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Practice Problems with Hints and Answers - College Algebra | MATH 121, Study notes of Algebra

Material Type: Notes; Class: COLLEGE ALGEBRA; Subject: MATHEMATICS; University: La Sierra University; Term: Unknown 1989;

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Pre 2010

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Math 121, Practice for Chapter 3
Hints and Answers
1. Use long division to find (6x4+ 8x3โˆ’47x2+ 19x+ 5) รท(2x2+ 6xโˆ’5).
Answer.
3x2โˆ’5xโˆ’1
2x2+ 6xโˆ’5 6x4+ 8x3โˆ’47x2+ 19x+ 5
โˆ’(6x4+ 18x3โˆ’15x2)
โˆ’10x3โˆ’37x2+ 19x
โˆ’(โˆ’10x3โˆ’30x2+ 25x)
โˆ’2x2โˆ’6x+ 5
โˆ’(โˆ’2x2โˆ’6x+ 5)
0
Thus the answer is 3x2โˆ’5xโˆ’1; notice there is no remainder.
2. Use synthetic division to find (x4+ 9x3โˆ’5x+ 10) รท(x+ 3)
Answer. The synthetic division with c=โˆ’3 is as follows
โˆ’3 1 9 0 โˆ’5 10
โˆ’3โˆ’18 54 โˆ’147
1 6 โˆ’18 49 โˆ’137
Therefore, the answer is:
x3+ 6x2โˆ’18x+ 49 + โˆ’137
x+ 3.
3. Let P(x) = 6x4+ 8x3โˆ’8x2โˆ’31x+ 4, use the remainder theorem to find P(6).
Answer. Use synthetic division with c= 6 as follows
6 6 8 โˆ’8โˆ’31 4
36 264 1536 9030
6 44 256 1505 9034
The remainder is 9034, and so P(6) = 9034.
4. Determine the far right and far left behavior of the polynomials:
(a) P(x) = โˆ’x6+ 20x5+ 7x2โˆ’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) = x3+ 5x+ 7.
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Math 121, Practice for Chapter 3 Hints and Answers

  1. Use long division to find (6x^4 + 8x^3 โˆ’ 47 x^2 + 19x + 5) รท (2x^2 + 6x โˆ’ 5).

Answer.

3 x^2 โˆ’ 5 x โˆ’ 1 2 x^2 + 6x โˆ’ 5 6 x^4 + 8x^3 โˆ’ 47 x^2 + 19x + 5 โˆ’(6x^4 + 18x^3 โˆ’ 15 x^2 ) โˆ’ 10 x^3 โˆ’ 37 x^2 + 19x โˆ’(โˆ’ 10 x^3 โˆ’ 30 x^2 + 25x) โˆ’ 2 x^2 โˆ’ 6 x + 5 โˆ’(โˆ’ 2 x^2 โˆ’ 6 x + 5) 0

Thus the answer is 3x^2 โˆ’ 5 x โˆ’ 1; notice there is no remainder.

  1. Use synthetic division to find (x^4 + 9x^3 โˆ’ 5 x + 10) รท (x + 3)

Answer. The synthetic division with c = โˆ’3 is as follows

Therefore, the answer is:

x^3 + 6x^2 โˆ’ 18 x + 49 +

x + 3

  1. Let P (x) = 6x^4 + 8x^3 โˆ’ 8 x^2 โˆ’ 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.

  1. Determine the far right and far left behavior of the polynomials:

(a) P (x) = โˆ’x^6 + 20x^5 + 7x^2 โˆ’ 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 โˆ’ 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 โˆ’ 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.

  1. (a) Find the x-intercepts of P (x) = (x โˆ’ 3)^2 (x โˆ’ 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 โˆ’ 3 , โˆ’ 1 , 2 , 3. The graph crosses x-axis at (โˆ’ 1 , 0) and (โˆ’ 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 = โˆ’3 and then smoothly turns and crosses (going up) the x-axis at x = โˆ’1, 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 โˆ’ 3000)^3 (x โˆ’ 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), (โˆ’ 1000 , 0) and (โˆ’ 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 = โˆ’1000 and x = 3000, whereas it merely touches the x-axis at x = โˆ’3000 and x = 2000.

  1. Consider the polynomial P (x) = 2x^5 โˆ’ 3 x^4 + 12x^2 + 13x โˆ’ 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 + (โˆ’2 + 3i)x^2 + x โˆ’ 2 + 3i) รท (x โˆ’ (2 โˆ’ 3 i)).

