Factoring Polynomials This section will cover the following topics Factoring the Greatest Common Factor Factoring Trinomials by Trial and Error Solving Equations by Factoring Factoring the Greatest Common Factor The most basic type of factoring for polynomials is to factor out the Greatest Common Factor (GCF). The goal of factoring is to undo multiplication. Letβs take a look at what multiplying a single term into a polynomial looks like, and then we will work backwards. Example of Multiplication of a Polynomial by a Single Term ππ ππππππ (2π₯π₯ 3 β 3π₯π₯ 2 + 5π₯π₯ β 4) 3 ππ 2 ππ ππ ππππ · 2π₯π₯ β ππππ · 3π₯π₯ + ππππ · 5π₯π₯ β ππππ · 4 6π₯π₯ 5 β 9π₯π₯ 4 + 15π₯π₯ 3 β 12π₯π₯ 2 We must first distribute the ππππππ to each term inside the parentheses, and then multiply term by term. Working backwards, letβs start with the polynomial 6π₯π₯ 5 β 9π₯π₯ 4 + 15π₯π₯ 3 β 12π₯π₯ 2 . When factoring the GCF deal with the numbers and each variable separately to determine the overall GCF. Finding the GCF of ππππππ β ππππππ + ππππππππ β ππππππππ GCF of the Coefficients (dealing with the numbers) 6, 9, 15, and 12 are the coefficients All of these numbers are divisible by 1 and 3 only. Always take the highest number, which is this case is 3 These include π₯π₯ 5 , π₯π₯ 4 , π₯π₯ 3 , and π₯π₯ 2 GCF of the Variable (dealing with the x- To find the GCF of variables, take the variable raised to the variable) lowest exponent. In this case, that is π₯π₯ 2 The Overall GCF Factor the GCF Start by factoring 3π₯π₯ 2 from each term. Then factor 3π₯π₯ 2 outside parentheses and the remaining terms inside Putting the GCF of the numbers and variables together, we get 6π₯π₯ 5 ππππππ · 2π₯π₯ 3 β β 9π₯π₯ 4 GCF = 3π₯π₯ 2 ππππππ · 3π₯π₯ 2 + + 15π₯π₯ 3 ππππππ · 5π₯π₯ ππππππ (2π₯π₯ 3 β 3π₯π₯ 2 + 5π₯π₯ β 4) β β 12π₯π₯ 2 ππππππ · 4 Factoring Trinomials by Trial and Error Once again, we will start with the idea that factoring will undo multiplication. For trinomials (polynomials with three terms), this means we will be undoing FOIL-ing (see the review on Polynomials for details). Example of FOIL-ing First Term · First Term Outside Term · Outside Term Inside Term · Inside Term Last Term · Last Term Finally, combine any like terms (ππ + 3)(ππ β 7) = ππ · ππ (ππ + 3)(π₯π₯ β ππ) = π₯π₯ 2 β ππ · ππ (π₯π₯ + ππ)(ππ β 7) = π₯π₯ 2 β 7π₯π₯ + ππ · ππ (π₯π₯ + ππ)(π₯π₯ β ππ) = π₯π₯ 2 β 7π₯π₯ + 3π₯π₯ β ππππ (π₯π₯ + 3)(π₯π₯ β 7) = π₯π₯ 2 β 4π₯π₯ β 21 To work backwards, we will start by considering the possible ways to factor the first term π₯π₯ 2 and the last term β21. We will then write all possible factorizations based on those Example: Factor π₯π₯ 2 β 4π₯π₯ β 21 Possible Factors of First Term Possible Factors of Last Term Possible Factorization Check by FOIL-ing ππππ = ππ · ππ βππππ = βππ · ππ (ππ β ππ)(ππ + ππ) π₯π₯ 2 + 4π₯π₯ β 21 ππππ = ππ · ππ βππππ = βππ · ππππ (ππ β ππ)(ππ + ππππ) π₯π₯ 2 + 20π₯π₯ β 21 Possible Factorization Check by FOIL-ing (ππππ + ππ)(ππ + ππ) 2π₯π₯ 2 + 7π₯π₯ + 3 (ππππ β ππ)(ππ β ππ) 2π₯π₯ 2 β 7π₯π₯ + 3 ππππ = ππ · ππ βππππ = ππ · βππ (ππ + ππ)(ππ β ππ) π₯π₯ 2 β 4π₯π₯ β 21* (ππ + ππ)(ππ β ππππ) βππππ = ππ · βππππ π₯π₯ 2 β 20π₯π₯ β 21 ππππ = ππ · ππ *Note that we could have stopped at the second row because we found the factorization. Example 2: Factor 2π₯π₯ 2 β 5π₯π₯ + 3 Possible Factors of First Term Possible Factors of Last Term ππ ππ = ππ · ππ ππ ππ = βππ · βππ ππππ = ππππ · ππ ππππ = ππππ · ππ (ππππ + ππ)(ππ + ππ) (ππππ β ππ)(ππ β ππ) 2π₯π₯ 2 + 5π₯π₯ + 3 2π₯π₯ 2 β 5π₯π₯ + 3 A Quick Tip Factoring can get complicated very quickly, and so can factoring techniques. On the placement exam, keep it simple. These represent the difficulty level you will find on the exam. Solving Equations by Factoring A very important point about solving equations by factoring is that one side of the equation must be equal to zero. Once you have that, solving equations by factoring is easy; simply factor and then set each factor equal to zero. Example 1: 2π₯π₯ 2 β 6π₯π₯ = 0 Factor out the GCF Set each factor equal to zero Solve each equation 2π₯π₯(π₯π₯ β 3) = 0 2π₯π₯ = 0 π₯π₯ β 3 = 0 2π₯π₯ = 0 π₯π₯ = 3 Example 2: 3π₯π₯ 2 β 5π₯π₯ = 2 3π₯π₯ 2 β 5π₯π₯ β 2 = 0 Write equation with = 0 Factor Set each factor equal to zero Solve each equation (3π₯π₯ + 1)(π₯π₯ β 2) = 0 3π₯π₯ + 1 = 0 π₯π₯ = β Practice Problems 1 3 π₯π₯ β 2 = 0 π₯π₯ = 2 Factor the following expressions, or factoring to solve the following equations 1. 6π₯π₯ 5 + 9π₯π₯ 4 β 24π₯π₯ 3 + 18π₯π₯ 2 4. 3π₯π₯ 2 β 4π₯π₯ = 0 2. π₯π₯ 2 β 4π₯π₯ β 32 5. π₯π₯ 2 β 3π₯π₯ β 28 = 0 3. 3π₯π₯ 2 + 14π₯π₯ β 5 6. 2π₯π₯ 2 β 5π₯π₯ = 7 Answers 1. 3π₯π₯ 2 (2π₯π₯ 3 + 3π₯π₯ 2 β 8π₯π₯ + 6) 4 4. π₯π₯ = 0, 3 2. (π₯π₯ β 8)(π₯π₯ + 4) 5. π₯π₯ = 7, β4 3. (3π₯π₯ β 1)(π₯π₯ + 5) 7 6. π₯π₯ = β1, 2 Additional Help You can also search YouTube.com for βfactoring the GCFβ, βfactoring trinomialsβ, or βsolving quadratic equations by factoringβ
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