fundamental theorem of algebra

1. The Fundamental Theorem of Algebra
Example A. b. Find the real 2nd degree polynomial Q(x)
with roots z = 3 + 2i and z* = 3 – 2i and with Q(1) = 2.
From part a.
Q(x) = k(x – z)(x – z*) = k(x2 – 6x + 13)
Since Q(1) = k(x2 – 6x + 13)
= k((1)2 – 6(1) + 13) = 2
Hence 8k = 2 or k = 1/4.
So Q(x) = ¼ (x2 – 6x + 13),
this is the specific equation given that Q(1) = 2.

2. The Fundamental Theorem of Algebra
Example B.
Factor 3x6 – 192 completely into real factors and list
all its roots.
3x6 – 192 = 3(x6 – 64)
= 3(x3 – 8)(x3 + 8)
= 3(x – 2)(x2 + 2x +4)(x + 2)(x2 – 2x + 4)
(x – 2) and (x + 2) give the real roots x = 2 and –2.
(x2 + 2x +4) and (x2 – 2x + 4) are irreducible
quadratics with conjugate complex roots.
The roots of x2 + 2x + 4 are x = = -1 ± i3-2 ±-12
2
The roots of x2 – 2x + 4 are x = = 1 ± i32 ±-12
2

3. The Fundamental Theorem of Algebra
P(x) is a real polynomial so the complex roots are in
conjugate pairs which are x = i, -i , (2 – i), (2 + i)
i + (-i) = 0, i * (-i) = 1, so the quadratic with roots i, -i
is (x2 + 1).
(2 + i) + (2 – i) = 4, (2 + i) * (2 – i) = 5, so the
quadratic with roots (2 + i), (2 – i) is (x2 – 4x + 5).
Therefore P(x) = k(x – 1)(x2 + 1)(x2 – 4x + 5) for
some constant k.
But P(0) = 10 = k(0 – 1)(0 + 1)(0 – 0 + 5) = -5k,
so k = -2. Hence P(x) = -2(x – 1)(x2 + 1)(x2 – 4x + 5).
Example C.
Given that x = 1, i , 2 – i are roots of a degree 5 real
polynomial P(x) and that P(0) = 10. Find P(x).

4. The Fundamental Theorem of Algebra
Exercise A. Factor the following polynomials completely into
real factors and list all its real and complex roots.
1. x3 – 1 2. x3 – 8 3. 8x3 + 27 4. 27x3 + 125
5. x4 – 16 6. 16x4 – 81
11. x6 + 1 12. x6 – 1
7. x4 – x2 – 2 8. 4x4 + 3x2 – 1
9. x4 + 3x2 + 2 10. 3x4 + 4x2 + 1
B. Find the real polynomial P(x) given its roots, the degree,
and its value at one point (the initial condition).
1. roots: x = 1 + i, degree 2 with P(0) = 5.
2. roots: x = 2 – i, degree 2 with P(0) = –2.
3. roots: x = 2, 1 + 3i, degree 3 with P(1) = –4.
4. roots: x = –1, 2 – i, degree 3 with P(–1) = 3.
5. roots: x = –2 + i , 1 + 2i, degree 4 with P(1) = –3.
6. roots: x = –1 – i , 3 + i, degree 4 with P(–1) = 1.

5. The Fundamental Theorem of Algebra
B. Find the real polynomial P(x) given its roots, the degree,
and its value at one point (the initial condition).
7. roots: x = 0 (ord = 2), i, degree 4, P(1) = 2.
8. roots: x = 1, 1 + i, (ord = 2), degree 5, P(2) = 1.
9. roots: x = –1, 2, i – 2 (ord = 2), degree 6, P(1) = 2.
10. roots: x = 1 (ord = 2), i√2 (ord = 2), degree 6, P(–1) = 2.
11. roots: x = 0, –1, –2 + i√3, degree 4, P(1) = 1.
12. roots: x = 0 (ord = 2), 3 + i√5 (ord = 2), degree 6, P(1) = 2.
13. What may we conclude from the Fundamental Theorem of
Algebra about the roots of polynomials with
only even degrees of x’s? only odd degrees of x’s?

6. The Fundamental Theorem of Algebra
(Answers to the odd problems) Exercise A.
1. (x – 1)(x2 + x +1), x = 1, (–1)2/3, –√–1
3. (2x + 3)(4x2 – 6x + 9), x = - , (1 – √3 i), (1 + √3 i)
5. (x – 2)(x + 2)(x2 + 4), x = – 2, 2, –2i, 2i
11. (x2 +1)(x2 + √3x + 1)(x2 – √3x + 1), x = – i, i, –(–1), –√–1, √–1
7. (x2 + 1)(x2 – 2), x = –√2, √2, – i, i
9. (x2 + 2)(x2 +1), x = –√2 i, √2 i, – i, i
Exercise B.
1. (x2 – 2x + 2) 3. (x3 – 4x2 + 14x – 20)
5. – (x4 + 2x3 + 2x2 + 10x + 25)
3
3
2
3
4
3
4
5/6 6 6
5
2
4
9
3
40
7. x4 – x2
9. (–1/10)(x + 1)(x – 2)(x2 + 4x + 5) 11. (x4 + 5x3 + 11x2 + 7x)1
24