Polynomial Functions
At the end of this lecture, a student must be able to:
Use synthetic division in dividing a polynomial function by a linear function
Relate factoring a polynomial to finding its remainder
Polynomial Functions
Definition
Letn∈N∪ {0}. A polynomial function of degree n is a function of the form
p(x) = anxn+an−1xn−1+an−2xn−2+. . .+a1x+a0
wherea0, a1, ..., an−2, an−1, an∈R, and an 6= 0.
Note:
1. We refer to an as the leading coefficient of p.
Polynomial Functions
Definition
Letn∈N∪ {0}. A polynomial function of degree n is a function of the form
p(x) = anxn+an−1xn−1+an−2xn−2+. . .+a1x+a0
wherea0, a1, ..., an−2, an−1, an∈R, and an 6= 0.
Note:
1. We refer to an as the leading coefficient of p.
Polynomial Functions
Definition
Letn∈N∪ {0}. A polynomial function of degree n is a function of the form
p(x) = anxn+an−1xn−1+an−2xn−2+. . .+a1x+a0
wherea0, a1, ..., an−2, an−1, an∈R, and an 6= 0.
Note:
1. We refer to an as the leading coefficient of p.
Examples:
1. p(x) = 3x− 1
2 is a polynomial function of degree 1 (linear).
2. p(x) = 4x−3x3+√3 is a polynomial function of degree 3 (cubic).
3. p(x) = −3(x2−4)2(x+ 1)3 is a polynomial function of degree 7.
4. p(x) = x12 −
√
Examples:
1. p(x) = 3x− 1
2 is a polynomial function of degree 1 (linear).
2. p(x) = 4x−3x3+√3 is a polynomial function of degree 3 (cubic).
3. p(x) = −3(x2−4)2(x+ 1)3 is a polynomial function of degree 7.
4. p(x) = x12 −
√
Examples:
1. p(x) = 3x− 1
2 is a polynomial function of degree 1 (linear).
2. p(x) = 4x−3x3+√3 is a polynomial function of degree 3 (cubic).
3. p(x) = −3(x2−4)2(x+ 1)3 is a polynomial function of degree
7.
4. p(x) = x12 −
√
Examples:
1. p(x) = 3x− 1
2 is a polynomial function of degree 1 (linear).
2. p(x) = 4x−3x3+√3 is a polynomial function of degree 3 (cubic).
3. p(x) = −3(x2−4)2(x+ 1)3 is a polynomial function of degree 7.
4. p(x) = x12 −
√
Examples:
1. p(x) = 3x− 1
2 is a polynomial function of degree 1 (linear).
2. p(x) = 4x−3x3+√3 is a polynomial function of degree 3 (cubic).
3. p(x) = −3(x2−4)2(x+ 1)3 is a polynomial function of degree 7.
4. p(x) = x12 −
√
Long Division
Use long division to find (2x3 −x2 +x−1)÷(x−2)
2x2 + 3x + 7 x−2
2x3 −x2 +x −1
−2x3+ 4x2 3x2 +x
−3x2 + 6x 7x −1
−7x+ 14 13
Long Division
Use long division to find (2x3 −x2 +x−1)÷(x−2)
2x2 + 3x + 7
x−2 2x3 −x2 +x −1
−2x3+ 4x2 3x2 +x
−3x2 + 6x 7x −1
−7x+ 14 13
Long Division
Use long division to find (2x3 −x2 +x−1)÷(x−2)
2x2 + 3x + 7
x−2 2x3 −x2 +x −1
−2x3+ 4x2 3x2 +x
−3x2 + 6x 7x −1
−7x+ 14 13
Division Algorithm
Theorem (Division Algorithm)
If p(x) is a polynomial and r∈R then there exists a unique polynomial q(x) of degree less than that of p(x) and a unique R ∈R such that
p(x) = (x−r)q(x) +R.
Note:
1. We callq the quotient and R the remainder.
p(x)
x−r =q(x) + R x−r
Division Algorithm
Theorem (Division Algorithm)
If p(x) is a polynomial and r∈R then there exists a unique polynomial q(x) of degree less than that of p(x) and a unique R ∈R such that
p(x) = (x−r)q(x) +R.
Note:
1. We callq the quotient and R the remainder.
p(x)
x−r =q(x) + R x−r
Division Algorithm
Theorem (Division Algorithm)
If p(x) is a polynomial and r∈R then there exists a unique polynomial q(x) of degree less than that of p(x) and a unique R ∈R such that
p(x) = (x−r)q(x) +R.
