 A method for breaking up a quadratic equation in the form ax 2 + bx + c into factors (expressions which multiply to give you the original trinomial).

Slides:



Advertisements
Similar presentations
AC Method of factoring ax2 + bx +c
Advertisements

Algebra I B. MacDonald.  Standard form ax 2 + bx + c  a, b, c are real numbers.
2.6 Factor x2 + bx + c.
4.3 Solve x2 + bx +c = 0 by Factoring
6.3 Factoring Trinomials II Ax 2 + bx + c. Factoring Trinomials Review X 2 + 6x + 5 X 2 + 6x + 5 (x )(x ) (x )(x ) Find factors of 5 that add to 6: Find.
Using the Zero-Product Property to Solve a Quadratic
Objective Factor quadratic trinomials of the form x2 + bx + c.
Factoring a Quadratic Expression
CRASH COURSE IN QUADRATICS In preparation for the Algebra CST -b + b 2 – 4ac 2ac √ (x+4)(x-3)=0 (x+1)(x+2) X 2 – 5x +4 F O I L Complete The Square.
Solving Quadratic Equations By Keith Rachels And Asef Haider.
.   Learn the definition of quadratic equation.  Multiply two binomials using the FOIL method.  Factor trinomials.  Solve quadratic equation by.
Solving Quadratic Equations Algebraically Lesson 2.2.
Solving Quadratic Equations by Factoring Algebra I.
Introduction A trinomial of the form that can be written as the square of a binomial is called a perfect square trinomial. We can solve quadratic equations.
Objective Solve quadratic equations by completing the square.
Factoring Polynomials
Copyright © Cengage Learning. All rights reserved.
Factoring Polynomials
EXAMPLE 1 Factor ax 2 + bx + c where c > 0 Factor 5x 2 – 17x + 6. SOLUTION You want 5x 2 – 17x + 6 = (kx + m)(lx + n) where k and l are factors of 5 and.
5.3.2 – Quadratic Equations, Finding Zeroes. Recall, we went over how to factor quadratics that are trinomials Example. Factor the expression x 2 + 7x.
Factoring Polynomials
8-4 Factoring ax2 + bx + c Warm Up Lesson Presentation Lesson Quiz
1.3 Solving Quadratic Equations by Factoring (p. 18) How can factoring be used to solve quadratic equation when a=1?
MM218 - Unit 7 Seminar Topics
6.6 Quadratic Equations. Multiplying Binomials A binomial has 2 terms Examples: x + 3, 3x – 5, x 2 + 2y 2, a – 10b To multiply binomials use the FOIL.
Section 4.4 – Factoring Quadratic Expressions Factors of a given number are numbers that have a product equal to the given numbers. Factors of a given.
Copyright © 2015, 2011, 2007 Pearson Education, Inc. 1 1 Chapter 6 Factoring.
Objective: 6.4 Factoring and Solving Polynomial Equations 1 5 Minute Check  Simplify the expression
2.6 Solving Quadratic Equations with Complex Roots 11/9/2012.
Solve x 2 + bx + c = 0 by factoring Section 4.3. What is a trinomial????? It has 3 terms connected by addition or subtraction Example : 3x 2 – 6x + 7.
Section 5.3 Factoring Quadratic Expressions
Slide Copyright © 2009 Pearson Education, Inc. 6.9 Solving Quadratic Equations by Using Factoring and by Using the Quadratic Formula.
5 – 2: Solving Quadratic Equations by Factoring Objective: CA 8: Students solve and graph quadratic equations by factoring, completing the square, or using.
Solving Quadratic Equations. Review of Solving Quadratic Equations ax 2 +bx +c = 0 When the equation is equal to zero, solve by factoring if you can.
Chapter 5.2 Solving Quadratic Equations by Factoring.
Holt Algebra Solving Quadratic Equations by Graphing and Factoring A trinomial (an expression with 3 terms) in standard form (ax 2 +bx + c) can be.
Example 1A Solve the equation. Check your answer. (x – 7)(x + 2) = 0
WARM UP Find the product. 1.) (m – 8)(m – 9) 2.) (z + 6)(z – 10) 3.) (y + 20)(y – 20)
5.3 Factoring Quadratic Function 12/7/2012. are the numbers you multiply together to get another number: 3 and 4 are factors of 12, because 3x4=12. 2.
5-4 Factoring Quadratic Expressions M11.A.1.2.1: Find the Greatest Common Factor and/or the Least Common Multiple for sets of monomials M11.D.2.1.5: Solve.
Solving Quadratic Equations by Factoring. Martin-Gay, Developmental Mathematics 2 Zero Factor Theorem Quadratic Equations Can be written in the form ax.
2.4 part 1 - Basic Factoring I can... - Factor using GCF -Factor a difference of two perfect squares -Factor basic trinomials.
Solving Quadratic Equations by Factoring Lesson 5.2.
Solving Quadratic Equations. Factor: x² - 4x - 21 x² -21 a*c = -21 b = -4 x + = -21 = x 3x3x x 3 (GCF) x-7 (x – 7)(x + 3)
§ 6.6 Solving Quadratic Equations by Factoring. Martin-Gay, Beginning and Intermediate Algebra, 4ed 22 Zero Factor Theorem Quadratic Equations Can be.
Factoring – Day 4 Factoring Trinomials Objective: To factor trinomials whose quadratic coefficient is 1.
3.4 Notes: Factoring p. 74 in your book. FACTORING We’ll use the Sum and Product Technique We’ll use the Sum and Product Technique Our job is to find.
ALGEBRA 2 – CHAPTER 5 QUADRATICS. 5-2 PROPERTIES OF PARABOLAS.
9.5 Solving by Factoring Algebra 14.0, Predicting with Evidence What are the zeros for y = x 2 – x – 6? Now factor x 2 – x – 6 = 0 and solve. What.
Chapter 9 Final Exam Review. Add Polynomials (2x² + x³ – 1) (2x² + x³ – 1) Like Terms terms that have the same variable (2x³ – 5x² + x) + (2x³ – 5x² +
Factoring Quadratic Trinomials a = 1 Chapter 10.5.
Lesson 2-3 The Quadratic Equation Objective: To learn the various ways to solve quadratic equations, including factoring, completing the square and the.
Algebra 2 cc Section 2.2 Solve quadratic equations by factoring
Factoring Day 1 I can factor a quadratic expression. x 2 + 3x + 2 Rewrite as (x + 1)(x + 2)
A factored form of x 2 + 5x - 24 is — A (x − 4)(x + 6) B (x − 3)(x + 8) C (x − 2)(x + 12) D (x − 6)(x + 4) Which of the following equals when factored.
April 6, 2009 You need:textbook calculator No Fantastic Five warm ups this week. Take notes and/or read section Work together if you need help –
Polynomial Equations and Factoring
Objectives Solve quadratic equations by factoring.
5.3 Factoring Quadratics.
Section 5.3 Factoring Quadratic Expressions
Warm up.
Solving Quadratic Equations
Factoring Quadratic Functions if a ≠ 1 (Section 3.5)
Factoring Special Cases
5.4 Factor and Solve Polynomial Equations
4.3 Solving Quadratic Equations by Factoring
Solve
You can find the roots of some quadratic equations by factoring and applying the Zero Product Property. Functions have zeros or x-intercepts. Equations.
4.5: Completing the square
Section 9.1 “Properties of Radicals”
Presentation transcript:

