Simplify Polynomial Factoring with These Step-by-Step Methods

Greatest Common Factor and Factoring by Grouping Section R.3 Obj. 4 Factoring Out the Greatest Common Factor Factoring Out the Greatest Common Factor Step 1 Identify the greatest common factor of all terms of the polynomial. Step 2 Write each term as the product of the GCF and another factor. Step 3 Use the distributive property to factor out the greatest common factor. Note: To check the factorization, multiply the polynomials. 1. Factor: 8x8 − 20 x6 − 12 x4 − 20 x3 2. Factor: y3 − 13 yn3 + 11y5n7 − 19 y3n3 3. Factor: 30 x4 y 6 z 2 − 12 x7 y3 z 4 Factoring Out a Negative Factor If the first term is negative, we usually factor out a negative common factor, or just a –1. 4. Factor: −3x2 + 12 x − 18 5. Factor: − y3 + 9 y 2 + 21y + 18 6. Factor: −15u 4 − 10u3 + 25u Factoring Out a Binomial Factor 7. Factor: 5x(2 x + 3) − 2(2 x + 3) 8. Factor: 5m(3 − m) + 9n(3 − m) 9. Factor: 2 x( y − 3) + 4c(3 − y) Note: Recall that b − a = −(a − b) Factoring by Grouping Factoring by Grouping To factor a four-term polynomial by grouping: Step 1: Identify and factor out the GCF from all four terms Step 2: Factor out the GCF from the first pair of terms and the GCF from the second pair of terms. Step 3: If the two terms share a common binomial factor, factor out the binomial factor. 10. Factor: 20 x2 − 25xy + 24 xy − 30 y 2 11. Factor: x3 − 5x2 + 4 x − 20 12. Factor: x3 + 8x2 − 5x − 40 13. Factor: 20 x6 −15x3 −16 x3 + 12 Factoring Trinomials Section R.3 Obj 5 Factoring Trinomials: AC-Method The AC-Method to Factor ax 2 + bx + c, a  0 Step 1 After factoring out the GCF, multiply the coefficients of the first and last terms, ac. Step 2 Find two integers whose product is ac and whose sum is b. (If no pair of integers can be found, then the trinomial cannot be factored further and is called a prime polynomial.) Step 3 Rewrite the middle term bx as the sum of two terms whose coefficients are the integers found in step 2. Step 4 Factor by grouping. Example: Factor 6 x2 + 11x − 10 Step 1: ac = (6)(−10) = −60 Step 2: Factors of –60: Step 3: 6 x2 + 15x − 4 x −10 Step 4: 3x(2 x + 5) − 2(2 x + 5) 1. Factor: 12 x2 − 17 x − 5 (–60)(1) (60)(–1) (–30)(2) (30)(–2) (–20)(3) (20)(–3) (–15)(4) (15)(–4) This pair will have a sum of 11. (–12)(5) (12)(–5) (–10)(6) (6)(–10) (2 x + 5)(3x − 2) 2. Factor: 8x2 + 6 x + 1 3. Factor: 6 x2 − 11x − 7 4. Factor: 24 x2 − 4 x − 4 5. Factor: 6 x2 + 37 x + 6 6. Factor: 6 x2 + 5x − 6 7. Factor: −6a2 − 25a − 25 8. Factor: 5x2 −18x + 9 Factoring Perfect Square Trinomials Factored Form of a Perfect Square Trinomial a2 + 2ab + b2 = (a + b)(a + b) = (a + b)2 a2 − 2ab + b2 = (a − b)(a − b) = (a − b)2 Example: Factor: 9 x2 + 12 x + 4 (3𝑥)2 (2)2 2(3𝑥)(2) Therefore, this is factored as (3𝑥 + 2)2 9. Factor: 25x2 − 60 x + 36 10. Factor: 4 x 2 + 11. Factor: 4 1 x+ 3 9 36 x2 − 84 x + 49 Factoring by Using Substitution Example: Factor 3( x − 1)2 − 2( x − 1) − 21 Note that