Math 590 LINEAR ALGEBRA
REVIEW OF LECTURES – VIII SUPPLEMENT
February 9 (Tue), 2016
Instructor:
Yasuyuki Kachi
Line #: 52658.
This is a supplement to “Review of Lectures – VIII”, a crash course on polynomials.
I’m sure you are familiar with polynomials. As elementary as it seems, dealing with
polynomials requires care. What’s included in this supplement is a bare minimum,
so I expect you to be proficient with these.
Polynomials are made of monomials. So you have to know what monomials are.
Spellings:
“polynomial ,”
“monomial ”.
First, we call each of
1, x, x2 , x3 , x4 , x5 , ··· ,
a monomial in x. Here, we include 1 in the list because 1 = x0 . Next, we also
call each of
√ 3
1 4
3, −2x, 4x2 ,
x , 7x5 , ··· ,
2x ,
4
a monomial in x.
More generally, suppose a is a constant real number, and n is an integer with
n ≥ 0. Then we call
a xn
a monomial in x.
Note that neither of
x
−1
nor
x
1
2
is a monomial, because the exponent is either negative or a non-integer.
1
Next, a polynomial is a finite sum of monomials. Thus
1 + x,
2 x + x2 ,
−
√
4
+ 3 x3 ,
5
x + x4 + x7 ,
−6 + 2x − 8x3 + x5 ,
are examples of polynomials in x. On the other hand, none of
√
3
x,
1 + x2 ,
√
1
,
x
3
,
4−x
2 + x2 ,
2x−2 + x2
is a polynomial in x.
• Note. There is no general rule that we must obey when it comes to the order
of terms in polynomial expressions. So, you may write either
2 + x,
or
x + 2.
These two are one and the same. Similarly, you may write either
4 − 3x + x2 ,
− 3x + x2 + 4,
4 + x2 − 3x,
x2 − 3x + 4,
x2 + 4 − 3x,
or
− 3x + 4 + x2 .
These six are all one and the same. Usually, though, we prefer to write polynomials
either in the ascending order or in the descending order of exponents. So,
4 − 3x + x2
ascending order ,
x2 − 3x + 4
descending order ,
and
are equally preferable.
2
Exercise 1.
Permute the order of terms, if necessary, to make each of the given
polynomials in the ascending order.
1 2
x .
2
(1)
2x + x4 −
(3)
x8 + 5x6 − 10x9 + 3.
(5)
h
4 3
x − 5x2 − 4x5 .
3
(2)
−
(4)
x+
√
5 x5 +
√
3 x3 −
√
2 x2 .
√
5 x5 .
x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1.
i
Answers :
1 2
x + x4 .
2
(1)
2x −
(3)
3 + 5x6 + x8 − 10x9 .
(5)
(2)
− 5x2 −
(4)
x−
√
4 3
x − 4x5 .
3
2 x2 +
√
3 x3 +
1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 .
Exercise 2.
Permute the order of terms, if necessary, to make each of the given
polynomials in the descending order.
(1)
6x2 + x3 .
(2)
(3)
x7 + 5x5 − 10x8 + 4x6 .
(4)
h
1 3
1 4
x +
x − x.
2
3
1 − x2 + x4 − x6 + x8 − x10 .
i
Answers :
(1)
x3 + 6x2 .
(2)
(3)
−10x8 + x7 + 4x6 + 5x5 .
(4)
3
1 3
1 4
x +
x − x.
3
2
−x10 + x8 − x6 + x4 − x2 + 1.
•
Polynomial addition.
We may add two polynomials and the outcome is a polynomial. Let me use some
examples to illustrate it.
Example 1.
Let’s calculate
3
x + 2 + x + 4x .
This is just uncover
the parentesis and reorder the terms in either the ascending or
descending order . So,
x2 + 2 + x3 + 4x = x2 + 2 + x3 + 4x
= x3 + x2 + 4x + 2.
This answer is clearly written in the descending order.
You can write it in the
ascending order instead. You don’t have to give both.
Example 2.
