Polynomials #5 - Questions and Answers

Question 1 Find all zeros of 𝑓 ( 𝑥 ) = 𝑥 3 − 4 𝑥 2 + 𝑥 + 6 f(x)=x 3 −4x 2 +x+6 Answer: Given Function: 𝑓 ( 𝑥 ) = 𝑥 3 − 4 𝑥 2 + 𝑥 + 6. f(x)=x 3 −4x 2 +x+6. Find the first zero (Trial Method): Substitute 𝑥 = − 1 x=−1: 𝑓 ( − 1 ) = ( − 1 ) 3 − 4 ( − 1 ) 2 + ( − 1 ) + 6 = − 1 − 4 − 1 + 6 = 0. f(−1)=(−1) 3 −4(−1) 2 +(−1)+6=−1−4−1+6=0. So, 𝑥 = − 1 x=−1 is a zero. Factorize 𝑓 ( 𝑥 ) f(x): Divide 𝑓 ( 𝑥 ) f(x) by ( 𝑥 + 1 ) (x+1): 𝑓 ( 𝑥 ) = ( 𝑥 + 1 ) ( 𝑥 2 − 5 𝑥 + 6 ) . f(x)=(x+1)(x 2 −5x+6). Factor the quadratic 𝑥 2 − 5 𝑥 + 6 x 2 −5x+6: 𝑥 2 − 5 𝑥 + 6 = ( 𝑥 − 3 ) ( 𝑥 − 2 ) . x 2 −5x+6=(x−3)(x−2). Complete Factorization: 𝑓 ( 𝑥 ) = ( 𝑥 + 1 ) ( 𝑥 − 3 ) ( 𝑥 − 2 ) . f(x)=(x+1)(x−3)(x−2). Zeros of 𝑓 ( 𝑥 ) f(x): 𝑥 = − 1 ,   𝑥 = 3 ,   𝑥 = 2. x=−1,x=3,x=2. Question 2 Find 𝑓 + 𝑔 f+g, 𝑓 − 𝑔 f−g, 𝑓 𝑔 fg, and 𝑓 𝑔 g f ​ . Then determine the domain for each function. Given: 𝑓 ( 𝑥 ) = 3 𝑥 2 − 14 𝑥 + 16 ,   𝑔 ( 𝑥 ) = 𝑥 − 2. f(x)=3x 2 −14x+16,g(x)=x−2. Answer: Addition ( 𝑓 + 𝑔 f+g): 𝑓 + 𝑔 = ( 3 𝑥 2 − 14 𝑥 + 16 ) + ( 𝑥 − 2 ) = 3 𝑥 2 − 13 𝑥 + 14. f+g=(3x 2 −14x+16)+(x−2)=3x 2 −13x+14. Subtraction ( 𝑓 − 𝑔 f−g): 𝑓 − 𝑔 = ( 3 𝑥 2 − 14 𝑥 + 16 ) − ( 𝑥 − 2 ) = 3 𝑥 2 − 15 𝑥 + 18. f−g=(3x 2 −14x+16)−(x−2)=3x 2 −15x+18. Multiplication ( 𝑓 𝑔 fg): 𝑓 𝑔 = ( 3 𝑥 2 − 14 𝑥 + 16 ) ( 𝑥 − 2 ) = 3 𝑥 3 − 14 𝑥 2 + 16 𝑥 − 6 𝑥 2 + 28 𝑥 − 32. fg=(3x 2 −14x+16)(x−2)=3x 3 −14x 2 +16x−6x 2 +28x−32. Combine terms: 𝑓 𝑔 = 3 𝑥 3 − 20 𝑥 2 + 44 𝑥 − 32. fg=3x 3 −20x 2 +44x−32. Division ( 𝑓 𝑔 g f ​ ): 𝑓 𝑔 = 3 𝑥 2 − 14 𝑥 + 16 𝑥 − 2 . g f ​ = x−2 3x 2 −14x+16 ​ . Domain: 𝑓 + 𝑔 f+g, 𝑓 − 𝑔 f−g, and 𝑓 𝑔 fg are polynomials, so their domain is all real numbers ( 𝑅 R). For 𝑓 𝑔 g f ​ , the domain excludes 𝑥 = 2 x=2 (where 𝑔 ( 𝑥 ) = 0 g(x)=0): Domain of  𝑓 𝑔 = 𝑅 ∖ { 2 } . Domain of  g f ​ =R∖{2}. Question 3 Factor 1 − 27 𝑎 3 1−27a 3 using the sum/difference of cubes. Answer: Given Expression: 1 − 27 𝑎 3 . 1−27a 3 . Apply Difference of Cubes Formula: 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏 ) ( 𝑎 2 + 𝑎 𝑏 + 𝑏 2 ) . a 3 −b 3 =(a−b)(a 2 +ab+b 2 ). Here: 1 = 1 3 ,   27 𝑎 3 = ( 3 𝑎 ) 3 . 