Name:
Solutions
153
153
Extra Credit 3
Directions: Show all work to receive full credit. Just answers with no work will be given no credit.
Use desmos and wolfram to check your answers.
• For problems 1-2 : Use polynomial long division to check if the indicated polynomials are factors of the given
polynomial. If they are rewrite the polynomial in the partially factored form. (i.e. f(x) = p(x)q(x))
1. (4pts total) f(x) = x4 − 10x3 + 35x2 − 50x + 24
(a) (x − 4)
Solution.
Since
x−4
)
x3 − 6x2 + 11x − 6
we have that (x−4) is a factor with f(x) = (x − 4)(x3 − 6x2 + 11x − 6)
x4 − 10x3 + 35x2 − 50x + 24
− x4 + 4x3
− 6x3 + 35x2
6x3 − 24x2
11x2 − 50x
− 11x2 + 44x
− 6x + 24
6x − 24
0
(b) (x + 1)
Solution.
Since
x+1
)
x3 − 11x2 + 46x − 96
4
3
the remainder is not 0 indicating that (x + 1) is not a factor of f.
2
x − 10x + 35x − 50x + 24
− x4 − x3
− 11x3 + 35x2
11x3 + 11x2
46x2 − 50x
− 46x2 − 46x
− 96x + 24
96x + 96
120
2. (4pts total) f(x) = 3x3 − x2 − 54x + 18
(a) (x − 2)
1
Solution.
Since
x−2
)
3x2 + 5x − 44
3
the remainder is not 0 indicating that (x − 2) is not a factor of f.
2
3x − x − 54x + 18
− 3x3 + 6x2
5x2 − 54x
− 5x2 + 10x
− 44x + 18
44x − 88
− 70
√
(b) (x − 3 2)
Solution.
√
Since long division
results √
in a remainder
of 0 we have that (x − 3 2) is factor of f(x) with
√
√
f(x) = (x − 3 2)(3x2 + (9 2 − 1)x − 3 2)
• For problems 3-5: (15pts each)
(a) (8pts) Find all the zeros of the polynomial (SHOW YOUR WORK!)
(b) (2pts) State their multiplicities
(c) (3pts) Determine the end behavior
(d) (2pts) Sketch a graph
3. f(x) = 6x3 − x2 − 57x + 70
Solution.
(a) Choosing different potential rational zeros until we land on a solution we could proceed as follows
6 − 1 − 57
70
2
12
22
− 70
which implies f(x) = (x − 2)(6x2 + 11x − 35) using our factoring skills for the
6
11 − 35
0
remaining quadratic we get
{
Thus the zeros of f are
7 5
2, − ,
2 3
}
f(x) = (x − 2)(2x + 7)(3x − 5)
(b) The multiplicities of each are 1.
(c) E.B. as x → ∞ we have f(x) → ∞, and as x → −∞ we have f(x) → −∞
(d)
4. g(x) = 3x5 − 16x4 + 5x3 + 90x2 − 138x + 36
2
Solution.
(a) Choosing different potential rational zeros until we land on a solution we could proceed as follows
3 − 16
5
90 − 138
36
3
3
9
− 21
− 48
126
− 36
−7
− 16
42
− 12
0
3
Choosing wisely again may result in 2
3
which tell us that f(x) = (x − 3)(3x4 − 7x3 − 16x2 + 42x − 12).
−7
− 16
42
− 12
6
−2
− 36
12
−1
− 18
6
0
which tells us that f(x) = (x − 3)(x −
3
2)(3x3 − x2 − 18x + 6). One more rational zero choice could result in
1
3
−1
− 18
6
1
0
−6
leaving us
3
0 − 18
0
√
√
1
with a quadratic that we can factor to get f(x) = 3(x − 3)(x − 2)(x − )(x + 6)(x − 6). The zeros of f are
3
1 √
then {3, 2, , ± 6}.
3
(b) The multiplicty of each of these is 1
(c) E.B. as x → ∞ f(x) → ∞, and as x → −∞ f(x) → −∞.
(d)
5. f(x) = −x3 + 5x2 − 11x + 15
Solution.
(a) Choosing different potential rational zeros until we land on a solution we could proceed as follows
−1
5 − 11
15
3
−3
6
− 15
telling us that f(x) = −(x − 3)(x2 − 2x + 5). Factoring the remaining quadratic
−1
2
−5
0
using the quadratic formula we get f(x) = −(x − 3)(x − (1 − 2i))(x − (1 + 2i)). This tells us that the zeros of f
are {3, 1 + 2i, 1 − 2i}.
(b) The multiplicities of each are 1
(c) E.B. as x → ∞ f(x) → −∞, and as x → −∞ f(x) → ∞.
(d)
3
• For problems 6-10 : (SHOW YOUR WORK!)(20pts each)
(a) (2 pts) Determine the y intercept
(b) (6 pts) Determine the x intercepts (if they exist)
(c) (2 pts) Determine the holes (if they exist)
(d) (2 pts) Determine the vertical asymptotes (if they exist)
(e) (4 pts) Determine the end behavior including the ”slant”/horizontal asymptote if it exists
(f) (4 pts) Sketch a graph of the function (label all the properties you found in parts a-e)
6. f(x) =
x−3
x−1
Solution.
(a) (0, 3)
(b) (3, 0)
(c) No holes
(d) x = 1
(e) Since the degree of the top and bottom are the same the HA is y = 1.
(f)
7. h(x) =
4
x2 − 16
Solution.
1
(a) (0, − )
4
(b) No x-int.s
4
(c) No holes
(d) x = 4 and x = −4
(e) Since the degree of the bottom is bigger the HA is y = 0.
(f)
8. q(x) =
x2 − 4x + 4
x
Solution.
(a) No y-int
(b) (2, 0)
(c) No holes
(d) x = 0
(e) Since the degree of the top is greater than the bottom we have a non-horizontal asymptote or ”slant” asymptote
given by the quotient when we use long division as follows:
x − 4 thus we have y = x − 4
)
2
x
x − 4x + 4
− x2
− 4x
4x
(f)
9. m(x) =
−3x2
x2 + 1
Solution.
5
(a)
(b)
(c)
(d)
(e)
(f)
(0, 0)
(0, 0)
no holes
no real VAs
Since the degrees of the top and bottom are the same the HA asymptote is given by y = −3.
3x4 − 15x3 + 3x2 + 63x − 54
3x2 − 12
Solution.
10. n(x) =
Using the rational zero theorem we can factor the top and bottom as n(x) =
3(x − 3)2 (x − 1)(x + 2)
which helps
3(x + 2)(x − 2)
us to determine the following
9
(a) (0, )
2
(b) (3, 0), (1, 0)
(c) hole is at x = −2
(d) VA is x = 2
(e) Since the degree of the top is greater than the bottom we perform long division to find the non-horizontal or
”slant” asymptote as follows:
x2 − 5x + 5 thus we have y = x2 − 5x + 5
)
2
4
3
3x − 12
3x − 15x + 3x2 + 63x − 54
− 3x4
+ 12x2
− 15x3 + 15x2 + 63x
15x3
− 60x
15x2 + 3x − 54
− 15x2
+ 60
3x + 6
(f)
6