Let’s calculate
1 2
4
+ 2x4 − 6x3 + x .
x
x +
5
This one looks pretty similar to Example 1 above, but this one involves more than
what was required in Example 1. Indeed, let’s just first uncover the parenthesis, and
reorder terms:
1 2
4
x +
+ 2x4 − 6x3 + x
x
5
= x4 +
1 2
x + 2x4 − 6x3 + x
5
= x4 + 2x4 − 6x3 +
4
1 2
x + x.
5
Realize that there are two monomials that involve x4 . You have to combine them.
x4 and 2x4 make 3x4 . So the above is simplified as
3x4 − 6x3 +
1 2
x + x.
5
This is the final answer.
⋆
If you like, you can do the above as follows:
x4
+)
+
2x4 − 6x3
+ x
3x4 − 6x3 +
⋆
1 2
x
5
1 2
x + x.
5
Let’s do a similar example, but in a different format.
Example 3.
Let’s find
= x4 +
= 6x4 +
f x
g x
f x + g x , where
7 3
5 2
x −
x − x,
2
2
and
1 3
x − x2 − 3x + 2.
2
This is essentially the same type of a problem as the previous
ones, but the difference
is that two polynomials are given names as f x and g x . Still, the same method
works. Here is how it goes:
5
f x +g x
7 3
5 2
1 3
4
4
2
= x +
x −
x − x + 6x +
x − x − 3x + 2
2
2
2
= x4 +
5 2
1 3
7 3
x −
x − x + 6x4 +
x − x2 − 3x + 2
2
2
2
uncovered parentheses
= x4 + 6x4 +
7 3
1 3
5 2
x +
x −
x − x2 − x − 3x + 2
2
2
2
re-ordered terms
7 3
5 2
1 3
4
4
2
= x + 6x +
x +
x + −
x − x + − x − 3x + 2
2
2
2
4
3
= 7x + 4x +
= 7x4 + 4x3 −
⋆
7 2
+ − 4x + 2
x
−
2
7 2
x − 4x + 2.
2
If you like, you can do the above as follows:
x4 +
7 3
x −
2
6x4 +
1 3
x −
2
7x4 +
4 x3
+)
5 2
x −
2
x
x2 − 3x + 2
7 2
x − 4x + 2
2
−
6
Exercise 3.
Do
(1)
x7 + 3 x5 + 2 x3 + − x6 − x4 − 2 x2 .
(2)
x4 + 9 x3 + 1 + − x4 − x3 − 5 x2 + 2x + 3 .
(3)
(4)
1 3
1
1 3
1
x +
x +
x −
x .
2
3
3
4
f x +g x ,
where
= x6 + 8x5 + 12x4 + 36x3 + 9x2 ,
= x8 − 3x6 − 8x4 − 24x3 + 45x − 120.
f x
g x
(5)
f x +g x ,
where
= x7 + x5 + x3 + x,
= x8 + x6 + x4 + x2 + 1.
f x
g x
h
i
Answers :
(1)
(3)
x7 − x6 + 3x5 − x4 + 2x3 − 2 x2 .
(2)
5 3
1
x +
x.
6
12
(4)
x8 − 2x6 + 8x5 + 4x4 + 12x3 + 9x2 + 45x − 120.
(5)
x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1.
7
8 x3 − 5x2 + 2x + 4.
•
Polynomial subtraction.
We may subtract one polynomial from another polynomial. The outcome is a
polynomial. This is very similar to polynomial additions. Let’s do some example.
Example 4.
Let’s do
3
2
x + 2x − 3x + 4 .
When you uncover the parenteses, you have to be careful. Namely:
x3 + 2x2 − 3x + 4 = x3 + 2x2 − 3x − 4.
In the above, notice that all the terms inside the second parenthesis got negated after
uncovering that parenthesis. That’s the correct way to do it.
Example 5.
Let’s do
x6 − 8x4 + 3 x2 − 2x6 − 3x4 .
