MA1210 College Assignment: Factoring Polynomials And Solving Rational Expressions Week 2 Lab

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Page1MA1210 College Assignment: Factoring Polynomials and SolvingRational Expressions Week 2 LabFactorizing Polynomials and Solving Rational ExpressionsAnswer the following questions and show all your calculations. There are differentmethods forfactoring. All methods involve some sort of trial and error. That’s why it isalways important to check your answer by foiling.1.Factor a trinomial whose leading coefficient is 1.Pick any one of the problems andsolve the trinomial. If the trinomial is prime, state this and explain why.a.x2+8x+15= (x + 3)(x + 5)b.x2–4x–5= (x-5)(x + 1)c.x2–14x+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 + 15We need to factor thetrinomial 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, thefactored 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-5Now, 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: Thefactors 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

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Page2Lastly, 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)Eachtrinomial can be factored, and none of them are prime because they have validfactorizations.2.Factor a trinomial whose leading coefficient is not 1.Pick any three problems andfactor the trinomial. If the trinomial is prime, state this and explain why.a.2x2+5x–3= (2x-1)(x+ 3)b.3x2–2x–5= (3x-5)(x + 1)c.6x2–17x+12=17±√(−17)2−4(6)(12)2(6)=17±112=1812𝑜𝑟1612This implies that this trinomial is prime, since solutions are not integers.d.8x2+33x+4=−33±√(33)2−4(8)(4)2(8)=−33±3116=−216𝑜𝑟−6416=−4This implies that this trinomial is prime, since solutions are not integers.e.9x2+5x–4=−5±√(5)2−4(9)(−4)2(9)=−5±1318=818𝑜𝑟−1818=−1This implies that this trinomial is prime, since solutions are notintegers.f.15x2–19x+6=19±√(−19)2−4(15)(6)2(15)=19±130=2030𝑜𝑟1830This implies that this trinomial is prime, since solutions are not integers.

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Page3Answer:Let's work through each trinomial and factor them, if possible, or state whether they areprimebased on the quadratic formula.2. a. 2x2+5x−32x^2 + 5x-3For 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-5For 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 + 12Let'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

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