Dividing Polynomials How do we use long division

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Dividing Polynomials • How do we use long division and synthetic division to divide

Dividing Polynomials • How do we use long division and synthetic division to divide polynomials? Holt Mc. Dougal Algebra 2

Dividing Polynomials You can use synthetic division to evaluate polynomials. This process is called

Dividing 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. Holt Mc. Dougal Algebra 2

Dividing Polynomials Example 1: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial

Dividing Polynomials Example 1: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. Holt Mc. Dougal Algebra 2

Dividing Polynomials Example 2: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial

Dividing Polynomials Example 2: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. Holt Mc. Dougal Algebra 2

Dividing Polynomials Example 3: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial

Dividing Polynomials Example 3: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. Holt Mc. Dougal Algebra 2

Dividing Polynomials Example 4: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial

Dividing Polynomials Example 4: Using Synthetic Substitution Use synthetic substitution to evaluate the polynomial for the given value. Holt Mc. Dougal Algebra 2

Dividing Polynomials Example 5: Geometry Application Write an expression that represents the area of

Dividing Polynomials Example 5: 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 x 3 – x 2 – 6 x. The volume V is related to the area A and the height h by V the equation V = A h. Rearranging for A gives A = h. 3 – x 2 – 6 x x A(x) = x+2 The area of the face of the rectangular prism can be represented by A(x)= x 2 – 3 x. Holt Mc. Dougal Algebra 2

Dividing Polynomials Lesson 3. 4 Practice C Holt Mc. Dougal Algebra 2

Dividing Polynomials Lesson 3. 4 Practice C Holt Mc. Dougal Algebra 2