Once again, when you uncover the parenteses,
6
4
6
4
2
x − 8x + 3 x − 2x − 3x
= x6 − 8x4 + 3x2 − 2x6 + 3x4
all the terms in the second parenthesis got negated
= x6 − 2x6 − 8x4 + 3x4 + 3x2
re-ordered terms
=
x6 − 2x6 + − 8x4 + 3x4 + 3x2
= − x6 − 5x4 + 3x2 .
8
⋆
If you like, you can do it like
x6 − 8x4 + 3x2
2x6 − 3x4
−)
− x6 − 5x4 + 3x2
Do
Exercise 4.
(1)
x3 + 11 x2 + 21 x − − x2 − x + 4 .
(2)
(3)
(4)
6
4
3
− x + 5x + x − 6 − − 2x − 3x + 7 x + 2x + 5 .
7
6
3
3 4
1 4
7 2
1 2
−
x +
x
x −
x +1 .
2
4
6
4
f x −g x ,
where
f x = x5 + 4x4 + 16x3 + 22x2 + 18x,
g x = x6 − 31x4 − 62x2 + 72x + 56.
(5)
f x −g x ,
f x
h
where
= x13 + x9 + x5 + x,
g x
= x11 + x7 + x3 .
i
Answers :
(1)
(3)
(5)
x3 + 12x2 + 22x − 4.
4 4
x + 2x2 − 1.
3
− x7 + 7x6 + 3x4 − 6x3 − 2x − 11.
(2)
(4)
−x6 + x5 + 35x4 + 16x3 + 84x2 − 54x − 56.
x13 − x11 + x9 − x7 + x5 − x3 + x.
9
•
Constsant multiplication.
We may multiply a constant with a polynomial.
Example 6.
Let’s do
10 2 x3 + 3x2 + 4 x + 5 .
This is simply multiply 10 to each of the terms. So
10 2 x3 + 3x2 + 4 x + 5 = 20 x3 + 30x2 + 40 x + 50.
Example 7.
Let’s do
−2 12 x5 − 21x3 + 48 x .
This is simply multiply −2 to each of the terms. So
5
3
−2 12 x − 21x + 48 x = −24 x5 + 42x3 − 96 x.
Example 8.
Sometimes you see something like
2x4 − 11x2 + 14x − 9
.
2
This is the same as
1 4
2x − 11x2 + 14x − 9 .
2
The answer is, of course,
x4 −
11 2
9
x + 7x −
.
2
2
10
Exercise 5.
Simplify:
(1)
6 x7 + 7 x6 + 21 x5 .
(3)
8x10 − 20x8 + 24x6 − 12x4
.
4
(5)
3f x
h
i
Answers :
where
(1)
f x
(2)
−4
− x2 + 5x + 3 .
(4)
1
3
3 4
3 3
3 2
3
x +
x +
x +
x .
5
7
25
65
= x5 + 4x4 + 16x3 + 22x2 + 18x.
6x7 + 42x6 + 126x5 .
(3)
2x10 − 5x8 + 6x6 − 3x4 .
(5)
3x5 + 12x4 + 48x3 + 66x2 + 54x.
•
(4)
(2)
4x2 − 20x − 12.
1 4
1 3
1 2
1
x +
x +
x +
x.
5
7
25
65
A monomial multipled to a polynomial.
More generally, we may multiply a monomial with a polynomial. The outcome is
a polynomial. This process is called an expansion .
Example 9.
Let’s expand
4
3
2
x x + 3x + 5x .
This is simply raise the exponent in each x-to-the-power inside the parenthesis by
1. So
4
3
2
= x5 + 3x4 + 5x3 .
x x + 3x + 5x
Example 10.
Let’s expand
3
2
2
x 8x + 24x + 6x + 3 .
This time you raise the exponent in each x-to-the-power inside the parenthesis by
2. So
2
3
2
x 8x + 24x + 6x + 3 = 8x5 + 24x4 + 6x3 + 3x2 .
11
Example 11.
Let’s expand
2x x2 + 8x + 5 .
This requires you to do two things at once, namely, (i) raise the exponent in each
x-to-the-power inside the parenthesis by 1, and then (ii) multiply 2 to each of the
terms. So
2x x2 + 8x + 5 = 2x3 + 16x2 + 10x.
