Download Practice Problems with Hints and Answers - College Algebra | MATH 121 and more Study notes Algebra in PDF only on Docsity!
Math 121, Practice for Chapter 3 Hints and Answers
- Use long division to find (6x^4 + 8x^3 โ 47 x^2 + 19x + 5) รท (2x^2 + 6x โ 5).
Answer.
3 x^2 โ 5 x โ 1 2 x^2 + 6x โ 5 6 x^4 + 8x^3 โ 47 x^2 + 19x + 5 โ(6x^4 + 18x^3 โ 15 x^2 ) โ 10 x^3 โ 37 x^2 + 19x โ(โ 10 x^3 โ 30 x^2 + 25x) โ 2 x^2 โ 6 x + 5 โ(โ 2 x^2 โ 6 x + 5) 0
Thus the answer is 3x^2 โ 5 x โ 1; notice there is no remainder.
- Use synthetic division to find (x^4 + 9x^3 โ 5 x + 10) รท (x + 3)
Answer. The synthetic division with c = โ3 is as follows
Therefore, the answer is:
x^3 + 6x^2 โ 18 x + 49 +
x + 3
- Let P (x) = 6x^4 + 8x^3 โ 8 x^2 โ 31 x + 4, use the remainder theorem to find P (6).
Answer. Use synthetic division with c = 6 as follows
The remainder is 9034, and so P (6) = 9034.
- Determine the far right and far left behavior of the polynomials:
(a) P (x) = โx^6 + 20x^5 + 7x^2 โ 12.
Answer. This polynomial has even degree with a negative leading coefficient, therefore its graph is down to the far left and down to the far right.
(b) P (x) = x^3 + 5x + 7.
Answer. This polynomial has odd degree with positive leading coefficient. Therefore, its graph is down to the far left and up to the far right.
(c) P (x) = 22x^8 + 2000000x^6 โ 4 x^9.
Answer. This polynomial has odd degree with negative leading coefficient. Therefore, the graph is up to the far left and down to the far right.
(d) P (x) = 22x^8 + 2000000x^6 โ 4 x^7.
Answer. This polynomial has even degree with positive leading coefficient, therefore its graph is up to the far left and up to the far right.
- (a) Find the x-intercepts of P (x) = (x โ 3)^2 (x โ 2)^4 (x + 1)^3 (x + 3). For each intercept, determine whether the graph of P (x) crosses or merely touches the x-axis.
Answer. The x-intercepts are โ 3 , โ 1 , 2 , 3. The graph crosses x-axis at (โ 1 , 0) and (โ 3 , 0) because the corresponding zeros have odd multiplicities, while the graph intersects but does not cross the x-axis at (3, 0) and (2, 0) because the corresponding zeros have even multiplicities.
(b) What is the degree of P (x)? Determine the far right and far left behavior of P (x).
Answer. The degree of P (x) is 10, and its leading coefficient is positive. Therefore, the graph goes up to the far left, and to the far right.
(c) Using the information above, sketch a very crude graph of P (x).
Answer. Try sketching a very crude graph as follows: starting from the left, the graph comes down and crosses the x-axis at x = โ3 and then smoothly turns and crosses (going up) the x-axis at x = โ1, then the graph will go up for a while and smoothly bend down to touch the x-axis at x = 2, then it will bend back up for a short period, and then bend down again to touch the x-axis at x = 3, where it bends up and continues go up after that. Then see how your graph matches with a graphing utility. It may miss some subtleties and even turning points, but it should give a rough idea of what the graph looks like.
(d) Find the x-intercepts of P (x) = (x โ 3000)^3 (x โ 2000)^2 (x + 1000)^1 (x + 3000)^2000. For each intercept, determine whether the graph of P (x) crosses or merely touches the x-axis.
Answer. The x-intercepts are (3000, 0), (2000, 0), (โ 1000 , 0) and (โ 3000 , 0). The graph touches but does not cross at zeros of even multiplicity, and crosses at zeros of odd multiplicity. Thus the graph crosses the x-axis at x = โ1000 and x = 3000, whereas it merely touches the x-axis at x = โ3000 and x = 2000.
- Consider the polynomial P (x) = 2x^5 โ 3 x^4 + 12x^2 + 13x โ 6.
(a) Use the Rational Zero Theorem to list the possible rational zeros of P (x)
Answer. The possible rational zeros are plus or minus all factors of 6 divided by all factors of 2:
ยฑ 1 , ยฑ 2 , ยฑ 3 , ยฑ 6 , ยฑ
Now use synthetic division to find (x^3 + (โ2 + 3i)x^2 + x โ 2 + 3i) รท (x โ (2 โ 3 i)).
