Algebra 2 Divide x 2 + 2x – 30 by x – 5. Lesson 6-3 Dividing Polynomials – 30Subtract: (x 2 + 2x) – (x 2 – 5x) = 7x. Bring down –30. xDivide = x. x – 5 x 2 + 2x – 30 x2xx2x x 2 – 5xMultiply: x(x – 5) = x 2 – 5x 7x Repeat the process of dividing, multiplying, and subtracting. 5Subtract: (7x – 30) – (7x – 35) = 5. The quotient is x + 7 with a remainder of 5, or simply x + 7, R 5. 7x – 35Multiply: 7(x – 5) = 7x – 35. x + 7Divide = 7. x – 5 x 2 + 2x – 30 x 2 – 5x 7x – 30 7xx7xx Additional Examples

Algebra 2 Determine whether x + 2 is a factor of each polynomial. Lesson 6-3 Dividing Polynomials a.x x + 16b.x 3 + 7x 2 – 5x – 6 Since the remainder is zero, x + 2 is a factor of x x x + 8 x + 2 x x + 16 x 2 + 2x 8x x 2 + 5x – 15 x + 2 x 3 + 7x 2 – 5x – 6 x 3 + 2x 2 5x 2 – 5x 5x x –15x – 6 –15x – Since the remainder = 0, x + 2 is not a factor of x 3 + 7x 2 – 5x – 6. / Additional Examples

Algebra 2 Use synthetic division to divide 5x 3 – 6x 2 + 4x – 1 by x – 3. Lesson 6-3 Dividing Polynomials Step 1:Reverse the sign of the constant term in the divisor. Write the coefficients of the polynomial in standard form. Step 2: Bring down the coefficient. Bring down the 5. 35–64–1This begins the quotient. 5 – Write x 5x 3 – 6x 2 + 4x – as 5 – 6 4 –1 Additional Examples

Algebra 2 (continued) Lesson 6-3 Dividing Polynomials Multiply 3 by 5. Write 35–64–1the result under –6. x15 5 9Add –6 and 15. Step 3: Multiply the first coefficient by the new divisor. Write the result under the next coefficient. Add. Step 4: Repeat the steps of multiplying and adding until the remainder is found. The quotient is 5x 2 + 9x + 31, R –64– x 2 + 9x + 31 Remainder Additional Examples

Algebra 2 The volume in cubic feet of a shipping carton is V(x) = x 3 – 6x 2 + 3x The height is x – 5 feet. Lesson 6-3 Dividing Polynomials a.Find linear expressions for the other dimensions. Assume that the length is greater than the width. b.If the width of the carton is 4 feet, what are the other two dimensions? x 2 – x – 2 = (x – 2)(x + 1) Factor the quotient. The length and the width are x + 1 and x – 2, respectively. x – 2 = 4 Substitute 4 into the expression for width. Find x. x = 6 Since the height equals x – 5 and the length equals x + 1, the height is 1 ft. and the length is 7 ft. 51–6 3 10Divide. 5–5–10 1–1–2 0 x 2 – x – 2 Remainder Additional Examples

Algebra 2 Use synthetic division to find P(3) for P(x) = x 4 – 2x 3 + x – 9. Lesson 6-3 Dividing Polynomials By the Remainder Theorem, P(3) equals the remainder when P(x) is divided by x – 3. The remainder is 21, so P(3) = –20 1 – Additional Examples