Example 12.
Let’s expand
5x4 6x5 + 12x3 + 8x2 + 7 .
This is similar, do two things at once: namely, (i) raise the exponent in each
x-to-the-power inside the parenthesis by 4, and then (ii) multiply 5 to each of the
terms. So
5
3
2
5x 6x + 12x + 8x + 7 = 30x9 + 60x7 + 40x6 + 35x4 .
4
Expand
Exercise 6.
(1)
(3)
(4)
h
x x2 − x + 4 .
6x
x3 2x4 + 8x3 + 5x .
(2)
1 3
1 2
1
−
x +
x −
x .
3
2
6
1 2 4
3
2
x x + 2x + 3x + 2x + 1 .
4
i
Answers :
(1)
x3 − x2 + 4x.
(2)
(3)
− 2x4 + 3x3 − x2 .
(4)
2x7 + 8x6 + 5x4 .
1 5
3 4
1 3
1 2
1 6
x +
x +
x +
x +
x .
4
2
4
2
4
12
•
Multiplying out two polynomials.
Even more generally, we may multiply out two polymials. The outcome is a polynomial. This process is once again called an expansion .
Example 13.
Let’s expand
x + 1 x2 + 1 .
It goes as follows.
x
+
1
x2 + 1
↑ ↑
multiply
multiply
2
= x x +1
3
= x +x
2
1 x +1
+
+
2
x +1
= x3 + x + x2 + 1
= x3 + x2 + x + 1.
In short,
x + 1 x2 + 1 = x3 + x2 + x + 1.
⋆ As you can see, the above consists of two multiplications, both “monomial times
polynomial” type, followed by one polynomial addition.
13
Let’s expand
Example 14.
2x2 + 3 x2 − 3x + 4 .
It goes as follows.
+
3
multiply
x2 −
↑
multiply
3x + 4
↑
x2 − 3x + 4
+
3
2x3 − 6x2 + 8x
+
3x2 − 9x + 12
= 2x
=
2x
x2 − 3x + 4
= 2x3 − 6x2 + 8x + 3x2 − 9x + 12
uncovered parentheses
= 2x3 − 6x2 + 3x2 + 8x − 9x + 12
re-ordered terms
= 2x3 − 3x2 − x + 12.
In short,
2x + 3 x2 − 3x + 4 = 2x3 − 3x2 − x + 12.
14
⋆ Once again, the procedure consists of two multiplications, both “monomial times
polynomial” type, followed by one polynomial addition. The following recaptures the
same algorithm:
x2
−
3x
+
4
2x
+
3
+ 12
×)
3 x2
−
9x
2 x3 −
6 x2
+
8x
2 x3 −
3 x2
−
x
+ 12
Let’s dissect. Take a look at the following which is a mere duplication of the above,
but with some highlighted boxes:
x2
−
3x
+
4
2x
+
3
+ 12
×)
3 x2
−
9x
2 x3 −
6 x2
+
8x
2 x3 −
3 x2
−
x
+ 12
Ignore the parts that are not enclosed by the boxes. Only look at those highlighted
by the boxes. This is just constant 3 which is a monomial multiplied to the
polynomial x2 − 3x + 4. The outcome is highlighted in the lowest box. Next,
look at the following, I have simply changed the locations of the boxes:
15
x2
−
3x
+
4
2x
+
3
+ 12
×)
3 x2
−
9x
2 x3 −
6 x2
+
8x
2 x3 −
3 x2
−
x
+ 12
Again, ignore everything else but only look at the highlighted parts. This is just
monomial 2x multiplied to the polynomial x2 − 3x + 4. The outcome is highlighted in the lowest box. Finally, just one more time I’m going to change the locations
of the boxes:
x2
−
3x
+
4
2x
+
3
+ 12
×)
3 x2
−
9x
2 x3 −
6 x2
+
8x
2 x3 −
3 x2
−
x
+ 12
Only look at the last three lines as highlighted. This is just polynomial additions.
