MATH 140 -- Intermediate Algebra -- Exercise Solutions
Section: 5.3

2)  The degree of 7 is 0.      (Because it is 7x⁰)

4) The degree of  - z³ is 3           (Because the exponent on the variable is 3)

6)  The degree of 12x³z is 4      (Because the sum of the exponents on the variables is 3 + 1 =4)

8) The polynomial 7x - 8 is a binomial of degree 1.   (It is a binomial because it has two terms and it has degree 1 because the term with largest degree has degree 1 -- the other term has degree 0).

10) The polynomial 5x² - 3x²y - 2x³ is a trinomial with degree 3. (It is a trinomial because it has three terms and it has degree 3 because the term with largest degree has degree 3 -- the first term has degree 2, the second term has degree 2+1 =3 and the third term has degree 3).

12)  The polynomial - 9 is a monomial (one term) and it has degree 0 (because it is -9x⁰).

14)  The polynomial - 2x²y - 3y² + 4x + y⁵ is neither a monomial, binomial, nor a trinomial(it has four terms).
The degree of - 2x²y - 3y² + 4x + y⁵ is 5. (The degree of the term terms with the largest degree -- The first term has degree 3, the second term has degree 2, the third term has degree1 and the last term has degree 5).

15) In your own words, describe how to find the degree of a term.
      Dispite what the author might say, there is ONLY ONE correct response to this question and it MUST be the Definition of the degree of a term.
      So the correct response is:
      Definition: The degree of a term is the sum of the exponents on the variables.

16) In your own words, describe how to find the degree of a polynomial.
      Dispite what the author might say, there are ONLY TWO correct responses to this question.
      The most direct response MUST be the Definition of the degree of a polynomial.
      Definition: The degree of a a polynomial is the degree of the leading term.
      However, contained in this definition is the definition of leading term.
       Definition: The leading term of a polynomial is the term with largest degree.
      In view of the definiton of leading term one might say that the degree of a polynomial is the degree of the term with largest degree.

18)  If the rule for a function named Q is given by the equation Q(x) = 5x² - 1, find Q(4).
       Solution: Note that Q(4) is the unique range element associated with the domain element 4 by the function named Q.
        Note that the rule for Q states that one finds the range element associated with a particular domain element by squaring the domain element, multiplying that square by 5 and then subtracting 1. All of this can be easily accomplished by substituting the domain element 4 into the rule so that we obtain
       Q(4) = 5(4²) - 1 = 5(16) - 1 = 79

20)  If the rule for a function named P is given by the equation Q(x) = x² + x + 1, find P(- 4).
       Solution: Note that P(- 4)is the unique range element associated with the domain element - 4 by the function named P.
        Note that the rule for P states that one finds the range element associated with a particular domain element by squaring the domain element, adding the domain element to that square and then adding 1. All of this can be easily accomplished by substituting the domain element - 4 into the rule so that we obtain
       P(- 4) = (- 4)² + (- 4) + 1 = 16 - 4 + 1= 13

22)  If the rule for a function named Q is given by the equation Q(x) = 5x² - 1, find Q(0).
       Solution: Note that Q(0) is the unique range element associated with the domain element 0 by the function named Q. Note also that the point (0, Q(0)) is the y-intercept of the graph of Q.
        Note that the rule for Q states that one finds the range element associated with a particular domain element by squaring the domain element, multiplying that square by 5 and then subtracting 1. All of this can be easily accomplished by substituting the domain element 4 into the rule so that we obtain
       Q(0) = 5(0²) - 1 = 5(0) - 1 = -1

24)  If the rule for a function named P is given by the equation Q(x) = x² + x + 1, find P(1/2).
       Solution: Note that P(1/2)is the unique range element associated with the domain element 1/2 by the function named P.
        Note that the rule for P states that one finds the range element associated with a particular domain element by squaring the domain element, adding the domain element to that square and then adding 1. All of this can be easily accomplished by substituting the domain element - 4 into the rule so that we obtain
       P(1/2) = (1/2)² + (1/2) + 1 = 1/4 +1/2 + 1= 1/4 + 2/4 +4/4 = 7/4

26) If the rule for a function named h is given by the equation h(x) = -16t² + 1053, find h(4).
      Solution: Note that h(4) is the unique range element associated with the domain element 4 by the function h.
     Note that the rule for h states that one finds the range element associated with a particular domain element by squaring the domain element, multiplying that square by - 16 and then adding 1053. All of this can be easily accomplished by substituting the domain element - 4 into the rule so that we obtain
       h(4) = - 16(4²) + 1053 = - 256 + 1053 = 797