Polynomial factorization #11 - Questions and Answers

Question 1 Factor the terms in each set, then identify the greatest common factor. α) 6, 15n b) 4m, m² Answer a) To find the GCF, list the factors and multiply those factors both numbers have in common. 6 → 2 · 3 15n → 3 · 5 · n So, the GCF is: 3 b)To find the GCF, list the factors and multiply those factors both numbers have in common. 4m → 2 · 2 · m m²→ m · m So, the GCF is: m Question 2 What is the Greatest Common Factor (GCF) in this trinomial? 12x² - 32x - 56 Answer Step 1 Given 12x² - 32x - 56 We have to find the greatest common factor in the trinomial. Step 2 We have 12x² - 32x - 56 ⇒ 4 × 3 × x² - 4 × 8 × x - 4 × 14 (done factorization) ⇒ 4 x (3 x x² - 8 x x - 14) (taken 4 as common) ⇒ 4x (3x² - 8x – 14) ⇒ 4(3x² - 8x - 14) So, The greatest common factor is 4 Answer Question 3 Find a polynomial f(x) of degree 4 that has the following zeros. 5(multiplicity 2), -6 0 f(x)= Answer 1 Step 1 If a, b, c and d are the zeroes of polynomila f(x) of degree 4 then the polynomial f(x) = (x a) (x - b) (x - c) (x - d) Step 2 In this case, the roots are 5 (multiplicity of 2), -6, 0 hence a = 5, b = 5, c = -6 and d = 0 Hence, f(x) in the factored form = (x - 5)(x - 5)(x + 6)(x - 0) = x(x - 5)²(x+6) Answer 2 Explanation: To find a polynomial f(x) of degree 4 with the given zeros, we can start by considering the factored form of a polynomial. The factored form is given by: f(x) = a(x - r₁)(x - r₂)(x - r₃)(x - r₄), where r₁, r₂, r₃, and r₄ are the zeros of the polynomial and a is a constant. Given the zeros 5 (multiplicity 2), -6, and 0, we can substitute these values into the factored form. Using this information, the polynomial f(x) becomes: f(x) = a(x - 5) (x - 5)(x + 6)(x - 0). To determine the value of a, we need additional information. Let's assume that the leading coefficient of the polynomial is 1. This means that a = 1. Therefore, the polynomial f(x) can be expressed as: f(x) = (x - 5)(x - 5)(x + 6) (x − 0). Expanding this expression, we get: f(x) = (x - 5)²(x + 6)x. Answer 3 To find a polynomial f(x) of degree 4 with the given zeros, we can use the fact that the polynomial can be expressed as the product of linear factors corresponding to each zero. The zeros are: 5 with multiplicity 2, -6, and 0. The linear factors are: x - 5 (for the zero 5 with multiplicity 1) (x-5)² (for the zero 5 with multiplicity 2) x + 6 (for the zero -6) x (for the zero 0) Now we can multiply these factors together to get the polynomial f(x): f(x) = (x - 5)(x - 5)(x + 6)(x) Simplifying further: f(x) = (x² - 10x + 25)(x² + 6x) Expanding this expression: f(x) = x⁴ + 6x³ - 10x³ - 60x² + 25x² + 150x f(x) = x⁴ - 4x³ - 35x² + 150x Therefore, the polynomial f(x) of degree 4 with the given zeros is: f(x) = x⁴ - 4x³ - 35x² + 150x Answer 4 Using the zero-product property, we can write the polynomial f(x) as the product of its linear factors: f(x) = a(x - r₁)(x - r₂)(x - r₃)(x - r₄), where a is a constant and r₁, r₂, r₃, and r₄ are the zeros of the polynomial. In this case, we have the zeros 5 (with a multiplicity of 2), -6, and 0. Therefore, we can write: f(x) = a(x - 5)²(x + 6) (x – 0). To determine the value of the constant a, we can use one of the given zeros. Let's use x = 0 and substitute it into the polynomial: f(0) = a(0 - 5)²(0 + 6)(0 - 0). Since f(0) is equal to 0 (one of the given zeros), we can solve for a: 0 = a(-5)²(6) (0). Simplifying, we get: 0 = a(25)(6)(0). Since any number multiplied by 0 is equal to 0, we have: 0 = 0. This equation is always true, regardless of the value of a. Therefore, we can choose any value for a. Let's choose a = 1. Then the polynomial f(x) becomes: f(x) = (x - 5)²(x + 6)x. Thus, the polynomial f(x) of degree 4 with the given zeros is: f(x) = (x - 5)²(x + 6)x. Question 4 Find a polynomial f(x) of degree 3 that has the following zeros. 2 (multiplicity 2), -6 f(x)= Answer The polynomial of degree 3 is computed. The factors of the polynomial = (x-2)(x-2)(x+6). The polynomial of degree 3, f(x)=(x-2)(x-2)(x+6) = (x-2)²(x+6) = (x²+4-4x) (x+6) = x³ + 4x - 4x² + 6x² + 24 - 24x = x³ + 2x² - 20x + 24 Question 5 Factor the polynomial completely, and find all its zeros. State the multiplicity of each zero. P(x) = x³ + x² + 9x +9 Answer Step 1 The Given Polynomial P(x) = x³ + x² + 9x +9 P(x) = x³ + x² + 9x +9 P(x) = x²(x+1) + 9(x+1) P(x) = (x² + 9)(x+1) find the zeros of P by setting each factor equal to 0 P(x) = (x² + 9)(x+1) let, x+1=0 x=-1 Step 2 SO, (x² + 9) (x + 1) = 0 (x²+9)(-1+1) = 0 x²+9=0 x² = -9 x = ±3i therefore, the zero of P are (-1, ±3i) by factorization theorem: p(x)=(x+1)(x-3i)(x+3i) thus polynomial P(x) has the following zeros -1 (multiplicity of 1) 3i (multiplicity of 1) -3i (multiplicity of 1)