2 โˆ’ 3 i 1 โˆ’2 + 3i 1 โˆ’2 + 3i 2 โˆ’ 3 i 0 2 โˆ’ 3 i 1 0 1 0

Now x^2 + 1 has zeros ยฑi. Thus the zeros of Q(x) are 2 + 3i, 2 โˆ’ 3 i, i, โˆ’i. Therefore,

Q(x) = (x โˆ’ (2 + 3i))(x โˆ’ (2 โˆ’ 3 i))(x โˆ’ i)(x + i)

  1. (a) Find a polynomial with real coefficients of smallest degree that has roots 2 + 3i, โˆ’2 and

Answer. The roots of the polynomial of smallest degree would be 2 + 3i, 2 โˆ’ 3 i, โˆ’2 and 3. Therefore, we let

P (x) = (x โˆ’ (2 + 3i))(x โˆ’ (2 โˆ’ 3 i))(x + 2)(x โˆ’ 3) = (x^2 โˆ’ 4 x + 13)(x^2 โˆ’ x โˆ’ 6) = x^4 โˆ’ 5 x^3 + 11x^2 + 11x โˆ’ 78.

(b) With help from (a), find a polynomial P (x) with real coefficients of smallest degree that has roots 2 + 3i, โˆ’2 and 3 such that P (1) = 120.

Answer. The polynomial P (x) will have the form

P (x) = k(x โˆ’ (2 + 3i))(x โˆ’ (2 โˆ’ 3 i))(x + 2)(x โˆ’ 3)

where we will find k so that P (1) = 120. Therefore, P (x) = k(x^4 โˆ’ 5 x^3 + 11x^2 + 11x โˆ’ 78), and so P (1) = โˆ’ 60 k. Now โˆ’ 60 k = 120 and so k = โˆ’2. Consequently, P (x) = โˆ’ 2 x^4 + 10x^3 โˆ’ 22 x^2 โˆ’ 22 x + 156.

  1. 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 โˆ’ 2 x^2 โˆ’ 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 = โˆ’4. Also, f (x) โ†’ +โˆž as x โ†’ โˆ’ 4 โˆ’^ and f (x) โ†’ โˆ’โˆž as x โ†’ โˆ’ 4 +; f (x) โ†’ โˆ’โˆž as x โ†’ 4 โˆ’^ and f (x) โ†’ +โˆž as x โ†’ 4 +. There is no slant asymptote.

(b) g(x) =

2 x^2 โˆ’ 2 x + 3

Answer. The x-intercepts are x = ยฑ1, the y-intercept is y = โˆ’ 2 /3, there is no horizontal asymptote, there is a vertical asymptote x = โˆ’3, and f (x) โ†’ โˆ’โˆž as x โ†’ โˆ’ 3 โˆ’^ and f (x) โ†’

+โˆž as x โ†’ โˆ’ 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 โˆ’ 6 +

x + 3

Therefore, the slant asymptote is y = 2x โˆ’ 6.

(c) h(x) =

x^4 โˆ’ 16 x^2 โˆ’ 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 = โˆ’3, and f (x) โ†’ +โˆž as x โ†’ โˆ’ 3 โˆ’, f (x) โ†’ โˆ’โˆž as x โ†’ โˆ’ 3 +, f (x) โ†’ โˆ’โˆž as x โ†’ 3 โˆ’^ and f (x) โ†’ +โˆž as x โ†’ 3 +. There is no slant asymptote.

  1. With the help of information from the previous question, graph the rational functions:

(a) f (x) =

2 x^2 โˆ’ 2 x^2 โˆ’ 16

(b) g(x) =

2 x^2 โˆ’ 2 x + 3

Answer. See link to Graphs for #10 under review answers.

  1. 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 โˆ’ 1 = 3(2x + 2) โˆ’ 1 = 6x + 5. Therefore, V = x(2x + 2)(6x + 5).

  1. (a) Is x โˆ’ 1 a factor of xn^ โˆ’ 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^ โˆ’ 1, then P (1) = 1n^ โˆ’ 1 = 0. Therefore the factor theorem implies x โˆ’ 1 is a factor of P (x).

(b) Let P (x) = xn^ + 1, when n is odd, (โˆ’1)n^ = โˆ’1, thus P (โˆ’1) = (โˆ’1)n^ + 1 = โˆ’1 + 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 (โˆ’1)n^ = 1, thus (โˆ’1)n^ + 1 = 2. According to the factor theorem, x + 1 is not a factor of xn^ + 1 when n is even.

  1. Let F (x) =

anxn^ + anโˆ’ 1 xnโˆ’^1 +... + a 1 x + a 0 bmxm^ + bmโˆ’ 1 xmโˆ’^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.