Factoring Trinomials SWBAT: Factor Trinomials by Grouping

Factoring Using Distributive Property Use the distributive property to completely factor the polynomial First, find the GCF of both terms Then write each term as the product of the GCF and its remaining factors Finally, use the distributive property to factor out the GCF

Ex 1 Factor:

Ex 2: Factor

Try Some 1)3) 2)4)

Factoring by Grouping We must have 4 terms Write as a sum of two binomials. Then we factor out the GCF of each grouping Finally we apply the distributive property we used with multiplying binomials.

Factoring by Grouping Factor 1)Group terms with like factors 1)Factor out GCF from each grouping 2)Use Distributive property

Factoring by Grouping Factor

Factor by Grouping Factor Completely

Examples 1) 3) 2) 4)

Factoring Trinomials by Grouping We know that two integers multiplied together are both factors of the product. Similarly, when two binomials are multiplied, each binomial is a factor of the product When we factor trinomials, we are just reversing the distributive property and using grouping to completely factor We will be factoring trinomials in the form

Factoring Trinomials by Grouping Factor: Our first step is to multiply the coefficient of “a” with the value for “c” 1 (6) = 6 Next we want to get all the factors of our product 1(6), 2(3)

Continued.. Now that we have the factors of our product, we need to choose the factors that add up to our middle coefficient of our polynomial: which is 5 The factors that will work are 3 and 2, and now we can rewrite our polynomial Now we can group our terms and factor as we did before

Continued Grouping and factoring gives us: Which are the original two binomials we started with

Factoring Trinomials Factor the polynomial Multiply the first coefficient by the last coefficient 1(8) = 8 Find factors of 8: 1(8) and 2(4) Which two factors add up to our “b” value of 6? We are going to use the factors 2 and 4 to rewrite out equation

Continued Our new equation becomes

Factoring Trinomials We need to remember that the values for “b” and “c” might not always be positive. When we look at which factors sum up to “b”, we have to look as the negative factors as well. Lets try some together…

Example

Try some examples 1) 2) 3) 4)

Factoring Trinomials (a is not 1) When we factored our previous trinomials, the coefficient of our x 2 term was always one We will now look at trinomials where the x 2 coefficient is greater than one. Our trinomials will be in the form ax 2 + bx + c We will still use the steps we used before, but now we will have more factors to deal with.

Example Factor 6x x + 5 First multiply our a term and our c term together (6)(5) = 30 Now lets find our factors of 30 (1)(30)(2)(15),(3)(10),(5)(6) Now, same as we did previously, we need to pick the factors that add up to our b term, which is = 17 Now let us rewrite our equation using the products we picked.

Example Factor 6x x + 5 We rewrite our equation into: 6x 2 + 2x + 15x + 5 Now we can solve by factoring with grouping (6x 2 + 2x) + (15x + 5) 2x(3x + 1) + 5(3x + 1) (2x + 5)(3x + 1)

Example Factor 2x 2 - 3x – 20 (a)(c) : (2)(-20) = -40 Factors of (-40) (1)(-40) (2)(-20)(4)(-10) (5)(-8) (-1)(40) (-2)(20)(-4)(10) (-5)(8) Which factors add up to – 3 ? 5 – 8 = -3

Example Rewrite with our factors and un-distribute 2x 2 – 8x + 5x – 20 (2x 2 - 8x) + (5x – 20) 2x(x – 4) + 5(x – 4) (2x + 5)(x - 4)

Try Some on your own 1)3x 2 + 5x + 2 2)5d 2 + 6d – 8 3)9g 2 – 12g + 4

Prime Trinomials A polynomial that cannot be written as a product of two polynomials with integral coefficients is called a prime polynomial Lets try to factor 2x 2 + 5x – 2 (a)(c) : (2)(-2) = - 4 Factors of – 4: – (1)(-4) (-1)(4) (2)(-2) We can see that none of these factors add up to our b term, which is 5 Therefore, since there are no factors that sum to 5, the polynomials cannot be factored using integers, making the polynomial prime