Questions and Answers #91 Polynomial Factorization

Questions and Answers Sheet 91 Polynomial Factorization Question #1 How do you multiply (2𝑥 + 5)3 ? Answer: If there is (𝑎 + 𝑏)(𝑐 + 𝑑) then: (𝑎 + 𝑏)(𝑐 + 𝑑) = 𝑎 ∗ 𝑐 + 𝑎 ∗ 𝑑 + 𝑏 ∗ 𝑐 + 𝑏 ∗ 𝑑 (2𝑥 + 5)3 = (2𝑥 + 5) ∗ (2𝑥 + 5) ∗ (2𝑥 + 5) = [4𝑥 2 + 10𝑥 + 10𝑥 + 25] ∗ (2𝑥 + 5) If there is (𝑎 + 𝑏 + 𝑐)(𝑑 + 𝑒) then: (𝑎 + 𝑏 + 𝑐)(𝑑 + 𝑒) = 𝑎𝑑 + 𝑎𝑒 + 𝑏𝑑 + 𝑏𝑒 + 𝑐𝑑 + 𝑐𝑒 = (4𝑥 2 + 20𝑥 + 25) ∗ (2𝑥 + 5) = 8𝑥 3 + 20𝑥 2 + 40𝑥 2 + 100𝑥 + 50𝑥 + 125 The answer is: = 8𝑥 3 + 60𝑥 2 + 150𝑥 + 15 Question #2 How do you multiply (𝑥 + 4)3 ? Answer: Given: (𝑥 + 4)3 (𝑥 + 4)3 = (𝑥 + 4)(𝑥 + 4)(𝑥 + 4) Using foil and (𝑎 + 𝑏)2 = 𝑎2 + 2𝑎𝑏 + 𝑏 2 : (𝑥 + 4)3 = (𝑥 2 + 8𝑥 + 16)(𝑥 + 4) = 𝑥 3 + 8𝑥 2 + 16𝑥 + 4𝑥 2 + 32𝑥 + 64 Add like terms: (𝑥 + 4)3 = 𝑥 3 + 12𝑥 2 + 48𝑥 + 64 Question #3 How do you multiply (2𝑥 + 4)(3𝑥 2 − 5𝑥 + 7)? Answer: Explanation: (2𝑥 + 4)(3𝑥 2 − 5𝑥 + 7) = 2𝑥(3𝑥 2 − 5𝑥 + 7) + 4(3𝑥 2 − 5𝑥 + 7) = 6𝑥 3 − 10𝑥 2 + 14𝑥 + 12𝑥 2 − 20𝑥 + 28 = 6𝑥 3 + 2𝑥 2 − 6𝑥 + 28 Question #4 How do you multiply (𝑥 + 4)(3𝑥 2 − 8𝑥 − 7)? Answer: To multiply two polynomials requires distribution. Each term of one polynomial must multiple each of the terms in the other polynomial. We start with: (𝑥 + 4)(3𝑥 2 − 8𝑥 − 7) = 𝑥(3𝑥 2 − 8𝑥 − 7) + 4(3𝑥 2 − 8𝑥 − 7) First we multiply x from the first polynomial by each term of the second polynomial: = 3𝑥 3 − 8𝑥 2 − 7𝑥 + 4(3𝑥 2 − 8𝑥 − 7) Now we multiply 4 from the first polynomial by each term of the second polynomial: = 3𝑥 3 − 8𝑥 2 − 7𝑥 + 12𝑥 2 − 32𝑥 − 28 Now we simplify: = 3𝑥 3 − 8𝑥 2 − 7𝑥 + 12𝑥 2 − 32𝑥 − 28 = 3𝑥 3 + 4𝑥 2 − 39𝑥 − 28 ⇒ 3𝑥 3 + 4𝑥 2 − 39𝑥 − 28 Question #5 How do you simplify (1 − 2𝑥)(𝑥 2 + 6𝑥 + 3)? Answer: Explanation: We must ensure that each term in the 2nd bracket is multiplied by each term in the 1st bracket.This can be achieved as follows. (1 − 2𝑥)(𝑥 2 + 6𝑥 + 3) = 1(𝑥 2 + 6𝑥 + 3) − 2𝑥(𝑥 2 + 6𝑥 + 3) This shows each term in the 1st