Solve 𝑥 2 +𝑏𝑥+𝑐=0 by Factoring Lesson 1.3 Honors Algebra 2 Solve 𝑥 2 +𝑏𝑥+𝑐=0 by Factoring Lesson 1.3

Goals Goal Rubric Factor trinomials of the form 𝑥 2 +𝑏𝑥+𝑐=0. Factor the Difference of Two Squares. Factor the Perfect Square Trinomial. Solve quadratic equations by factoring and using the Zero Product Property. Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

Vocabulary Monomial Binomial Trinomial Quadratic Equation Root of an Equation Zero of an Equation

Quadratic Expressions Quadratic expressions can have one, two or three terms, such as –16t2, –16t2 + 25t, or –16t2 + 25t + 2. Quadratic expressions with one term are monomials (–16t2). A monomial is an expression that is either a number, a variable, or the product of a number and one or more variables. Quadratic expressions with two terms are binomials (–16t2 + 25t). A binomial is the sum of two monomials. Quadratic expressions with three terms are trinomials (–16t2 + 25t + 2). A trinomial is the sum of three monomials.

Factoring a Trinomial of the form x2 + bx + c To factor a simple trinomial of the form x 2 + bx + c (leading coefficient is 1), express the trinomial as the product of two binomials. Factoring a Trinomial of the Form x2 + bx + c Step 1: Find the pair of integers whose product is c and whose sum is b. That is, determine m and n such that mn = c and m + n = b. Step 2: Write x2 + bx + c = (x + m)(x + n). Step 3: Check your work by multiplying the binomials. The coefficient of x is the sum of the two numbers. x2 – 11x + 24 = (x – 3)(x – 8) The last term is the product of the two numbers.

EXAMPLE 1 Factor trinomials of the form x2 + bx + c Factor the expression. a. x2 – 9x + 20 b. x2 + 3x – 12 SOLUTION a. You want x2 – 9x + 20 = (x + m) (x + n) where mn = 20 and m + n = – 9. ANSWER Notice that m = – 4 and n = – 5. So, x2 – 9x + 20 = (x – 4)(x – 5).

EXAMPLE 1 Factor trinomials of the form x2 + bx + c b. You want x2 + 3x – 12 = (x + m) (x + n) where mn = – 12 and m + n = 3. ANSWER Notice that there are no factors m and n such that m + n = 3. So, x2 + 3x – 12 cannot be factored.

1. x2 – 3x – 18 (x – 6) (x + 3) 2. n2 – 3n + 9 3. r2 + 2r – 63 Your Turn: for Example 1 Factor the expression. If the expression cannot be factored, say so. 1. x2 – 3x – 18 ANSWER (x – 6) (x + 3) 2. n2 – 3n + 9 ANSWER cannot be factored 3. r2 + 2r – 63 ANSWER (r + 9)(r –7)

Conjugate Pairs (3x + 6) (3x - 6) (r - 5) (r + 5) (2b - 1) (2b + 1) The following pairs of binomials are called conjugates. Notice that they all have the same terms, only the sign between them is different. (3x + 6) (3x - 6) and (r - 5) (r + 5) and (2b - 1) (2b + 1) and (x2 + 5) (x2 - 5) and

Multiplying Conjugates When we multiply any conjugate pairs, the middle terms always cancel and we end up with a binomial. (3x + 6)(3x - 6) = 9x2 - 36 (r - 5)(r + 5) = r2 - 25 (2b - 1)(2b + 1) = 4b2 - 1

Only TWO terms (a binomial) Difference of Two Squares Binomials that look like this are called a Difference of Two Squares: Only TWO terms (a binomial) 9x2 - 36 A MINUS between! The first term is a Perfect Square! The second term is a Perfect Square!

Difference of Two Squares A binomial is the difference of two square if both terms are squares and the signs of the terms are different. 9x 2 – 25y 2 – c 4 + d 4

Factoring the Difference of Two Squares A Difference of Squares! A Conjugate Pair!

Perfect Square Trinomials Factor the polynomial 25x 2 + 20x + 4. The result is (5x + 2)2, an example of a binomial squared. Any trinomial that factors into a single binomial squared is called a perfect square trinomial.

Perfect Square Trinomials (a + b)2 = a 2 + 2ab + b 2 (a – b)2 = a 2 – 2ab + b 2 So if the first and last terms of our polynomial to be factored can be written as expressions squared, and the middle term of our polynomial is twice the product of those two expressions, then we can use these two previous equations to easily factor the polynomial. a 2 + 2ab + b 2 = (a + b)2 a 2 – 2ab + b 2 = (a – b)2

Factoring a Prefect Square Trinomial When you have to factor a perfect square trinomial, the patterns make it easier Product Doubled Example: Factor 36x2 + 60x + 25 Perfect Square 6x 30x 5 Perfect Square First you have to recognize that it’s a perfect square trinomial Square Root Product Square Root And so, the trinomial factors as: Check: (6x + 5)2

Factoring Special Products

a. x2 – 49 = x2 – 72 = (x + 7) (x – 7) b. d 2 + 12d + 36 EXAMPLE 2 Factor with special patterns Factor the expression. a. x2 – 49 = x2 – 72 Difference of two squares = (x + 7) (x – 7) b. d 2 + 12d + 36 = d 2 + 2(d)(6) + 62 Perfect square trinomial = (d + 6)2 c. z2 – 26z + 169 = z2 – 2(z) (13) + 132 Perfect square trinomial = (z – 13)2

