Page 1 MA1210 College Assignment: Factoring Polynomials and Solving Rational Expressions Week 2 Lab Factorizing Polynomials and Solving Rational Expressions Answer the following questions and show all your calculations. There are different methods for factoring. All methods involve some sort of trial and error. That’s why it is always important to check your answer by foiling. 1. Factor a trinomial whose leading coefficient is 1 . Pick any one of the problems and solve the trinomial. If the trinomial is prime, state this and explain why. a. x 2 + 8x + 15 = (x + 3)(x + 5) b. x 2 – 4x – 5 = (x - 5)(x + 1) c. x 2 – 14x + 45 = (x - 9)(x - 5) Answer: Let's go through each trinomial and factor them one by one. 1. a. x2+8x+15x^2 + 8x + 15 We need to factor the trinomial x2+8x+15x^2 + 8x + 15. • Step 1: The leading coefficient is 1, so the factors of 15 need to add up to 8. • Step 2: The factors of 15 are (1, 15), (3, 5), ( - 1, - 15), and ( - 3, - 5). • Step 3: The pair that adds up to 8 is 3 and 5. Thus, the factored form of the trinomial is: x2+8x+15=(x+3)(x+5)x^2 + 8x + 15 = (x + 3)(x + 5) 1. b. x2−4x−5x^2 - 4x - 5 Now, let's factor x2−4x−5x^2 - 4x - 5. • Step 1: The leading coefficient is 1, so we look for factors of - 5 that add up to - 4. • Step 2: The factors of - 5 are (1, - 5) and ( - 1, 5). • Step 3: The pair that adds up to - 4 is - 5 and 1. Thus, the factored form of the trinomial is: x2−4x−5=(x−5)(x+1)x^2 - 4x - 5 = (x - 5)(x + 1) 1. c. x2−14x+45x^2 - 14x + 45 Page 2 Lastly, let's factor x2−14x+45x^2 - 14x + 45. • Step 1: The leading coefficient is 1, so we look for factors of 45 that add up to - 14. • Step 2: The factors of 45 are (1, 45), (3, 15), (5, 9), ( - 1, - 45), ( - 3, - 15), ( - 5, - 9). • Step 3: The pair that adds up to - 14 is - 9 and - 5. Thus, the factored form of the trinomial is: x2−14x+45=(x−9)(x−5)x^2 - 14x + 45 = (x - 9)(x - 5) Conclusion • a. x2+8x+15=(x+3)(x+5)x^2 + 8x + 15 = (x + 3)(x + 5) • b. x2−4x−5=(x−5)(x+1)x^2 - 4x - 5 = (x - 5)(x + 1) • c. x2−14x+45=(x−9)(x−5)x^2 - 14x + 45 = (x - 9)(x - 5) Each trinomial can be factored, and none of them are prime because they have valid factorizations. 2. Factor a trinomial whose leading coefficient is not 1 . Pick any three problems and factor the trinomial. If the trinomial is prime, state this and explain why. a. 2 x 2 + 5x – 3 = (2x - 1 )(x + 3 ) b. 3x 2 – 2x – 5 = (3x - 5)(x + 1) c. 6x 2 – 17x + 12 = 17 ± √ ( − 17 ) 2 − 4 ( 6 ) ( 12 ) 2 ( 6 ) = 17 ± 1 12 = 18 12 𝑜𝑟 16 12 This implies that this trinomial is prime, since solutions are not integers. d. 8x 2 + 33x + 4 = − 33 ± √ ( 33 ) 2 − 4 ( 8 ) ( 4 ) 2 ( 8 ) = − 33 ± 31 16 = − 2 16 𝑜𝑟 − 64 16 = − 4 This implies that this trinomial is prime, since solutions are not integers. e. 9 x 2 + 5x – 4 = − 5 ± √ ( 5 ) 2 − 4 ( 9 ) ( − 4 ) 2 ( 9 ) = − 5 ± 13 18 = 8 18 𝑜𝑟 − 18 18 = − 1 This implies that this trinomial is prime, since solutions are not integers. f. 15x 2 – 19x + 6 = 19 ± √ ( − 19 ) 2 − 4 ( 15 ) ( 6 ) 2 ( 15 ) = 19 ± 1 30 = 20 30 𝑜𝑟 18 30 This implies that this trinomial is prime, since solutions are not integers. Page 3 Answer: Let's work through each trinomial and factor them, if possible, or state whether they are prime based on the quadratic formula. 2. a. 2x2+5x−32x^2 + 5x - 3 For this trinomial, the leading coefficient is 2, so we apply factoring by splitting the middle term. • Step 1: Multiply the leading coefficient (2) by the constant term ( - 3), giving - 6. • Step 2: Find two numbers that multiply to - 6 and add up to 5. These numbers are 6 and - 1. • Step 3: Split the middle term: 2x2+6x−x−32x^2 + 6x - x - 3. • Step 4: Factor by grouping: 2x(x+3)−1(x+3)2x(x + 3) - 1(x + 3). • Step 5: Factor out the common binomial: (2x−1)(x+3)(2x - 1)(x + 3). Thus, the factored form of 2x2+5x−32x^2 + 5x - 3 is: 2x2+5x−3=(2x−1)(x+3)2x^2 + 5x - 3 = (2x - 1)(x + 3) 2. b. 3x2−2x−53x^2 - 2x - 5 For this trinomial, the leading coefficient is 3, so we apply factoring by splitting the middle term. • Step 1: Multiply the leading coefficient (3) by the constant term ( - 5), giving - 15. • Step 2: Find two numbers that multiply to - 15 and add up to - 2. These numbers are - 5 and 3. • Step 3: Split the middle term: 3x2−5x+3x−53x^2 - 5x + 3x - 5. • Step 4: Factor by grouping: x(3x−5)+1(3x−5)x(3x - 5) + 1(3x - 5). • Step 5: Factor out the common binomial: (3x−5)(x+1)(3x - 5)(x + 1). Thus, the factored form of 3x2−2x−53x^2 - 2x - 5 is: 3x2−2x−5=(3x−5)(x+1)3x^2 - 2x - 5 = (3x - 5)(x + 1) 2. c. 6x2−17x+126x^2 - 17x + 12 Let's solve this using the quadratic formula, as given in the problem: The quadratic formula is: x=−b±b2−4ac2ax = \ frac{ - b \ pm \ sqrt{b^2 - 4ac}}{2a} For 6x2−17x+126x^2 - 17x + 12, we have: • a=6a = 6