The outcome is highlighted in the lowest line in the box. This is exactly the final
outcome.
Let’s do another example.
16
Let’s expand
Example 15.
− 2x2 + 5x − 3 x2 + 2x − 7 .
It goes as follows.
− 2x2
+ 5x
2
= − 2x x + 2x − 7
2
=
− 2 x4 − 4x3 + 14 x2
− 3
x2 + 2x
↑ ↑
multiply multiply
multiply
+
2
5x x + 2x − 7
+
5 x3 + 10x2 − 35 x
−
+
−7
↑
2
3 x + 2x − 7
− 3x2 − 6x + 21
= − 2 x4 − 4x3 + 14 x2 + 5 x3 + 10x2 − 35 x − 3x2 − 6x + 21
uncovered parentheses
= − 2x4 − 4x3 + 5x3 + 14x2 + 10x2 − 3x2 − 35x − 6x + 21
re-ordered terms
= − 2x4 + x3 + 21x2 − 41x + 21.
In short,
− 2x2 + 5x − 3 x2 + 2x − 7 = − 2x4 + x3 + 21x2 − 41x + 21.
17
⋆
Just like the previous example, we can do it the following way:
×)
x2
+
2x
−
7
− 2 x2
+
5x
−
3
3 x2
−
6x
+
21
+
21
−
5 x3 + 10 x2
− 2 x4 −
4 x3 + 14 x2
− 2 x4 +
x3 + 21 x2
− 35 x
− 41 x
You can do it either way, whichever way suits you better. Let’s do more examples.
Example 16.
Let’s expand
2
2
x −x+1 x +x+1 .
It goes as follows.
x
2
− x
+ 1
x
multiply
multiply
multiply
18
2
↑
+x+1
↑ ↑
= x2
x2 + x + 1
4
3
2
= x +x +x
−
x
+
x2 + x + 1
−x −x −x
3
2
x2 + x + 1
+
1
+
2
x +x+1
= x4 + x3 + x2 − x3 − x2 − x + x2 + x + 1
uncovered parentheses
= x4 + x3 − x3 + x2 − x2 + x2 − x + x + 1
re-ordered terms
= x4 + x2 + 1.
In short,
2
2
x − x + 1 x + x + 1 = x4 + x2 + 1.
⋆
The following is an alternative way:
x2
+
x
+
1
x2
−
x
+
1
x2
+
x
+
1
−
x
+
1
×)
x4
x4
− x3
−
x2
x3
+
x2
+
x2
+
19
•
Let’s do
x
x
x
x
x
x
x
(0)
(1)
(2)
(3)
(4)
(5)
(6)
− 1 · 1,
−1 x+1 ,
− 1 x2 + x + 1 ,
3
2
−1 x +x +x+1 ,
− 1 x4 + x3 + x2 + x + 1 ,
− 1 x5 + x4 + x3 + x2 + x + 1 ,
− 1 x6 + x5 + x4 + x3 + x2 + x + 1 ,
··
Let’s try (5). Let’s do it the second way:
x5
+
x4
+
x3
+
x2
+
x
+
1
x
−
1
−
1
−
1
×)
x6
+
− x5
−
x4
−
x3
−
x2
−
x
x5
+
x4
+
x3
+
x2
+
x
x6
In short,
(5)
x − 1 x5 + x4 + x3 + x2 + x + 1 = x6 − 1.
