MAC 1147
Homework Help
WebAssign
Write in standard form:
9i
(3 + 4i)2
9i
9i
=
(3 + 4i)2
(3 + 4i)(3 + 4i)
9i
=
9 + 12i + 12i + 16i2
9i
=
9 + 24i − 16
9i
=
−5 + 24i
−5 − 24i
9i
∗
=
−5 + 24i −5 − 24i
−45i − 216i2
=
25 − 576i2
−45i + 216
=
25 + 576
216 − 45i
=
601
216
45
=
−
i
601 601
Section 2.2
63. Find the polynomial that has the given zeros.
√
√
1 + 3, 1 − 3
Let’s evaluate each zero that we were given,
x=1+
√
x−1− 3=0
√
x − (1 + 3) = 0
√
3
Similarly,
x=1−
√
x−1+ 3=0
√
x − (1 − 3) = 0
Now let’s find our polynomial,
√
3
√
3))(x − (1 − 3)) = 0
√
√
√
√
x2 − x + x 3 − x + 1 − 3 − x 3 + 3 − 3 = 0
(x − (1 +
√
x2 − 2x − 2 = 0
So, our polynomial with the given zeros is f (x) = x2 − 2x − 2.
71. Find a polynomial of degree n that has the given zeros.
x = −5, 1, 2 and n = 4
So we are given three zeros, but asked to find a polynomial with degree four. When given a
degree that is more than the number of zeros they give us, we want to increase the number of
zeros we already have. There are many different ways we can do this. We can have our zeros be
x = −5, −5, 1, 2 or x = −5, 1, 1, 2 or x = −5, 1, 2, 2. I’m going to let our zeros be x = −5, 1, 1, 2.
(x + 5)(x − 1)2 (x − 2) = 0
(x + 5)(x − 1)(x − 1)(x − 2) = 0
(x2 + 4x − 5)(x − 1)(x − 2) = 0
(x3 + 3x2 − 9x + 5)(x − 2) = 0
x4 + x3 − 15x2 + 23x − 10 = 0
So, my polynomial f (x) = x4 + x3 − 15x2 + 23x − 10 has zeros x = −5, 1, 1, 2 and has degree 4.
A good indication that you made a mistake is if you end up with a polynomial that isn’t 4.
Section 2.4
69. Write the complex number in standard form.
√
5i)(7 − 10i)
√
√
√
= 21 − 3 10i + 7 5i + 5 2
√
√
√
= (21 + 5 2) + (7 5 − 3 10)i
√
√
√
√
√
√
(21+5 2)+(7 5−3 10)i is in standard form of (a+bi) where a = 21+5 2 and b = 7 5−3 10
(3 +
√
−5)(7 −
√
−10) = (3 +
√
Section 2.4
50. Find a polynomial with real coefficients and given zeros.
√
−5, −5, 1 + 3i
A very important note in problems like this are whenever you have complex zeros - THEY MUST
OCCUR IN PAIRS! Your book explains this on p.158. Basically, if you have a polynomial function
that has REAL COEFFICIENTS and a+bi, where b 6= 0, is a zero of the function, then the conjugate
a − bi MUST ALSO be a zero of the function.
√
√
We are given that 1 + 3i is one of our zeros. So, we MUST HAVE the conjugate 1 − 3i as a
zero, as well.
√
√
f (x) = (x + 5)2 (x − (1 + 3i))(x − (1 − 3i))
√
√
= (x + 5)2 (x − 1 − 3i)(x − 1 + 3i)
NOTICE THAT: (x − 1 −
√
3i) and (x − 1 +
√
3i) are conjugates of each other. So,
√
3i)(x − 1 + 3i)
√
√
= (x + 5)2 ((x − 1) − 3i)((x − 1) + 3i)
= (x + 5)2 (x − 1 −
√
= (x + 5)2 ((x − 1)2 − 3i2 )
= (x + 5)2 ((x − 1)2 + 3)
= (x + 5)2 ((x − 1)(x − 1) + 3)
= (x + 5)2 ((x2 − 2x + 1) + 3)
= (x + 5)2 (x2 − 2x + 4)
= (x + 5)(x + 5)(x2 − 2x + 4)
= (x2 + 10x + 25)(x2 − 2x + 4)
= x4 + 8x3 + 9x2 − 10x + 100
So, f (x) = x4 +√8x3 + 9x√2 − 10x + 100 is our polynomial with REAL COEFFICIENTS and with
zeros: −5, −5, 1 + 3i, 1 − 3i.
52. Let f (x) = x4 − 2x3 − 3x2 + 12x − 18. (Hint: one factor is x2 − 6).
a.) Write f (x) as the product of factors that are irreducible over the RATIONALS.
Use long division to divide x4 − 2x3 − 3x2 + 12x − 18 by x2 − 6. You should get:
x4 − 2x3 − 3x2 + 12x − 18
= x2 − 2x + 3
x2 − 6
This means that
f (x) = x4 − 2x3 − 3x2 + 12x − 18 = (x2 − 6)(x2 − 2x + 3)
Notice that x2 − 6 is irreducible over the rationals (but not the reals) and x2 − 2x + 3 are
irreducible over the reals so it is also irreducible over the rationals. So, f (x) = (x2 − 6)(x2 − 2x + 3)
is a product of irreducible factors over the rationals.
(p.159 in your book gives an explanation of the difference between irreducible factors over the
rationals and irreducible factors over the reals)
b.) Write f (x) as a product of linear and quadratic factors that are irreducible over the REALS.