bracket multiplying the contents of the 2nd bracket. now, distribute the brackets. ⇒ 𝑥 2 + 6𝑥 + 3 − 2𝑥 3 − 12𝑥 2 − 6𝑥 and collecting 'like terms' −2𝑥 3 + (𝑥 2 − 12𝑥 2 ) + (6𝑥 − 6𝑥) + 3 = −2𝑥 3 − 11𝑥 2 + 3 Question #6 How do you multiply (−𝑏𝑐 + 𝑎2 )(𝑏𝑐 + 𝑎2 )? Answer: Let's switch the terms around in the binomials: (𝑎2 + 𝑏𝑐)(𝑎2 − 𝑏𝑐) This is a difference of squares where: (𝑥 2 − 𝑦 2 ) = (𝑥 + 𝑦)(𝑥 − 𝑦) If we let 𝑥 = 𝑎2 and 𝑦 = 𝑏𝑐 ((𝑎2 )2 − (𝑏𝑐)2 ) = (𝑎2 + 𝑏𝑐)(𝑎2 − 𝑏𝑐) Question #7 How do you multiply [(2𝑧 + 𝑟) + 1]2 ? Answer: [(2𝑧 + 𝑟) + 1]2 = ((2𝑧 + 𝑟) + 1)((2𝑧 + 𝑟) + 1) expand factors using FOIL = (2𝑧 + 𝑟)2 + 2(2𝑧 + 𝑟) + 1 = 4𝑧 2 + 4𝑧𝑟 + 𝑟 2 + 4𝑧 + 2𝑟 + 1 = 4𝑧 2 + 𝑟 2 + 4𝑧𝑟 + 4𝑧 + 2𝑟 + 1 Question #8 How do you multiply (𝑥 + 4)2 ? Answer: (𝑥 + 4)2 means (𝑥 + 4) × (𝑥 + 4) By the Distributive Property this can be written 𝑥 × (𝑥 + 4) + 4 × (𝑥 + 4) Applying the Distributive Property again (but in the reverse form): (𝑥 2 + 4𝑥) + (4𝑥 + 16) Collecting like terms: 𝑥 2 + 8𝑥 + 16 Question #9 How do you find the product of (𝑥 + 1)(𝑥 2 + 𝑥 + 1)? Answer: To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis. (𝑥 + 1)(𝑥 2 + 𝑥 + 1) becomes: (𝑥 × 𝑥 2 ) + (𝑥 × 𝑥) + (𝑥 × 1) + (1 × 𝑥 2 ) + (1 × 𝑥) + (1 × 1) 𝑥3 + 𝑥2 + 𝑥 + 𝑥2 + 𝑥 + 1 We can now group and combine like terms: 𝑥 3 + 1𝑥 2 + 1𝑥 + 1𝑥 2 + 1𝑥 + 1 𝑥 3 + 1𝑥 2 + 1𝑥 2 + 1𝑥 + 1𝑥 + 1 𝑥 3 + (1 + 1)𝑥 2 + (1 + 1)𝑥 + 1 𝑥 3 + 2𝑥 2 + 2𝑥 + 1 Question #10 How do you find the product of (𝑥 + 5)2 ? Answer: We can use Binomial Expansion to find the product. The Binomial Theorem States that, For Any Integer 𝑛 > 0, (𝑥 + 𝑦)𝑛 =(n combination 0) 𝑥 𝑛 +(n combination 1) 𝑥 𝑛−1 𝑦1 + ⋯ + (n combination 𝑛 − 1) 𝑥 1 𝑦 𝑛−1 + (n combination n) 𝑦 𝑛 So, Here, use can use the formula. (𝑥 + 5)2 = (2 combination 0) 𝑥 2 + (2 combination 1) 𝑥 2−1 51 + (2 combination 1) 𝑥 1 52−1 + (2 combination 2) 52 = 1 × 𝑥 2 + 1 × 𝑥 ∗ 5 + 1 × 𝑥 ∗ 5 + 1 × 52 = 𝑥 2 + 5𝑥 + 5𝑥 + 25 = 𝑥 2 + 10𝑥 + 25