Factoring Using the Distributive Property

Slides:

Advertisements

Similar presentations

Solve an equation with variables on both sides

Advertisements

Solve an absolute value equation EXAMPLE 2 SOLUTION Rewrite the absolute value equation as two equations. Then solve each equation separately. x – 3 =

Solve an equation using subtraction EXAMPLE 1 Solve x + 7 = 4. x + 7 = 4x + 7 = 4 Write original equation. x + 7 – 7 = 4 – 7 Use subtraction property of.

Standardized Test Practice

Solving Equations with variables on both sides of the Equals Chapter 3.5.

Standardized Test Practice

Standardized Test Practice

© 2007 by S - Squared, Inc. All Rights Reserved.

Solve Equations with Variables on Both Sides

Factoring Using the Distributive Property Chapter 9.2.

Solving Equations Medina1 Variables on Both Sides.

I can solve one-step equations in one variable.. Equations that have the same solutions. In order to solve a one-step equation, you can use the properties.

Solve an equation by combining like terms EXAMPLE 1 8x – 3x – 10 = 20 Write original equation. 5x – 10 = 20 Combine like terms. 5x – =

Rational Equations Section 8-6.

Solve an absolute value equation EXAMPLE 2 SOLUTION Rewrite the absolute value equation as two equations. Then solve each equation separately. x – 3 =

Practice 2.2 Solving Two Step Equations.

Solve an equation using addition EXAMPLE 2 Solve x – 12 = 3. Horizontal format Vertical format x– 12 = 3 Write original equation. x – 12 = 3 Add 12 to.

EXAMPLE 1 Solve by equating exponents Rewrite 4 and as powers with base Solve 4 = x 1 2 x – 3 (2 ) = (2 ) 2 x – 3x – 1– 1 2 = 2 2 x– x + 3 2x =

Example 1 Solving Two-Step Equations SOLUTION a. 12x2x + 5 = Write original equation. 112x2x + – = 15 – Subtract 1 from each side. (Subtraction property.

5-Minute Check on Chapter 2 Transparency 3-1 Click the mouse button or press the Space Bar to display the answers. 1.Evaluate 42 - |x - 7| if x = -3 2.Find.

Transparency 2 Click the mouse button or press the Space Bar to display the answers.

Transparency 2 Click the mouse button or press the Space Bar to display the answers.

Solving Linear Equations Substitution. Find the common solution for the system y = 3x + 1 y = x + 5 There are 4 steps to this process Step 1:Substitute.

Systems of Equations: Substitution

Use the substitution method

Solve Linear Systems by Substitution January 28, 2014 Pages

© by S-Squared, Inc. All Rights Reserved.

Solve Linear Systems by Substitution Students will solve systems of linear equations by substitution. Students will do assigned homework. Students will.

Y=3x+1 y 5x + 2 =13 Solution: (, ) Solve: Do you have an equation already solved for y or x?

Multiply one equation, then add

© 2007 by S - Squared, Inc. All Rights Reserved.

2.2 Solving Two- Step Equations. Solving Two Steps Equations 1. Use the Addition or Subtraction Property of Equality to get the term with a variable on.

9-2 Factoring Using the Distributive Property Objectives: 1)Students will be able to factor polynomials using the distributive property 2)Solve quadratic.

CHAPTER 9 LESSON 4 SOLVE POLYNOMIAL EQUATIONS in FACTORED FORM.

Over Lesson 8–1 A.A B.B C.C D.D 5-Minute Check 1 2 ● 5 ● 7 ● x ● x ● y Factor 70x 2 y.

6-2 Solving Systems Using Substitution Hubarth Algebra.

Solving Polynomials Review of Previous Methods Factoring Cubic Expressions.

Solve Polynomial Equations in Factored Form March 25, 2014 Pages

Substitution Method: Solve the linear system. Y = 3x + 2 Equation 1 x + 2y=11 Equation 2.

Rewrite a linear equation

Solve Linear Systems By Multiplying First

Solving Equations with the Variable on Each Side

Solve for variable 3x = 6 7x = -21

2 Understanding Variables and Solving Equations.

Solve a quadratic equation

3-2: Solving Systems of Equations using Substitution

6-2 Solving Systems By Using Substitution

6-2 Solving Systems Using Substitution

Solving Equations by Factoring and Problem Solving

Solving Systems using Substitution

3-2: Solving Systems of Equations using Substitution

Solving Systems of Equations using Substitution

Do Now 1) t + 3 = – 2 2) 18 – 4v = 42.

3-2: Solving Systems of Equations using Substitution

Solve an equation by combining like terms

Equations: Multi-Step Examples ..

Example 1 b and c Are Positive

7.4 Solving factored polynomials

6.3 Solving Quadratic Equations by Factoring

Solving Multi-Step Equations

Day 136 – Common Factors.

Solve an inequality using subtraction

Solving Polynomials 1 Review of Previous Methods 2 Grouping.

3-2: Solving Systems of Equations using Substitution

Lesson 7-6 Multiplying a Polynomial by a Monomial

3-2: Solving Systems of Equations using Substitution

3-2: Solving Systems of Equations using Substitution

3-2: Solving Systems of Equations using Substitution

3-2: Solving Systems of Equations using Substitution

Definition of logarithm

Presentation transcript:

Factoring Using the Distributive Property 9.2 Factoring Using the Distributive Property

Greatest Common Factor (GCF)/Distributive Property Look for the largest common number and/or variables in each term Divide each term by the common terms

Use the Distributive Property to factor . First, find the CGF of 15x and . Factor each number. Circle the common prime factors. GFC: Write each term as the product of the GCF and its remaining factors. Then use the Distributive Property to factor out the GCF. Rewrite each term using the GCF. Simplify remaining factors. Distributive Property Example 2-1a

Answer: The completely factored form of is Example 2-1a

Use the Distributive Property to factor . Factor each number. Circle the common prime factors. GFC: or Rewrite each term using the GCF. Distributive Property Answer: The factored form of is Example 2-1a

Use the Distributive Property to factor each polynomial. a. Answer: Answer: Example 2-1b

Grouping Steps Arrange terms in descending order Separate into groups of 2 in () Factor the GCF from each group Factor an additional (-1) if necessary The terms in () must match Combine the two GCF into a binomial in () and place it next to the other binomial

Grouping Example

Group terms with common factors. Factor the GCF from each grouping. Answer: Distributive Property Example 2-2a

Factor Answer: Example 2-2b

Group terms with common factors. Factor GCF from each grouping. Answer: Distributive Property Example 2-3a

Factor Answer: Example 2-3b

Solve Then check the solutions. If , then according to the Zero Product Property either or Original equation or Set each factor equal to zero. Solve each equation. Answer: The solution set is Example 2-4a

Check Substitute 2 and for x in the original equation. Example 2-4a

Solve Then check the solutions. Answer: {3, –2} Example 2-4b

Solve Then check the solutions. Write the equation so that it is of the form Original equation Subtract from each side. Factor the GCF of 4y and which is 4y. Zero Product Property or Solve each equation. Example 2-5a

Answer: The solution set is Check by substituting 0 and for y in the original equation. Example 2-5a

Solve Answer: Example 2-5b

Practice Problems Page 484 Problems 4-15, 16-39, 48-59 (odd)

Practice Problems (Day 2) Page 484 Problems 4-15, 16-39, 48-59 (even)