20
By extrapolation:
(0)
x − 1 ·1
= x − 1,
(1)
x−1 x+1
= x2 − 1,
(2)
2
x−1 x +x+1
= x3 − 1,
(3)
x − 1 x3 + x2 + x + 1
= x4 − 1,
(4)
x − 1 x4 + x3 + x2 + x + 1
= x5 − 1,
(5)
x − 1 x5 + x4 + x3 + x2 + x + 1
= x6 − 1,
(6)
x − 1 x6 + x5 + x4 + x3 + x2 + x + 1 = x7 − 1,
··
⋆
A minor tweak:
(0)
1 − x ·1
= 1 − x,
(1)
1−x 1+x
= 1 − x2 ,
(2)
1 − x 1 + x + x2
= 1 − x3 ,
(3)
1 − x 1 + x + x2 + x3
= 1 − x4 ,
(4)
1 − x 1 + x + x2 + x3 + x4
= 1 − x5 ,
(5)
1 − x 1 + x + x2 + x3 + x4 + x5
= 1 − x6 ,
(6)
2
3
4
5
6
= 1 − x7 ,
1−x 1+x+x +x +x +x +x
··
21
⋆
Further tweak:
(0)
1 + x ·1
= 1 + x,
(1)
1+x 1−x
= 1 − x2 ,
(2)
1 + x 1 − x + x2
= 1 + x3 ,
(3)
1 + x 1 − x + x2 − x3
= 1 − x4 ,
(4)
1 + x 1 − x + x2 − x3 + x4
= 1 + x5 ,
(5)
1 + x 1 − x + x2 − x3 + x4 − x5
= 1 − x6 ,
(6)
2
3
4
5
6
1+x 1−x+x −x +x −x +x
= 1 + x7 ,
··
Exercise 7.
Expand
(1)
x+3 x+4 .
(2)
x+5 x−2 .
(3)
x2 − 7 x2 − 3x + 1 .
(4)
x3 − 4x2 + 2 x2 + 3x − 8 .
h
i
Answers :
(1)
x2 + 7x + 12.
(2)
x2 + 3x − 10.
(3)
x4 − 3x3 − 6x2 + 21x − 7.
(4)
x5 − x4 − 20x3 + 34 x2 + 6x − 16.
22
Expand
Exercise 8.
(1)
(2)
x−1
x−1
x20 + x19 + x18 + x17 + x16 + x15 + x14 + x13 + x12 + x11
+x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 .
x40 + x39 + x38 + x37 + x36 + x35 + x34 + x33 + x32 + x31
+x30 + x29 + x28 + x27 + x26 + x25 + x24 + x23 + x22 + x21
+x20 + x19 + x18 + x17 + x16 + x15 + x14 + x13 + x12 + x11
+x10 + x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1 .
(3)
1−x
1 + x + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10
+ x11 + x12 + x13 + x14 + x15 + x16 + x17 + x18 + x19 + x20
+ x21 + x22 + x23 .
(4)
1+x
1 − x + x2 − x3 + x4 − x5 + x6 − x7 + x8 − x9 + x10
− x11 + x12 − x13 + x14 .
(5)
1+x
1 − x + x2 − x3 + x4 − x5 + x6 − x7 + x8 − x9 + x10
− x11 + x12 − x13 + x14 − x15 + x16 − x17 + x18 − x19 + x20
− x21 + x22 − x23 + x24 − x25 + x26 − x27 + x28 − x29 + x30
− x31 + x32 − x33 + x34 − x35 + x36 − x37 + x38 − x39 + x40
− x41 + x42 − x43 + x44 − x45 + x46 − x47 .
h
Answers
i
:
(1)
x21 − 1.
(2)
x41 − 1.
(3)
1 − x24 .
(4)
1 + x15 .
(5)
1 − x48 .
23
•
Squaring Polynomials.
2
Next, squaring polynomials. For a polynomial f x , f x
means f x ·f x .
Example 17.
Let’s expand
2
x2 + 3 .
This is the same as
x2 + 3 x2 + 3 .
So
x2 + 3 x2 + 3 = x2 x2 + 3 + 3 x2 + 3
= x4 + 3 x2 + 3 x2 + 9
= x4 + 6 x2 + 9.
Now, some of you might say “why not use the binomial formula?”. That’s excellent
point. Yes, binomial formula is
2
a+b
= a2 + 2ab + b2 .
Substitute a with x2 , and substitute b with 3:
2
2
2
x +3
= x2
+ 2 · x2 · 3 + 3 2
= x4 + 6 x2 + 9.
Example 18.
Let’s expand
2
x2 + x + 2 .
This time you have to do it like
follows:
x2 + x + 2 x2 + x + 2 .