So, in part (a.) we found that f (x) = (x2 − 6)(x2 − 2x + 3) is a product of factors that are
irreducible over the rationals. Now we need to write f (x) as a product of factors that are irreducible
over the reals. Well, (x2 − 2x + 3) is irreducible over the reals. But (x2 − 6) is actually reducible
over the reals so we need to factor (x2 − 6) further so that it is irreducible over the reals.
(x2 − 6) = (x −
√
6)(x +
√
6)
These are now factors that are irreductive over the reals. So substitute what we just found into
f (x).
f (x) = (x2 − 6)(x2 − 2x + 3)
√
√
f (x) = (x − 6)(x + 6)(x2 − 2x + 3)
√
√
So f (x) = (x − 6)(x + 6)(x2 − 2x + 3) is a product of linear and quadratic factors that are
irreducible over the reals.
c.) Write f (x) in completely factored form.
So, now we are asked to completely factor f (x). (This means we will have complex numbers in
our solution.) Let’s do this by further breaking down what we found in part b.
√
√
√
In√part b. we found that f (x) = (x − 6)(x + 6)(x2 − 2x + 3). We can’t factor (x − 6) or
(x + 6) any further. But we can factor (x2 − 2x + 3) further.
Use the quadratic formula to factor (x2 − 2x + 3).
p
√
√
4 − 4(1)(3)
2 ± 2 2i
=
= 1 ± 2i
2
2
√
√
√
√
So, (x2 − 2x + 3) = (x − (1 + 2i)(x − (1 − 2i) = (x − 1 − 2i)(x − 1 + 2i). If we substitute
this into f (x) we get:
2±
f (x) = (x −
√
6)(x +
√
6)(x − 1 −
√
2i)(x − 1 +
√
2i)
So we have written f (x) in completely factored form.
62. Use the given zero to find all the zeros of the function.
f (x) = x3 + 4x2 + 14x + 20; zeros: −1 − 3i
A very important note in problems like this are whenever you have complex zeros - THEY MUST
OCCUR IN PAIRS! Your book explains this on p.158. Basically, if you have a polynomial function
that has REAL COEFFICIENTS and a+bi, where b 6= 0, is a zero of the function, then the conjugate
a − bi MUST ALSO be a zero of the function.
√
√
We are given that 1 + 3i is one of our zeros. So, we MUST HAVE the conjugate 1 − 3i as a
zero, as well.
√
Now use synthetic division to divide x3 + 4x2 + 14x + 20 by 1 − 3i. You should get
x3 + 4x2 + 14x + 20
√
= x2 + (3 − 3i)x + (2 − 6i)
1 − 3i
√
You can now use synthetic division to divide x2 + (3 − 3i)x + (2 − 6i) by 1 + 3i. You should get
x2 + (3 − 3i)x + (2 − 6i)
√
=x+2
1 + 3i
Notice that x + 2 = 0 implies that x = −2. So our final zero of f (x) is −2.
√
√
So, altogether the zeros of f (x) are 1 + 3i, 1 − 3i, −2.
Horizontal and Vertical Asymptotes
I was also asked to help explain how to find a horizontal asymptote. If you refer to p.170 in your
book you will find a helpful description of horizontal asymptotes and how to find them.
I’ll do a few homework problems to try and help your understanding.
Section 2.6
10. Find all vertical and horizontal asymptotes of the graph of the function.
f (x) =
1
(x − 2)3
The vertical asymptotes of f are the zeros of the denominator of f(x). Our denominator is (x−2)3 .
So,
(x − 2)3 = 0
x−2=0
x=2
So, the denominator has a real solution, x = 2, so the graph has the line x = 2 as a vertical
asymptote.
To find the horizontal asymptote let’s expand this denominator. So,
f (x) =
1
(x − 2)3
1
(x − 2)(x − 2)(x − 2)
1
= 3
x − 6x2 + 12x − 8
=
To find the horizontal asymptote(s) of f we need to compare the degrees of the numerator and
the denominator of f (x). The numerator of f (x) is 1, so the degree of the numerator is 0. The
denominator of f (x) is x3 − 6x2 + 12x − 8, so the degree of the numerator is 3.
the degree of the numerator = 0 < degree of the denominator = 3
So, the graph of f has the line y = 0 as a horizontal asymptote.
13. Find all vertical and horizontal asymptotes of the graph of the function.
f (x) =
x3
x2 − 1
The vertical asymptotes of f are the zeros of the denominator of f(x). Our denominator is x2 − 1.
So,
x2 − 1 = 0
x2 = 1
√
x=± 1
x = ±1
So, the denominator has real solutions, x = 1 and x = −1, so the graph has the lines x = 1 and
x = −1 as a vertical asymptote.
To find the horizontal asymptote(s) of f we need to compare the degrees of the numerator and
the denominator of f (x). The numerator of f (x) is x3 , so the degree of the numerator is 3. The
denominator of f (x) is x2 − 1, so the degree of the numerator is 2.
the degree of the numerator = 3 > degree of the denominator = 2
So, the graph of f has no as a horizontal asymptote.
15. Find all vertical and horizontal asymptotes of the graph of the function.
f (x) =
3x2 + 1
+x+9
x2
The vertical asymptotes of f are the zeros of the denominator of f(x). Our denominator is
x2 + x + 9. Using the quadratic formula we find that the denominator has no REAL zeros. So, f has
no vertical asymptote.
To find the horizontal asymptote(s) of f we need to compare the degrees of the numerator and
the denominator of f (x). The numerator of f (x) is 3x2 + 1, so the degree of the numerator is 2.
The denominator of f (x) is x2 + x + 9, so the degree of the numerator is 2.
the degree of the numerator = 2 = degree of the denominator = 2
So, the graph of f has the line y =
an
bn
=
3
1
= 3 as a horizontal asymptote.