Polynomial arithmetic #4 - Questions and Answers

Question 1 Determine whether 𝑔 ( 𝑥 ) = 𝑥 3 / 2 − 𝑥 2 + 2 g(x)=x 3/2 −x 2 +2 is a polynomial. If it is, state its degree. If not, say why it is not a polynomial. If it is a polynomial, write it in standard form. Identify the leading term and the constant term. Answer: Step 1: To determine the degree of polynomial. Step 2: Given: 𝑔 ( 𝑥 ) = 𝑥 3 / 2 − 𝑥 2 + 2 g(x)=x 3/2 −x 2 +2 Step 3: A polynomial is an expression of terms that involves mathematical operations like addition, subtraction, division, and multiplication where each term consists of variable and constant, and the variables have non-negative integer exponents. The exponents of the variable 𝑥 x are 3 / 2 3/2 and 2 2 in the function 𝑔 ( 𝑥 ) = 𝑥 3 / 2 − 𝑥 2 + 2 g(x)=x 3/2 −x 2 +2. Since 𝑥 x has a non-integer exponent ( 3 / 2 3/2), 𝑔 ( 𝑥 ) g(x) is not a polynomial. Question 2 DISCUSS DISCOVER: How Many Real Zeros Can a Polynomial Have? Give examples of polynomials that have the following properties, or explain why it is impossible to find such a polynomial. (a) A polynomial of degree 3 that has no real zeros (b) A polynomial of degree 4 that has no real zeros (c) A polynomial of degree 3 that has three real zeros, only one of which is rational (d) A polynomial of degree 4 that has four real zeros, none of which is rational What must be true about the degree of a polynomial with integer coefficients if it has no real zeros? Answer: Step 1: (a) Let a 3rd-degree polynomial 𝑃 ( 𝑥 ) P(x): If the leading term is positive, the end behavior is: 𝑦 → + ∞  as  𝑥 → + ∞  and  𝑦 → − ∞  as  𝑥 → − ∞ . y→+∞ as x→+∞ and y→−∞ as x→−∞. If the leading term is negative, the end behavior is: 𝑦 → − ∞  as  𝑥 → + ∞  and  𝑦 → + ∞  as  𝑥 → − ∞ . y→−∞ as x→+∞ and y→+∞ as x→−∞. This means the graph of a third-degree polynomial must cross the x-axis at least once. Thus, every polynomial 𝑃 ( 𝑥 ) P(x) of degree 3 has at least one real zero. Step 2: (b) Let a 4th-degree polynomial 𝑃 ( 𝑥 ) P(x): 𝑃 ( 𝑥 ) = 𝑥 4 + 𝑥 2 + 3. P(x)=x 4 +x 2 +3. The terms 𝑥 4 x 4 and 𝑥 2 x 2 are non-negative as 𝑥 x and the constant term is positive. 𝑃 ( 𝑥 ) P(x) is strictly positive. So, the polynomial 𝑃 ( 𝑥 ) P(x) has no real zeros. Step 3: (c) Let the 3rd-degree polynomial 𝑃 ( 𝑥 ) = ( 𝑥 − 2 ) ( 𝑥 − 𝜋 ) ( 𝑥 − 1 ) P(x)=(x− 2 ​ )(x−π)(x−1): 𝑃 ( 𝑥 ) = 𝑥 3 − 𝑥 2 ( 2 + 𝜋 + 1 ) + 𝑥 ( 2 𝜋 + 𝜋 + 2 ) − 2 𝜋 . P(x)=x 3 −x 2 ( 2 ​ +π+1)+x( 2 ​ π+π+ 2 ​ )− 2 ​ π. A 3rd-degree polynomial with one rational zero ( 𝑥 = 1 x=1) and two irrational zeros ( 𝑥 = 2 , 𝜋 x= 2 ​ ,π) exists. Question 3 Find specific terms of an arithmetic sequence given other terms. How many terms are there in the following finite arithmetic sequence? 7 , 3 , − 1 , … , − 89 7,3,−1,…,−89 Options: 23 24 25 28 30 Answer: Given arithmetic sequence: 𝑎 1 = 7 , 𝑎 𝑛 = − 89 , 𝑑 = − 4 a 1 ​ =7,a n ​ =−89,d=−4. Using the 𝑛 n-th term formula: 𝑎 𝑛 = 𝑎 1 + ( 𝑛 − 1 ) 𝑑 . a n ​ =a 1 ​ +(n−1)d. Substitute values: − 89 = 7 + ( 𝑛 − 1 ) ( − 4 ) . −89=7+(n−1)(−4). − 96 = − 4 ( 𝑛 − 1 ) . −96=−4(n−1). 𝑛 − 1 = 24    ⟹    𝑛 = 25. n−1=24⟹n=25. Correct option: 3) 25. Question 4 To calculate: The coefficient and degree of each term of the polynomial 8 − 𝑥 2 𝑦 4 + 𝑦 7 8−x 2 y 4 +y 7 , and then find the degree of the polynomial. Answer: Given polynomial: 8 − 𝑥 2 𝑦 4 + 𝑦 7 8−x 2 y 4 +y 7 . Terms: 8 , − 𝑥 2 𝑦 4 , 𝑦 7 8,−x 2 y 4 ,y 7 . Coefficients: 8 , − 1 , 1 8,−1,1. Degree of terms: 8 → 0 8→0, − 𝑥 2 𝑦 4 → 6 −x 2 y 4 →6, 𝑦 7 → 7 y 7 →7. Degree of polynomial: 7. Question 5 Write an equation for the 𝑛 n-th term of the arithmetic sequence. Answer: The 𝑛 n-th term of an arithmetic sequence is: 𝑇 𝑛 = 𝑎 + ( 𝑛 − 1 ) 𝑑 . T n ​ =a+(n−1)d. Where 𝑎 a = first term, 𝑑 d = common difference.