Dividing Polynomials Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2

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

Example 1: 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

Check It Out! Example 1a Divide using long division. (15x2 + 8x – 12) ÷ (3x + 1) Step 1 Write the dividend in standard form, including 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

Check It Out! Example 1b Divide using long division. (x2 + 5x – 28) ÷ (x – 3) 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

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 and a coefficient for any missing terms, and the divisor must be in the form (x – a).

Example 2A: Using Synthetic Division to Divide by a Linear Binomial Divide using synthetic division. 1 3 (3x2 + 9x – 2) ÷ (x – ) Step 1 Find a. Then write the coefficients and a in the synthetic division format. 1 3 a = For (x – ), a = . 1 3 1 3 3 9 –2 Write the coefficients of 3x2 + 9x – 2.

Example 2B: Using Synthetic Division to Divide by a Linear Binomial 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.

Check It Out! Example 2a 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.

Check It Out! Example 2b 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.

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.

Example 3A: Using Synthetic Substitution 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 

Example 3B: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. 1 3 P(x) = 6x4 – 25x3 – 3x + 5 for x = – . – 1 3 6 –25 0 –3 5 Write the coefficients of the dividend. Use 0 for the coefficient of x2. Use a = . 1 3 –2 9 –3 2 6 –27 9 –6 7 P( ) = 7 1 3

Check It Out! Example 3a 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 

Check It Out! Example 3b Use synthetic substitution to evaluate the polynomial for the given value. 1 5 P(x) = 5x2 + 9x + 3 for x = . 1 5 5 9 3 Write the coefficients of the dividend. Use a = . 1 5 1 2 5 10 5 P( ) = 5 1 5

Example 4: Geometry Application Write an expression that represents the area of the top face of a rectangular prism when the height is x + 2 and the volume of the prism is x3 – x2 – 6x. The volume V is related to the area A and the height h by the equation V = A  h. Rearranging for A gives A = . V h x3 – x2 – 6x x + 2 A(x) = Substitute. –2 1 –1 –6 0 Use synthetic division. The area of the face of the rectangular prism can be represented by A(x)= x2 – 3x. –2 6 1 –3

Check It Out! Example 4 Write an expression for the length of a rectangle with width y – 9 and area y2 – 14y + 45. The area A is related to the width w and the length l by the equation A = l  w. y2 – 14y + 45 y – 9 l(x) = Substitute. 9 1 –14 45 Use synthetic division. 9 –45 1 –5 The length of the rectangle can be represented by l(x)= y – 5.