the parentheses are not present in all terms, therefore you can't factor it out. You never want to multiply when you are factoring, it would be like working backwards... Instead, think of the parentheses as a variable, say p (for parenthesis). p = ( x −1) Then the problem, in terms of p, would be: 3 p 2 − 2 p − 21 This is easily factored as (3 p + 7)( p − 3) Substitute the parentheses back: (3( x −1) + 7)(( x − 1) − 3) Simplify: (3x − 3 + 7)( x −1 − 3) = (3x − 4)( x − 4) 12. Factor: 5(m + 2)2 −14(m + 2) − 3 13. Factor: 8(3x − 2)2 + 6(3x − 2) + 1 14. Factor: ( x + 3)2 − ( x + 5)2 Hint: Difference of squares. Factoring Binomials Section R.3 Obj 6 Difference of Squares Factored Form of a Difference of Squares a2 − b2 = (a + b)(a − b) Any even power is a square. The sum of two squares is not factorable. Example: Factor: 25x2 − 9 y 2 (5x)2 − (3 y)2 Factored form: (5x + 3 y)(5x − 3 y) 1. Factor: 36 x 2 − 1 2. Factor: 1 2 9 y − 4 25 3. Factor: 49 z8 − 225x2 4. Factor: m4 n 4 − 4 Using a Difference of Squares in Grouping Example: Factor: xy 2 − 9 x + y 2 − 9 x( y 2 − 9) + 1( y 2 − 9) ( y 2 − 9)( x + 1) ( y + 3)( y − 3)( x + 1) 5. Factor: x2 y − 4 x2 − 25 y + 100 6. Factor: x2 y 2 − x2 − 4 y 2 + 4 7. Factor: u 2 − 4u + 4 − 9 y 2 Hint: Group the first three terms. Applying a General Strategy to Factor Polynomials Section R.3 Obj 7 Factoring Strategies First Step Number of Terms Technique 4 or More Factor by Grouping 3 (trinomial) Factoring If a perfect trinomial (𝑎2 + 2𝑎𝑏 + 𝑏 2 ) = (𝑎 + 𝑏)2 (𝑎2 − 2𝑎𝑏 + 𝑏 2 ) = (𝑎 − 𝑏)2 Pull out (factor) GCF 2 (binomial) Difference of Squares (𝑎2 − 𝑏 2 ) = (𝑎 − 𝑏)(𝑎 + 𝑏) Sum/ Difference of Cubes (𝑎3 + 𝑏 3 ) = (𝑎 + 𝑏)(𝑎2 − 𝑎𝑏 + 𝑏 2 ) (𝑎3 − 𝑏 3 ) = (𝑎 − 𝑏)(𝑎2 + 𝑎𝑏 + 𝑏 2 ) Note: A sum of squares (𝑎2 + 𝑏 2 ) cannot be factored over the real plane (numbers). Factoring a Four Term Polynomial Example: Factor: 𝑥 4 − 𝑥 2 − 64𝑥 2 + 64 Factor the terms in pairs = 𝑥 2 (𝑥 2 − 1) − 64(𝑥 2 − 1) Factor the (𝑥 2 − 1) out of each term = (𝑥 2 − 64)(𝑥 2 − 1) Factor the Difference of Perfect Squares = (𝑥 + 8)(𝑥 − 8)(𝑥 + 1)(𝑥 − 1) 1. 8𝑥 4 − 12𝑥 3 − 128𝑥 2 + 192𝑥 Solution: 4𝑥(2𝑥 3 − 3𝑥 2 − 32𝑥 + 48) Factor out the GCF = 4𝑥(𝑥 2 (2𝑥 − 3) − 16(2𝑥 − 3)) Factor the terms in pairs = 4𝑥(𝑥2 − 16)(2𝑥 − 3) Factor the Difference of Perfect Squares = 4𝑥(𝑥 + 4)(𝑥 − 4)(2𝑥 − 3) Factoring a Trinomial by Using Substitution Example: Factor(𝑥 2 − 2)2 + 5(𝑥 2 − 2) − 14 Let p = 𝑥 2 − 2 and rewrite the equation = 𝑝 2 + 5𝑝 − 14 Factor = (𝑝 + 7)(𝑝 − 2) Now substitute p = 𝑥 2 − 2 = ((𝑥 2 − 2) + 7)((𝑥 2 − 2) − 2) Remove parenthesis and combine terms = (𝑥 2 + 5)(𝑥 2 − 4) Factor the Difference of Perfect Squares = (𝑥 2 + 5)(𝑥 + 2)(𝑥 − 2) 2) (𝑥 2 + 1)2 + 3(𝑥 2 + 1) − 10 Solution: = 𝑝 2 + 3𝑝 − 10 Let p = 𝑥 2 + 1 and rewrite the equation = (𝑝 + 5)(𝑝 − 2) Factor = ((𝑥 2 + 1) + 5)((𝑥 2 + 1) − 2) Now substitute p = 𝑥 2 + 1 = (𝑥 2 + 6)(𝑥 2 − 1) Remove parenthesis and combine terms = (𝑥 2 + 6)(𝑥 + 1)(𝑥 − 1) Factor the Difference of Perfect Squares