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Factoring Polynomials. GCF. Factor by grouping. Factor a trinomial

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Presentation on theme: "Factoring Polynomials. GCF. Factor by grouping. Factor a trinomial"— Presentation transcript:

1 Factoring Polynomials. GCF. Factor by grouping. Factor a trinomial
Factoring Polynomials *GCF *Factor by grouping *Factor a trinomial *Special Cases: Difference of Squares Perfect Square Trinomials -Sum or Difference of Cubes

2 WARM UP - Your HW ?s HERE:

3 Factoring Factoring is the process of writing a polynomial as the product of two or more polynomials. Ex: 6x2 – x – 2 = (2x + 1)(3x – 2) Polynomials that cannot be factored using integer coefficients are called irreducible over the integers or prime. Ex: Not prime: 6x+3 = 3(2x+1) Prime: 2x+1 The goal in factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors is prime or irreducible. In this situation, the polynomial is said to be factored completely. In any factoring problem, the first step is to look for the greatest common factor.

4 Factor by Grouping x3 + 4x2 + 3x + 12 x2(x +4) + 3 (x+4) (x+4) (x2 +3)
1. Group first two and last two terms (x3 + 4x2) (+3x+12) 2. Pull out a GCF from each grouping. ___(x3 + 4x2) ___ (+3x+12) x2(x +4) (x+4) 3. Pull out the GCF of (x+4) for final answer (x+4) (x2 +3)

5 Factor by Grouping x3 + 5x2 – 2x - 10 x2 (x+5) -2 (x+5) (x+5) (x2 -2)
1. Group first two and last two terms (x3 + 5x2) (-2x+10) 2. Pull out a GCF from each grouping. ___(x3 + 5x2) ___ (-2x+10) x2 (x+5) (x+5) 3. Pull out the GCF of (x+4) for final answer (x+5) (x2 -2)

6 Factoring Methods Guess & Check BIG X - By Grouping Box Method

7 A Strategy for Factoring ax2 + bx + c GUESS & CHECK
(Assume, for the moment, that there is no greatest common factor.) 1. Find two First terms whose product is ax2: ( x + )( x + ) = ax2 + bx + c 2. Find two Last terms whose product is c: (x )(x ) = ax2 + bx + c I 3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx: ( x + )( x + ) = ax2 + bx + c O (sum of O + I) If no such combinations exist, the polynomial is prime.

8 Factoring Methods – GUESS & CHECK
(Easy with prime #s or when a =1) 3x2 + 7x + 2 x2+6x + 8 Possible with others too, but harder. 6x2 + 19x + 10

9 Now, follow steps to factor by grouping.
Factoring Methods Grouping- Big X You need 4 terms to factor by grouping, so we’re going to split the 19x: What multiplies to A*C (6x2 *10) and adds to 19? The X can help organize your thoughts. Now, follow steps to factor by grouping. 6x2 + 19x + 10 6x2 + 15x + 4x + 10 (6x2 +15x)(+ 4x +10) 3x(2x+5) +2 (2x+5) (3x+2)(2x+5) You need 4 terms to factor by grouping, so we’re going to split the 19x into something that multiplies to A*C (6x2 *10) and adds to 19. The X can help organize your thoughts.

10 Factoring Methods BOX METHOD 3x +2 2x + 5 6x2 + 19x + 10 A
A*C – 60x2 Find the factors that add to 19x 1* *30 3*20 4*15 5*12 6*10 FINAL ANSWER: (3x+2)(2x+5) BEFORE YOU BEGIN, YOU MUST PULL OUT ANY GCFs! FILL IN THE BOX: Fill in box with the factors: What multiplies to A*C (6x2 *10) and adds to 19? Now, factor rows and columns. 6x2 10 A C 6x2 15x 4x 10 3x +2 You need 4 terms to factor by grouping, so we’re going to split the 19x into something that multiplies to A*C (6x2 *10) and adds to 19. The X can help organize your thoughts. 2x A C

11 Special Cases – Difference of Squares
(a2 – b2)=(a+b)(a-b) Ex: 36x2 – 64y2 25x x4 – 81 x2 + 16

12 Special Cases – Perfect Square Trinomials
a2 ± 2ab + b2 = (a±b)2 Ex: 25x2 – 60x + 36 Sum or Difference of Cubes

13 Special Cases – Sum or Difference of Cubes
a3 + b3 = (a+b)(a2 – ab + b2) a3 - b3 = (a-b)(a2 + ab + b2) Ex: 27x3 + 81 Notice, each only has one negative sign and the first grouping matches the original sign. Sum or Difference of Cubes

14 Factoring Polynomials

15 Common Factors a. 18x3 + 27x2 b. x2(x + 3) + 5(x + 3)
In any factoring problem, the first step is to look for the greatest common factor. The greatest common factor is a n expression of the highest degree that divides each term of the polynomial. The distributive property in the reverse direction can be used to factor out the greatest common factor. ab + ac = a(b + c) a. 18x3 + 27x2 b. x2(x + 3) + 5(x + 3)

16 Text Example Factor: a. 18x3 + 27x2 b. x2(x + 3) + 5(x + 3) Solution
a. We begin by determining the greatest common factor. 9 is the greatest integer that divides 18 and 27. Furthermore, x2 is the greatest expression that divides x3 and x2. Thus, the greatest common factor of the two terms in the polynomial is 9x2. 18x3 + 27x2 = 9x2(2x) + 9x2(3) Express each term with the greatest common factor as a factor. = 9x2(2 x + 3) Factor out the greatest common factor. Skip slide, solve using last slide in class b. In this situation, the greatest common factor is the common binomial factor (x + 3). We factor out this common factor as follows. x2(x + 3) + 5(x + 3) = (x + 3)(x2 + 5) Factor out the common binomial factor.


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