3 + 5x2 – x + 7. P(2) = 2(2)3 + 5(2)2 – (2) + 7 P(2) = 41 

Check Substitute –3 for x in P(x) = x3 + 3x2 + 4. Use synthetic substitution to evaluate the polynomial for the given value. P(x) = x3 + 3x2 + 4 for x = –3. –3 1 3 0 4 Write the coefficients of the dividend. Use 0 for the coefficient of x2 Use a = –3. –3 1 4 P(–3) = 4 Check Substitute –3 for x in P(x) = x3 + 3x2 + 4. P(–3) = (–3)3 + 3(–3)2 + 4 P(–3) = 4 

The Factor Theorem Let 𝑓 𝑥 be a polynomial. If 𝑓 𝑐 =0, then 𝑥−𝑐=0 is a factor of 𝑓 𝑥 . If 𝑥−𝑐 is a factor of 𝑓(𝑥), then 𝑓 𝑐 =0. Example: Solve the equation 2 𝑥 3 −3 𝑥 2 −11𝑥+6=0 given that 3 is a zero of this function. Since 3 is a zero of the function, then 𝑓 3 =0. The Factor Theorem tells us that 𝑥−3 is a factor of this polynomial function.

Use synthetic division to divide 2 𝑥 3 −3 𝑥 2 −11𝑥+6 by 𝑥−3. 2 −3 −11 6 6 9 − 6 2 3 −2 The result is 2 𝑥 2 +3𝑥−2. The remainder, 0, verifies that 𝑥−3 is a factor of 2 𝑥 3 −3 𝑥 2 −11𝑥+6. 2 𝑥 3 −3 𝑥 2 −11𝑥+6=(𝑥−3)(2 𝑥 2 +3𝑥−2) Is this factored completely? No. We have to factor further or use the quadratic formula to solve.

We need to try factoring or use the quadratic formula to solve for 2 𝑥 2 +3𝑥−2=0. 2 𝑥 2 +3𝑥−2=(2𝑥−1)(𝑥+2) 2 𝑥 3 −3 𝑥 2 −11𝑥+6=(𝑥−3)(2𝑥−1)(𝑥+2) Now we can solve by setting up each expression = 0. 𝑥−3=0 𝑥=3 𝑥= 1 2 2𝑥−1=0 𝑥+2=0 𝑥=−2 The solution set is −2, 1 2 , 3 .