1=1 3 ,27a 3 =(3a) 3 . Substitute: 1 − 27 𝑎 3 = ( 1 − 3 𝑎 ) ( 1 2 + ( 1 ) ( 3 𝑎 ) + ( 3 𝑎 ) 2 ) . 1−27a 3 =(1−3a)(1 2 +(1)(3a)+(3a) 2 ). Simplify: 1 − 27 𝑎 3 = ( 1 − 3 𝑎 ) ( 1 + 3 𝑎 + 9 𝑎 2 ) . 1−27a 3 =(1−3a)(1+3a+9a 2 ). Final Answer: 1 − 27 𝑎 3 = ( 1 − 3 𝑎 ) ( 1 + 3 𝑎 + 9 𝑎 2 ) . 1−27a 3 =(1−3a)(1+3a+9a 2 ). Question 4 Use the Intermediate Value Theorem to show that there is a root of 𝑓 ( 𝑥 ) = sin ⁡ 𝑥 − 𝑥 2 + 𝑥 f(x)=sinx−x 2 +x in ( 1 , 2 ) (1,2). Answer: Given Function: 𝑓 ( 𝑥 ) = sin ⁡ 𝑥 − 𝑥 2 + 𝑥 . f(x)=sinx−x 2 +x. Continuity: 𝑓 ( 𝑥 ) f(x) involves sin ⁡ 𝑥 sinx, 𝑥 2 x 2 , and 𝑥 x, which are continuous functions. So, 𝑓 ( 𝑥 ) f(x) is continuous on [ 1 , 2 ] [1,2]. Evaluate 𝑓 ( 𝑥 ) f(x) at Endpoints: At 𝑥 = 1 x=1: 𝑓 ( 1 ) = sin ⁡ ( 1 ) − ( 1 ) 2 + 1 ≈ 0.841 − 1 + 1 = 0.841 > 0. f(1)=sin(1)−(1) 2 +1≈0.841−1+1=0.841>0. At 𝑥 = 2 x=2: 𝑓 ( 2 ) = sin ⁡ ( 2 ) − ( 2 ) 2 + 2 ≈ 0.909 − 4 + 2 = − 1.091 < 0. f(2)=sin(2)−(2) 2 +2≈0.909−4+2=−1.091<0. Conclusion: 𝑓 ( 1 ) > 0 , 𝑓 ( 2 ) < 0 f(1)>0,f(2)<0. By the Intermediate Value Theorem, 𝑓 ( 𝑥 ) f(x) has at least one root in ( 1 , 2 ) (1,2). Question 5 Find a cubic polynomial 𝑓 ( 𝑥 ) f(x) with real coefficients that has complex zeros 𝑥 = 2 ± 2 𝑖 x=2± 2 ​ i and 𝑥 x-intercept ( 5 , 0 ) (5,0). Answer: Given Zeros: 2 + 2 𝑖 ,   2 − 2 𝑖 2+ 2 ​ i,2− 2 ​ i (complex conjugates). Real root: 𝑥 = 5 x=5. Form the Polynomial: Complex roots form a quadratic factor: ( 𝑥 − ( 2 + 2 𝑖 ) ) ( 𝑥 − ( 2 − 2 𝑖 ) ) = ( 𝑥 − 2 ) 2 + 2. (x−(2+ 2 ​ i))(x−(2− 2 ​ i))=(x−2) 2 +2. Real root: ( 𝑥 − 5 ) (x−5). Combine Factors: 𝑓 ( 𝑥 ) = ( 𝑥 − 5 ) ( ( 𝑥 − 2 ) 2 + 2 ) . f(x)=(x−5)((x−2) 2 +2). Expand: Expand ( 𝑥 − 2 ) 2 + 2 (x−2) 2 +2: ( 𝑥 − 2 ) 2 + 2 = 𝑥 2 − 4 𝑥 + 4 + 2 = 𝑥 2 − 4 𝑥 + 6. (x−2) 2 +2=x 2 −4x+4+2=x 2 −4x+6. Multiply by ( 𝑥 − 5 ) (x−5): 𝑓 ( 𝑥 ) = ( 𝑥 − 5 ) ( 𝑥 2 − 4 𝑥 + 6 ) = 𝑥 3 − 4 𝑥 2 + 6 𝑥 − 5 𝑥 2 + 20 𝑥 − 30. f(x)=(x−5)(x 2 −4x+6)=x 3 −4x 2 +6x−5x 2 +20x−30. Combine terms: 𝑓 ( 𝑥 ) = 𝑥 3 − 9 𝑥 2 + 26 𝑥 − 30. f(x)=x 3 −9x 2 +26x−30. Final Answer: 𝑓 ( 𝑥 ) = 𝑥 3 − 9 𝑥 2 + 26 𝑥 − 30. f(x)=x 3 −9x 2 +26x−30.