 A method for breaking up a quadratic equation in the form ax 2 + bx + c into factors (expressions which multiply to give you the original trinomial).  Use Algebra tiles to help you develop the patterns in factoring.

 x 2 + 5x + 4  x 2 + 7x + 12  x 2 + 6x + 8

 x 2 – 6x + 8  x 2 - 9x + 20  x 2 – 4x + 4  Notice how the answer changes if the b term is negative.

 x 2 + 4x – 12  x 2 + 3x – 18  x 2 – 7x – 8  x 2 – 3x – 10

 Factor each of the following expressions.  x x + 48  x 2 – 22x – 48  x 2 – 13x + 42  x 2 + 5x - 24

 Difference of squares  a 2 – b 2 = (a + b)(a – b)  Examples  x 2 – 49 = (x + 7)(x – 7)  4x 2 – 1 = (2x + 1)(2x – 1)  Practice  Factor: x 2 – 64  Factor: 9x

 Check to see if each term in the expression has a common factor.  Examples:  8x 3 – 50x What factors do they have in common?  2x(4x 2 – 25)  2x(2x + 5)(2x – 5)  Factor the following  5x 4 – 10x 3  2x x + 96

 1. Try to factor out a common monomial factor from all terms first.  2. Then look to see if the difference of squares format exists.  3. Use your techniques for factoring a quadratic trinomial if applicable.  Do-Now: Factor the following….  2x 2 – 18x + 40  3x 3 – 9x 2 – 30x

 What makes this expression more difficult to factor than what we have done before?  2x 2 + 7x + 6  You can use guess and check using FOIL, but it can become tedious for more difficult problems. Try these using guess and check.  3x 2 – 10x + 8  6x 2 + x – 15

 You may want to use this method on more complicated examples, such as……..  6x 2 + x – 12  First identify the values of a and c. ▪ a = 6, c = -12  Next, multiply a times c. ▪ (6)(-12) = -72  Then find two numbers that multiply to be (a)(c) and add to be the value of b. ▪ (9)(-8) = -72, = 1

 Now rewrite the original expression with the x-term written as the sum of the two numbers you found.  6x 2 + x – 12 ▪ 6x 2 – 8x + 9x – 12  Finally, break the expression into two parts and factor twice.  (6x 2 – 8x) + (9x – 12)  2x(3x – 4) + 3(3x – 4)  (3x – 4)(2x + 3)

 2x 2 + 5x – 3  9x x + 4  12x x + 6  15x 2 + 8x – 16  20x 2 – 54x + 36

 It allows us to solve: ax 2 + bx + c = 0 using the “Zero Product Property”  If the product of two expressions is equal to zero, then………  one or both of the expressions must equal zero.  Examples. Solve the following equations.  x 2 – x – 42 = 0  x 2 = 7x

 Factor:  3x x + 10  8x 2 + 4x – 24  Solve the following equation.  4x x – 24 = 0

 When you solve y = ax 2 + bx + c when y = 0, you are finding the roots or zeros of the function. These are the points where the parabola crosses the x-axis.