University:
North Central College
Course:
MATH 152 | Calculus II
Academic year:

2024
Views:

106

Pages:

3
Author:

nesikhiyawl0

Question 1 Is the algebraic expression a polynomial? If it is, write the polynomial in standard form: x²-x³+x⁴-5. Answer 1 Step 1 Definition: The formula for the norm of a polynomial function of degree n is aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₁x + a₀ where aₙ ≠ 0. Step 2 x²-x³+x⁴-5 is the algebraic expression provided. Yes. The algebraic expression x² - x³ + x⁴ - 5 is a polynomial of degree 4. Its standard form is x⁴ - x³ + x²-5. Answer 2 This fits the definition of a polynomial. It has multiple terms as well as non-negative integers for exponents. To put it in standard form, we put the exponents in descending order. Result: x⁴ - x³ + x² - 5 Answer 3 A polynomial is a single term or the sum of two or more terms containing variables with whole-number exponents. Each term has a variable with whole-number exponent so the algebraic expression is a polynomial. Note that the constant term has a variable with exponent 0. A polynomial In standard form is written in the order of descending powers of the variable. So, we rewrite the given as: x⁴ - x³+x²-5 Result: yes; x⁴ - x³ + x² - 5 Question 2 Determine the maximum number of real zeros that each polynomial function may have. Then use Descartes' Rule of Signs to determine how many positive and how many negative real zeros each polynomial function may have. Do not attempt to find the zeros.; f(x) = - x³- x²+x+1 Answer 1 Step 1: Given f(x) = - x³- x²+x+1 Step 2: Solution Number of possible zeroes= degree =3 f(x) = - x³- x²+x+1 Number of times sign changes=2 So number of positive zeroes=2 f(-x) = x³- x²-x+1 Number of times sign changes=1 Number of negative zeroes=1 Answer 2 The maximum number of real zeros of a polynomial function f is equal to the degree of f. f(x) = -x³ - x² + x + 1 → The maximum number of real zeros is 3 since the degree is 3. f(x) = -x³ - x² + x + 1⇒ Since there is only 1 change of signs in the coefficients of f(x), therefore the number of positive zeros must be 1. f(-x) = x³- x² - x +1⇒ Since there are 2 changes of signs in the coefficients of f(-x), therefore the number of negative zeros can be 2 or 0. Question 3 How do you write y=4x-11 in standard form? Answer The standard form of a linear equation is (normally) Ax+By=C usually with the restriction that A>0 and A ∈ Z Rewriting y=4x-11 -4x+1y=-11 or 4x-1y=11 Question 4 What is the leading term, leading coefficient, and degree of this polynomial p(x)=2(2x+3) (x-1)(x+2)? Answer 1 You should first approach this problem by multiplying the entire function out through FOIL, and get p(x) = 4x³ + 10x² - 2x – 12. Then identify the leading term, leading coefficient, and degree of this polynomial by definition. Leading term: 4x³ Leading coefficient: 4 Degree of polynomial: 3 Answer 2 Find the degree, the leading coefficient, and the leading term of p(x)=2(2x+3)(x-1)(x+2). p(x)=2(2x+3)(x-1)(x+2) = (x-1)(x+2)(4x+6) Solution: The degree of a polynomial is the highest of the degrees of the polynomials Question 5 What is the standard form of y=(x-6)(x-4)(x-1)? Answer Explanation: To rewrite the equation in standard form, start by expanding the first two brackets: y=(x-6)(x-4)(x-1) y=(x(x)+x(-4)+x(-6)-6(-4))(x-1) Simplify. y = (x² - 4x - 6x + 24) (x - 1) y = (x² - 10x + 24) (x - 1) Expand the remaining two brackets: y = (x² - 10x + 24) (x - 1) y = x²(x) +x²(-1) - 10x(x) - 10x(-1)+24(x)+24(-1) Simplify. y = x³ - x² - 10x² + 10x + 24x - 24 y = x³ - 11x² + 34x - 24