Factoring by using different methods

Factoring a quadratic without a number in front of “x2”

The guess method Ex: x2+4x-5 Find two numbers that multiply to give you the last number and also add to give you the middle number. The second number is +4 and the last number is -5.

Ex: x2+4x-5 2. The two numbers would be “+5” and “-1”. Since 5•-1= -5 and 5+(-1)=4. 3. Make the two sets of parenthesis and insert the two numbers you found. When there is no number in front of “x2”, the parenthesis start out with “x” (x+5)(x-1)

Using the quadratic formula Ex: x2+4x-5 1. Label “a”,”b”, and “c” a=1, b=4, c=-5 2. Plug in the numbers and solve. (Note, this is only useful if you are finding x-intercepts or solving the equation.)

Factor by grouping Used when there are four terms in the polynomial. Goal: create two binomials out of the polynomial by using other factoring methods. Check: FOIL the two binomials and if the product is the beginning polynomial, the two binomials are correct.

Factor by Grouping Example of a polynomial: (10x2 + 5x – 2x – 1) Change to (10x2 +5x)(-2x -1) Take GCF of both binomials. 5x(2x + 1)-1(2x+1) (Notice that the remaining binomials are the same) Keep the identical binomial (5x-1)(2x+1)

Difference of squares Most will look like: (a2 – b2) Factored form: (a – b)(a + b) Hints: Be able to identify the squared forms of numbers, such as :22 = 4, 32 = 9, 42=16… etc.

Other special Factoring Cubic Factors: (a3 ± b3) Two factored forms: (a3 + b3) = (a + b)(a2 – ab + b2) (a3 – b3) = (a – b)(a2 + ab + b2) Helpful hint: know the cubed form of numbers: 23 = 8, 33 = 27, 43 = 64, etc.

Factoring quadratic equations with a number in front of the “x2” term. By the guess method By the a-c method With the quadratic formula

Using the guess method Ex. 4x2-12x+9 Find the factors of the first and last terms. Factors of the first term are (4x and x) or (2x and 2x), factors of the last term are (9 and 1) or (3 and 3) Use those factors to build the two sets of parenthesis. Foil the answer you got to see if it is correct. If it isn’t, choose a different combination.

3. Use those factors to build the two sets of parenthesis 3. Use those factors to build the two sets of parenthesis. For example, (2x+3)(2x+3) 4.Foil the answer you got to see if it is correct. If it isn’t, choose a different combination.

Using the a-c method Ex. 6x2-5x-6 Multiply “a” by “c”. a = 6 b = -5 c = -6

Using the a-c method Ex. 6x2-5x-6 Multiply “a” by “c”. a = 6 b = -5 c = -6 6•-6 = 36

2. Find two numbers that multiply to “a•c” and add up to give you “b” 2. Find two numbers that multiply to “a•c” and add up to give you “b”. An easy way to do this is to list the factors of “a•c”. a•c = -36 b = -5 What are the numbers?

2. Find two numbers that multiply to “a•c” and add up to give you “b” 2. Find two numbers that multiply to “a•c” and add up to give you “b”. An easy way to do this is to list the factors of “a•c”. a•c = -36 b = -5 What are the numbers? (-9)•(4) = -36 (-9)+(4) = -5

3. Take the two numbers that we found in part 2 and substitute them in for the “-5x” in the original equation, resulting in 6x2-9x+4x-6

6x2-9x+4x-6 Since there are now four terms, we will use factoring by grouping. So, break the equation apart. (6x2-9x)(4x-6)

(6x2-9x)(4x-6) Factor each set of parenthesis separately. The terms inside both parenthesis have to match.

(6x2-9x)(4x-6) Factor each set of parenthesis separately. The terms inside both parenthesis have to match. 3x(x-3) and +2(x-3)

3x(x-3) and +2(x-3) Combine the two outside terms in a parenthesis and keep one of the matching sets in another. (3x+2)(x-3)

Using the quadratic formula Ex. 6x2-5x-6 1. Locate the three numbers for “a”,”b”, and “c”. a = 6 b = -5 c = -6

2. Plug the values into the quadratic formula. 3. solve