4. x2 – 9 (x – 3) (x + 3) 5. q2 – 100 (q – 10) (q + 10) Your Turn: for Example 2 Factor the expression. 4. x2 – 9 (x – 3) (x + 3) ANSWER 5. q2 – 100 (q – 10) (q + 10) ANSWER 6. y2 + 16y + 64 (y + 8)2 ANSWER

7. w2 – 18w + 81 (w – 9)2 Your Turn: for Example 2 Factor the expression. 7. w2 – 18w + 81 (w – 9)2 ANSWER

Zeros Zero of a function – is a value of the input x that makes the output f(x) equal zero. The zeros of a function are the x-intercepts. Unlike linear functions, which have no more than one zero, quadratic functions can have up to two zeros, as shown at right. These zeros are always symmetric about the axis of symmetry.

Roots You can also find zeros by using algebra. For example, to find the zeros of f(x)= x2 + 2x – 3, you can set the function equal to zero. The solutions to the related equation x2 + 2x – 3 = 0 represent the zeros of the function. The solution to a quadratic equation of the form ax2 + bx + c = 0 are roots. The roots of an equation are the values of the variable that make the equation true.

Quadratic Equations Consider the zeros of function P(x) = 2x2 + 4x – 16 x-intercepts of function P(x) = 2x2 + 4x – 16 roots of equation 2x2 + 4x – 16 = 0 solution set of equation 2x2 + 4x – 16 = 0 Each is solved by finding the values of x that make 2x2 + 4x – 16 = 0 true. Quadratic Equation in One Variable An equation that can be written in the form ax2 + bx + c = 0 where a, b, and c are real numbers with a  0, is a quadratic equation in standard form.

Zero Product Property You can find the roots of some quadratic equations by factoring and applying the Zero Product Property. Functions have zeros or x-intercepts. Equations have solutions or roots. Reading Math

Procedure Solving quadratic equations by factoring: Write the original equation equal to zero. Factor the quadratic expression. Use the Zero Product Property (set each factor equal to zero). Solve for x.

x2 – 5x – 36 = 0 (x – 9)(x + 4) = 0 x – 9 = 0 or x + 4 = 0 x = 9 or EXAMPLE 3 Standardized Test Practice SOLUTION Write original equation. x2 – 5x – 36 = 0 Factor. (x – 9)(x + 4) = 0 x – 9 = 0 or x + 4 = 0 Zero product property x = 9 or x = – 4 Solve for x. ANSWER The correct answer is C.

EXAMPLE 4 Use a quadratic equation as a model Nature Preserve A town has a nature preserve with a rectangular field that measures 600 meters by 400 meters. The town wants to double the area of the field by adding land as shown. Find the new dimensions of the field.

EXAMPLE 4 Use a quadratic equation as a model SOLUTION 480,000 = 240,000 + 1000x + x2 Multiply using FOIL. 0 = x2 + 1000x – 240,000 Write in standard form. 0 = (x – 200) (x + 1200) Factor. x – 200 = 0 x + 1200 = 0 or Zero product property x = 200 or x = – 1200 Solve for x.

EXAMPLE 4 Use a quadratic equation as a model ANSWER Reject the negative value, – 1200. The field’s length and width should each be increased by 200 meters. The new dimensions are 800 meters by 600 meters.

8. Solve the equation x2 – x – 42 = 0. Your Turn: for Examples 3 and 4 8. Solve the equation x2 – x – 42 = 0. ANSWER x = – 6 or x = 7 9. WHAT IF ? In Example 4, suppose the field initially measures 1000 meters by 300 meters. Find the new dimensions of the field. ANSWER The new dimensions are 1200 meters by 500 meters.

The zeros of the function are –3 and 4. EXAMPLE 5 Find the zeros of quadratic functions Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 Check Graph y = x2 – x – 12. The graph passes through (–3, 0) and (4, 0). SOLUTION a. y = x2 – x – 12 Write original function. = (x + 3) (x – 4) Factor. The zeros of the function are –3 and 4.

Check Graph y = x2 + 12x + 36. The graph passes through ( – 6, 0). EXAMPLE 5 Find the zeros of quadratic functions Find the zeros of the function by rewriting the function in intercept form. a. y = x2 – x – 12 b. y = x2 + 12x + 36 Check Graph y = x2 + 12x + 36. The graph passes through ( – 6, 0). SOLUTION b. y = x2 + 12x + 36 Write original function. = (x + 6) (x + 6) Factor. The zero of the function is – 6.

The zeros of the function are – 7 and 2. GUIDED PRACTICE Your Turn: for Example for Example 5 Find the zeros of the function by rewriting the function in intercept form. 10. y = x2 + 5x – 14 ANSWER The zeros of the function are – 7 and 2. 11. y = x2 – 7x – 30 ANSWER The zeros of the function are – 3 and 10.

12. f(x) = x2 – 10x + 25 GUIDED PRACTICE Your Turn: for Example 5 Find the zeros of the function by rewriting the function in intercept form. 12. f(x) = x2 – 10x + 25 The zero of the function is 5. ANSWER