24
It goes as
x2 + x + 2 x2 + x + 2
= x2 x2 + x + 2 + x x2 + x + 2 + 2 x2 + x + 2
=
x4 + x3 + 2x2 + x3 + x2 + 2x + 2x2 + 2x + 4
= x4 + x3 + 2x2 + x3 + x2 + 2x + 2x2 + 2x + 4
= x4 + x3 + x3 + 2x2 + x2 + 2x2 + 2x + 2x + 4
= x4 + 2x3 + 5x2 + 4x + 4.
You can certainly do it this way:
x2
+
x
+
2
x2
+
x
+
2
2 x2
+
2x
+
4
x2
+
2x
+
4x
+
4
×)
x3 +
x4 +
x4 +
x3 + 2 x2
2 x3
+ 5 x2
To conclude,
2
x2 + x + 2
= x4 + 2x3 + 5x2 + 4x + 4.
Example 19.
Let’s expand
2
3
2
x + 4x − 3x + 2 .
Let’s just do it in the second way.
25
x3
+
4 x2
−
3x
+
2
x3
+
4 x2
−
3x
+
2
2 x3
+
8 x2
−
6x
+
4
− 3 x4 − 12 x3
+
9 x2
−
6x
+ 16 x4 − 12 x3
+
8 x2
+
25 x2 − 12 x
+
4
×)
4 x5
x6 + 4 x5
− 3 x4
+ 2 x3
x6 + 8 x5 + 10 x4 − 20 x3
To conclude,
2
x3 + 4x2 − 3x + 2
= x6 + 8x5 + 10x4 − 20x3 + 25x2 − 12x + 4.
Example 20.
Let’s expand
2
x4 − 5x3 − 6 x + 4 .
As before, we can handle it like
×)
x4 − 5 x3
−
6x
+
4
x4 − 5 x3
−
6x
+
4
4 x4 − 20 x3
− 6 x5 + 30 x4
− 5 x7
+ 25 x6
− 24 x + 16
+ 36 x2 − 24 x
+ 30 x4 − 20 x3
x8 − 5 x7
− 6 x5 + 4 x4
x8 − 10 x7 + 25 x6
− 12 x5 + 68 x4 − 40 x3 + 36 x2 − 48 x + 16
26
To conclude,
2
4
3
x − 5x − 6 x + 4
= x8 − 10x7 + 25x6 − 12x5 + 68x4 − 40x3 + 36x2 − 48x + 16.
Example 21.
Let’s expand
2
x4 + x3 + x2 + x + 1 .
The same deal:
x4 + x3 + x2 + x + 1
x4 + x3 + x2 + x + 1
×)
x4 + x3 + x2 + x + 1
x5 + x4 + x3 + x2 + x
x6 + x5 + x4 + x3 + x2
x7 + x6 + x5 + x4 + x3
x8 + x7 + x6 + x5 + x4
x8 + 2x7 + 3x6 + 4x5 + 5x4 + 4x3 + 3x2 + 2x + 1
To conclude,
2
4
3
2
x +x +x +x+1
= x8 + 2x7 + 3x6 + 4x5 + 5x4 + 4x3 + 3x2 + 2x + 1.
27
Exercise 9.
Expand
(1)
2
3
x − 8x .
(3)
2
x3 + x2 − 4 .
(5)
2
1 + x + x2 + x3 + x4 + x5 .
h
2
2
2x − 3x + 5 .
(2)
2
1
1
2
x +
.
x+
2
3
(4)
i
Answers :
(1)
x6 − 16x4 + 64x2 .
(3)
x6 + 2x5 + x4 − 8x3 − 8x2 + 16.
(4)
x4 + x3 +
(5)
1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + 5x6 + 4x7 + 3x8 + 2x9 + x10 .
•
4x4 − 12x3 + 29x2 − 30x + 25.
(2)
11
11
11 2
x +
x+
.
12
3
9
Product of three or more polynomials.
Example 22.
How about
2
4
2
x+1 x +x+1 x +x +1 ?
