Factoring Quadratic Expressions Lesson 4-4 Part 2 Algebra 2 Factoring Quadratic Expressions Lesson 4-4 Part 2

Goals Goal Rubric To find binomial factors of quadratic expressions To factor special quadratic expressions. 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 Perfect Square Trinomial Difference of Two Squares

Big Idea: Solving Equations and Inequalities Essential Question Big Idea: Solving Equations and Inequalities Why is it useful to recognize special quadratic expressions?

Factoring ax2 + bx + c In the previous lesson you factored trinomials of the form x2 + bx + c. Now you will factor trinomials of the form ax + bx + c, where a ≠ 0.

THE FUN WAY TO FACTOR TRINOMIALS X-BOX Method THE FUN WAY TO FACTOR TRINOMIALS

X-Box Method This is a guaranteed method for factoring trinomials of the form ax2 + bx +c — no guessing necessary! We will learn how to factor quadratic equations using the x-box method Background knowledge needed: Basic x-factor problems General form of a trinomial The factoring box

m n mn m n m+n The X-Factor Given m & n on the sides of the x factor m•n is the top of the x factor m n m+n m+n is the bottom of the x factor

The X-Factor mn n m m+n

What’s on Top & Bottom? 5 3 Given m & n on the sides of the x factor ? m•n is the top of the x factor 5 3 ? m+n is the bottom of the x factor

5 3 15 5 3 8 The X-Factor Given m & n on the sides of the x factor m•n is the top of the x factor 5 3 8 m+n is the bottom of the x factor

The X-Factor -3 9 ? -3 9 ?

The X-Factor -3 9 -27 -3 9 6

The X-Factor 12 Given m•n 7 Given m+n 12 ? ? Find m & n 7

The X-Factor 12 Given m•n 7 Given m+n 12 3 4 Find m & n 7

The X-Factor 8 Given m•n 9 Given m+n 8 ? ? Find m & n 9

The X-Factor 8 Given m•n 9 Given m+n 8 8 1 Find m & n 9

The X-Factor -84 Given m•n -5 Given m+n -84 Find m & n -5

To solve a difficult X-Factor START AT 1 AND CHECK THE SUM -84 -84 Given m•n 1 -84 = -83 -41 = -39 -28 = -25 -21 = -17 -14 = -8 7 -12 = -5 -5 Given m+n -84 Find m & n -12 7 -5

Why Know the “X” Factor?

Why Know the “X” Factor? The “X” factor is a Simple way to FACTOR TRINOMIALS

General Form of a Trinomial c is COEFFICIENT of last term b is COEFFICIENT of x a is COEFFICIENT of x2

Identify a, b & c b is 3 c is 9 a is 5

Factor the X-Box Method ax2 + bx + c Combine the x-factor and a 2⨯2 box to create the x-box method. Factor Col. 1 Factor Col. 2 Product ac=mn First and Last Coefficients 1st Term GCF Row 1 Factor n n m Middle Last term Factor m Factor Row 2 b=m+n Sum

X-Box Procedure ax2 + bx + c 1st Term Factor n Factor m Last term Create an x-factor with the product ac on the top, the middle term b on the bottom and the factors m & n on the sides. Create a 2x2 box. In the top left, put the 1st term. In the bottom right corner, put the last term. Put the two factors m and n times the variable x in the open boxes. Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 1st Term Factor n Factor m Last term

Factoring ax2+bx+c Factor 5x2+11x+2 5•2= 10 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 1 10 11 x 2 5x x 10x 5x2 1 2 (x+2)(5x+1)

Factor 3x2 – x – 4 3 -4 -1 3x -4 x 3x -4x 3x2 1 -4 3•-4= -12 Factoring ax2+bx+c Factor 3x2 – x – 4 3•-4= -12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 3 -4 -1 3x -4 x 3x -4x 3x2 1 -4 (x+1)(3x – 4)

Factor 12x2 +5x – 2 8 -3 5 4x -1 3x 12x2 8x -3x 2 -2 12•-2= -24 Factoring ax2+bx+c Factor 12x2 +5x – 2 12•-2= -24 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 8 -3 5 4x -1 3x 12x2 8x -3x 2 -2 (3x+2)(4x – 1)

Factor 15x2 +7x – 2 10 -3 7 5x -1 3x 10x -3x 15x2 2 -2 15•-2= -30 Factoring ax2+bx+c Factor 15x2 +7x – 2 15•-2= -30 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 10 -3 7 5x -1 3x 10x -3x 15x2 2 -2 (3x+2)(5x – 1)