2 โ 3 i 1 โ2 + 3i 1 โ2 + 3i 2 โ 3 i 0 2 โ 3 i 1 0 1 0
Now x^2 + 1 has zeros ยฑi. Thus the zeros of Q(x) are 2 + 3i, 2 โ 3 i, i, โi. Therefore,
Q(x) = (x โ (2 + 3i))(x โ (2 โ 3 i))(x โ i)(x + i)
- (a) Find a polynomial with real coefficients of smallest degree that has roots 2 + 3i, โ2 and
Answer. The roots of the polynomial of smallest degree would be 2 + 3i, 2 โ 3 i, โ2 and 3. Therefore, we let
P (x) = (x โ (2 + 3i))(x โ (2 โ 3 i))(x + 2)(x โ 3) = (x^2 โ 4 x + 13)(x^2 โ x โ 6) = x^4 โ 5 x^3 + 11x^2 + 11x โ 78.
(b) With help from (a), find a polynomial P (x) with real coefficients of smallest degree that has roots 2 + 3i, โ2 and 3 such that P (1) = 120.
Answer. The polynomial P (x) will have the form
P (x) = k(x โ (2 + 3i))(x โ (2 โ 3 i))(x + 2)(x โ 3)
where we will find k so that P (1) = 120. Therefore, P (x) = k(x^4 โ 5 x^3 + 11x^2 + 11x โ 78), and so P (1) = โ 60 k. Now โ 60 k = 120 and so k = โ2. Consequently, P (x) = โ 2 x^4 + 10x^3 โ 22 x^2 โ 22 x + 156.
- Find the x-intercepts, y-intercepts, horizontal, vertical and slant asymptotes (if they exist), and determine the left and right behavior near the vertical asymptotes for:
(a) f (x) =
2 x^2 โ 2 x^2 โ 16
Answer. The x-intercepts are x = ยฑ1, the y-intercept is y = 1/8, there is an horizontal asymptote y = 2, and there are vertical asymptotes x = 4 and x = โ4. Also, f (x) โ +โ as x โ โ 4 โ^ and f (x) โ โโ as x โ โ 4 +; f (x) โ โโ as x โ 4 โ^ and f (x) โ +โ as x โ 4 +. There is no slant asymptote.
(b) g(x) =
2 x^2 โ 2 x + 3
Answer. The x-intercepts are x = ยฑ1, the y-intercept is y = โ 2 /3, there is no horizontal asymptote, there is a vertical asymptote x = โ3, and f (x) โ โโ as x โ โ 3 โ^ and f (x) โ
+โ as x โ โ 3 +. There is a slant asymptote since the degree in the numerator is one larger than the denominator. To find the asymptote, we do the division and find
g(x) = 2x โ 6 +
x + 3
Therefore, the slant asymptote is y = 2x โ 6.
(c) h(x) =
x^4 โ 16 x^2 โ 9
Answer. The x-intercepts are x = ยฑ2, the y-intercept is y = 16/9, there is no horizontal asymptote, the vertical asymptotes are x = 3 and x = โ3, and f (x) โ +โ as x โ โ 3 โ, f (x) โ โโ as x โ โ 3 +, f (x) โ โโ as x โ 3 โ^ and f (x) โ +โ as x โ 3 +. There is no slant asymptote.
- With the help of information from the previous question, graph the rational functions:
(a) f (x) =
2 x^2 โ 2 x^2 โ 16
(b) g(x) =
2 x^2 โ 2 x + 3
Answer. See link to Graphs for #10 under review answers.
- A box is constructed so that the width is 2 units more than twice the height and the length is 1 unit less than three times the width. Write the volume of the box as a function of height x.
Answer. h = x, w = 2x + 2, l = 3w โ 1 = 3(2x + 2) โ 1 = 6x + 5. Therefore, V = x(2x + 2)(6x + 5).
- (a) Is x โ 1 a factor of xn^ โ 1 for each n?
(b) Show that x + 1 is a factor of xn^ + 1 if n is odd.
(c) Is x + 1 a factor of xn^ + 1 if n is even?
Hint: use the remainder theorem on all of these.
Answer. (a) Yes: let P (x) = xn^ โ 1, then P (1) = 1n^ โ 1 = 0. Therefore the factor theorem implies x โ 1 is a factor of P (x).
(b) Let P (x) = xn^ + 1, when n is odd, (โ1)n^ = โ1, thus P (โ1) = (โ1)n^ + 1 = โ1 + 1 = 0. According to the factor theorem x + 1 is a factor of xn^ + 1 when n is odd.
(c) No: when n is even (โ1)n^ = 1, thus (โ1)n^ + 1 = 2. According to the factor theorem, x + 1 is not a factor of xn^ + 1 when n is even.
- Let F (x) =
anxn^ + anโ 1 xnโ^1 +... + a 1 x + a 0 bmxm^ + bmโ 1 xmโ^1 +... + b 1 x^1 + b 0
. Desribe when F has horizontal or slant
asymptotes, and find a formula for the horizontal asymptote when it exists.