There are three polynomials involved. This one you have to do it step by step,
namely, you first do the boxed part:
x + 1 x2 + x + 1
x4 + x2 + 1
and then you multiply the outcome with the third factor
it:
28
x4 + x2 + 1.
Let’s do
x + 1 x2 + x + 1 :
Do
Step 1.
x2 + x + 1
x + 1
×)
x2 + x + 1
x3 + x2 + x
x3 + 2 x2 + 2 x + 1
In short,
x + 1 x2 + x + 1 = x3 + 2x2 + 2x + 1.
3
2
4
2
x + 2x + 2x + 1 x + x + 1 :
Do
Step 2.
x4
+
x2
+1
x3 + 2 x2 + 2 x + 1
×)
x4
2 x5
2x
x7
6
x2
+ 2 x3
+ 2x
+ x5
+
4
+ 2x
+ 2x
+
+1
2
x3
x7 + 2 x6 + 3 x5 + 3 x4 + 3 x3 + 3 x2 + 2 x + 1
In short,
x3 + 2x2 + 2x + 1 x4 + x2 + 1
= x7 + 2x6 + 3x5 + 3x4 + 3x3 + 3x2 + 2x + 1.
29
To conclude,
2
4
2
x+1 x +x+1 x +x +1
= x7 + 2x6 + 3x5 + 3x4 + 3x3 + 3x2 + 2x + 1.
Exercise 10.
Expand:
(1)
2
x−1 x+1 .
(2)
2
x−1 x−3 x −3 .
(3)
2
x−1 x+1 x +1 .
(4)
√
√
x2 − 2 x + 1 x2 + 2 x + 1 x2 − 1 .
(5)
√
√
√
√
x− 2 −1 x+ 2 −1 x− 2 +1 x+ 2 +1 .
h
i
Answers :
(1)
x3 + x2 − x − 1.
(2)
x4 − 4x3 + 12x − 9.
(3)
x4 − 1.
(4)
x6 − x4 + x2 − 1.
(5)
x4 − 6 x2 + 1.
30
•
Raising products.
The following has some bearings on certain types of polynomial multiplications:
(row 0)
1
1/
\1
1
1/
2\
/1 \2
1
1/
3
3\
1
4\
1
1/
1
1/
1
1/
/1
21
7\
/1
7\
3\
/1
4\
/1
5\
/1
/1
175
/1
4\
7\
/1 \4
50
5\
/1
/1
735
/1
7\
6\
(row 4)
24
5\
225
6\
(row 3)
6
35
85
6\
/1 \3
11
/1
15
6\
/1
10
5\
(row 2)
2
6
1/
(row 1)
1
/1 \5
274
/1
6\
/1 \6
1624
1764
/1
/1
7\
(row 5)
120
7\
720
(row 6)
/1 \7
I’m sure you are familiar with Pascal’s triangle. The above is a variation of it. This
can be used to get the expansions
x+1 ,
x+1 x+2 ,
x+1 x+2 x+3 ,
x+1 x+2 x+3 x+4 ,
x+1 x+2 x+3 x+4 x+5 ,
x+1 x+2 x+3 x+4 x+5 x+6 ,
···
31
··
·
Namely:
x+1
x+2
= x2 + 3 x + 2,
x+1
x+2
x+3
= x3 + 6 x2 + 11 x + 6,
x+1
x+2
x+3
x+4
= x4 + 10 x3 + 35 x2 + 50 x + 24,
x+1
x+2
x+3
x+4
x+5
= x5 + 15 x4 + 85 x3 + 225 x2 + 274 x + 120,
x+1
x+2
x+3
x+4
x+5
x+6
= x6 + 21 x5 + 175 x4 + 735 x3 + 1624 x2 + 1764 x + 720.
Exercise 11.
Expand:
x+1 x+2 x+3 x+4 x+5 x+6 x+7 .
⋆ As for this, the triangle in the previous page is apparently shown only up to the
sixth row. You have to extend it to the seventh row.
h
i
Answer :
x7 + 28x6 + 322x5 + 1960x4 + 6769x3 + 13132x2 + 13068x + 5040.
32