Factoring ax2+bx+c Factor 10x2 +9x +2 10•2= 20 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 5 4 9 5x 2 2x 5x 4x 10x2 1 2 (2x+1)(5x +2)

Your Turn: Factor: 2x² - 13x + 6

Factor 2x2 – 13x +6 -12 -1 -13 2x -1 x -12x -x 2x2 -6 6 2•6= 12 Factoring ax2+bx+c Factor 2x2 – 13x +6 2•6= 12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. -12 -1 -13 2x -1 x -12x -x 2x2 -6 6 (2x – 1)(x – 6)

Your Turn: Factor: 4x² + 11x – 3

Factor 4x²+11x – 3 12 -1 11 4x -1 x 12x -x 4x2 3 -3 4•-3= -12 Factoring ax2+bx+c Factor 4x²+11x – 3 4•-3= -12 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 12 -1 11 4x -1 x 12x -x 4x2 3 -3 (4x – 1)(x + 3)

Your Turn: Factor: 3x²+8x+4

Factor 3x²+8x+4 6 2 8 3x 2 x 6x 2x 3x2 2 4 4•3= 12 Factoring ax2+bx+c Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 6 2 8 3x 2 x 6x 2x 3x2 2 4 (3x + 2)(x + 2)

Your Turn: Factor: -5x²+6x-1

Factor -5x²+6x-1 5 1 6 -5x 1 x 5x x -5x2 -1 -1 -5•-1= 5 Factoring ax2+bx+c Factor -5x²+6x-1 -5•-1= 5 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. 5 1 6 -5x 1 x 5x x -5x2 -1 -1 (-5x + 1)(x – 1)

Factor Completely * Always Factor out the GCF First * To factor a polynomial completely, first factor at the GCF of the polynomial’s terms. Then factor the remaining polynomial until it is written as the product of polynomials that cannot be factored further. * Always Factor out the GCF First *

Example: Factor Completely Factor Completely: 18x2 – 33x + 12 First factor out the GCF 3 3(6x2 – 11x + 4) Then factor the remaining polynomial (6x2 – 11x + 4)

Factor 6x2 – 11x+4 -8 -3 -11 2x -1 3x -8x -3x 6x2 -4 4 6•4= 24 Factoring ax2+bx+c Factor 6x2 – 11x+4 6•4= 24 Set up & Solve X-factor: Put ac on top Put b on bottom Determine side factors Set up & Solve “BOX”: Put First Term in First Box Put Last Term in Last Box Put Side Factors Times x in Box Determine the GCF of the upper boxes and the remaining top & side factors. The sum of the factor’s for the top and the side are the factors of the polynomial. -8 -3 -11 2x -1 3x -8x -3x 6x2 -4 4 (3x – 4)(2x – 1)

Example: Factor Completely Factor Completely: 18x2 – 33x + 12 First factor out the GCF 3 3(6x2 – 11x + 4) Then factor the remaining polynomial (6x2 – 11x + 4) (3x – 4)(2x – 1) Completely Factored: 3(3x – 4)(2x – 1)

Your Turn: Completely factor 8x2 – 36x – 20 4(2x + 1)(x – 5)

Perfect Square Trinomial

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.

A perfect square trinomial results after squaring a binomial Example: (2x – 5)2 Multiply it out using FOIL (2x – 5)2 = (2x – 5) (2x – 5) = 4x2 – 10x – 10x + 25 F O I L = 4x2 – 20x + 25 The first and last terms are perfect squares. The middle term is double the product of the square roots of the first and last terms.

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

So how do we factor 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

Example – First verify it is a Perfect Square Trinomial = (3w + 4)2 = (4x – 9)2 = (5h – 16)(5h – 4) = (a + 3)2 m2 – 4m + 4 9w2 + 24w + 16 16x2 – 72x + 81 25h2 – 100h + 64 a2 + 6a + 9 m 2m 2 3w 12w 4 4x 36x 9 Not a perfect square trinomial! 5h 40h 8 a 3a 3

Your Turn: = (m – 3)2 m2 – 6m + 9 = (2w + 7)2 4w2 + 28w + 49 = (9x – 1)2 = (3a + 2)2 m2 – 6m + 9 4w2 + 28w + 49 81x2 – 18x + 1 9a2 + 12a + 4 m 3m 3 2w 14w 7 9x 9x 1 3a 6a 2

Review: Perfect-Square Trinomial

Difference of Two Squares D.O.T.S.

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 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!

Difference of Two Squares Example: Factor the polynomial x 2 – 9. The first term is a square and the last term, 9, can be written as 32. The signs of each term are different, so we have the difference of two squares Therefore x 2 – 9 = (x – 3)(x + 3). Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

Difference of Two Squares Example: Factor x2 - 64 = (x + 8)(x - 8) x2 = x • x 64 = 8 • 8 Example: Factor 9t2 - 25 = (3t + 5)(3t - 5) 9t2 = 3t • 3t 25 = 5 • 5

Difference of Two Squares Example: Factor x 2 – 16. Since this polynomial can be written as x 2 – 42, x 2 – 16 = (x – 4)(x + 4). Factor 9x2 – 4. Since this polynomial can be written as (3x)2 – 22, 9x 2 – 4 = (3x – 2)(3x + 2). Factor 16x 2 – 9y 2. Since this polynomial can be written as (4x)2 – (3y)2, 16x 2 – 9y 2 = (4x – 3y)(4x + 3y).

A Sum of Squares? A Sum of Squares, like x2 + 64, can NOT be factored! It is a PRIME polynomial.

Difference of Two Squares Example: Factor x 8 – y 6. Since this polynomial can be written as (x 4)2 – (y 3)2, x 8 – y 6 = (x 4 – y 3)(x 4 + y 3). Factor x2 + 4. Oops, this is the sum of squares, not the difference of squares, so it can’t be factored. This polynomial is a prime polynomial.

Difference of Two Squares Example: Factor 36x 2 – 64. Remember that you should always factor out any common factors, if they exist, before you start any other technique. Factor out the GCF. 36x 2 – 64 = 4(9x 2 – 16) Since the polynomial can be written as (3x)2 – (4)2, (9x 2 – 16) = (3x – 4)(3x + 4). So our final result is 36x 2 – 64 = 4(3x – 4)(3x + 4).

Recognizing D.O.T.S. Reading Math Recognize a difference of two squares: the coefficients of variable terms are perfect squares, powers on variable terms are even, and constants are perfect squares. Reading Math

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 3p 2 – 9q 4 3p 2 – 9q 4 3q 2  3q 2 3p2 is not a perfect square. 3p 2 – 9q 4 is not the difference of two squares because 3p 2 is not a perfect square.

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 100x 2 – 4y 2 100x 2 – 4y 2 2y 2y  10x 10x The polynomial is a difference of two squares. (10x)2 – (2y)2 a = 10x, b = 2y (10x + 2y)(10x – 2y) Write the polynomial as (a + b)(a – b). 100x2 – 4y2 = (10x + 2y)(10x – 2y)

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. x 4 – 25y 6 x 4 – 25y 6 5y 3 5y 3  x 2 x 2 The polynomial is a difference of two squares. (x 2)2 – (5y 3)2 a = x2, b = 5y3 Write the polynomial as (a + b)(a – b). (x 2 + 5y 3)(x 2 – 5y 3) x 4 – 25y 6 = (x 2 + 5y 3)(x 2 – 5y 3)

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 1 – 4x 2 1 – 4x 2 2x 2x  1 1 The polynomial is a difference of two squares. (1) – (2x)2 a = 1, b = 2x (1 + 2x)(1 – 2x) Write the polynomial as (a + b)(a – b). 1 – 4x 2 = (1 + 2x)(1 – 2x)

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. p 8 – 49q 6 p 8 – 49q 6 7q 3 7q 3 ● p 4 p 4 The polynomial is a difference of two squares. (p 4)2 – (7q 3)2 a = p4, b = 7q3 (p 4 + 7q 3)(p 4 – 7q 3) Write the polynomial as (a + b)(a – b). p 8 – 49q 6 = (p 4 + 7q 3)(p 4 – 7q 3)

Your Turn: Determine whether each binomial is a difference of two squares. If so, factor. If not, explain. 16x 2 – 4y 5 16x 2 – 4y 5 4x  4x 4y5 is not a perfect square. 16x 2 – 4y 5 is not the difference of two squares because 4y 5 is not a perfect square.

Big Idea: Solving Equations and Inequalities Essential Question Big Idea: Solving Equations and Inequalities Why is it useful to recognize special quadratic expressions? It is easier to factor a quadratic expression if you identify it as a perfect square trinomial or a difference of squares. Use the corresponding property to factor the quadratic expression.

Assignment Section 4-4 Part 2, Pg 234 – 236; #1 – 5 all, 8 – 44 even.