Note:
1. We callq the quotient and R the remainder.
p(x)
x−r =q(x) + R x−r
Synthetic Division
- a handy method of dividing a polynomial by a binomial of the formx−r, wherer is a constant
Synthetic Division
- a handy method of dividing a polynomial by a binomial of the formx−r, wherer is a constant
Consider the example earlier:
(2x3−x2+x−1)÷(x−2)
• Write the coefficients of p(x) in order of decreasing degree in a horizontal row. On the second row, write r one column to the left of an.
2 −1 1 −1
Consider the example earlier:
(2x3−x2+x−1)÷(x−2)
• Write the coefficients of p(x) in order of decreasing degree in a horizontal row. On the second row, write r one column to the left of an.
2 −1 1 −1
• Writean on the third row, on the same column.
2 −1 1 −1
2
• Multiply an by r and write the product in the row
directly belowan−1.
2 −1 1 −1
2 4
• Add the product andan−1 and write this in the third row, in the same column as an−1.
2 −1 1 −1
2 4
2 3?
• Repeat the process using the last number obtained in the third row as the multiplier of r.
2 −1 1 −1
2 4 6
• Repeat the process using the last number obtained in the third row as the multiplier of r.
2 −1 1 −1
2 4 6
2 3 7?
• Repeat the process using the last number obtained in the third row as the multiplier of r.
2 −1 1 −1
2 4 6 14
• Repeat the process using the last number obtained in the third row as the multiplier of r.
2 −1 1 −1
2 4 6 14
2 3 7 13?
• Starting from the leftmost, the entries in the third row (except the last) are the coefficients of the terms of the quotient in decreasing order of degrees. The last entry in the third row is the remainder.
2 −1 1 −1
2 4 6 14
2 3 7 13?
+
• Starting from the leftmost, the entries in the third row (except the last) are the coefficients of the terms of the quotient in decreasing order of degrees. The last entry in the third row is the remainder.
2 −1 1 −1
2 4 6 14
2 3 7 13?
+
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4 20
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4 20
5 19?
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4 20 76
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4 20 76
5 19 76?
Example: (5x3−x2+ 6)÷(x−4) Solution:
5 −1 0 6
4 20 76 304
Example: (5x3−x2+ 6)÷(x−4)
Solution:
5 −1 0 6
4 20 76 304
5 19 76 310?
+
Example: (5x3−x2+ 6)÷(x−4)
Solution:
5 −1 0 6
4 20 76 304
5 19 76 310?
+
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) + R mx+b
p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+ mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+mb)
To divide p(x) by mx+b:
1. Divide p(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) + R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+ mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+mb)
To divide p(x) by mx+b:
1. Divide p(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) + R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+mb)
To divide p(x) by mx+b:
1. Divide p(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) + R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+ mb)
To divide p(x) by mx+b:
1. Divide p(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose wedivide p(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) +
R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+ mb)
To divide p(x) by mx+b:
1. Divide p(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) + R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) + R (x+ mb)
To divide p(x) by mx+b:
1. Dividep(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x) +
R
mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x) +
R
(x+ mb)
To divide p(x) by mx+b:
1. Dividep(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Suppose we dividep(x) by mx+b, wherem6= 0:
p(x)
mx+b = q(x)+ R mx+b p(x)
m(x+ mb) = q(x) + R
m(x+mb) (factor out m)
m· p(x)
m(x+ mb) = m· q(x) + R m(x+mb )
!
(multiply by m)
p(x)
(x+ mb) = m·q(x)+ R (x+ mb)
To divide p(x) by mx+b:
1. Dividep(x) by x+ mb (in x−r, r=−b m)
2. Remainder is the same
Example: (4x5+x3−2x2+ 1)÷(2x+ 3)
Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3 2
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)(2x4−3x3+5x2−17 2x+
51 4 )+(−
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)
(2x4−3x3+5x2−17 2x+
51 4 )+(−
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)(2x4
−3x3+5x2−17 2x+
51 4 )+(−
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)(2x4−3x3
+5x2−17 2x+
51 4 )+(−
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)(2x4−3x3+5x2−17 2x+
51 4 )
Example: (4x5+x3−2x2+ 1)÷(2x+ 3) Solution:
Divide by x−r where r=−b m:
4 0 1 −2 0 1
− 3
2 −6 9 −15 51
2 − 153
4 4 −6 10 −17 512 − 149
4
Retaining the remainder and dividing the quotient by m:
4x5+x3−2x2+1 = (2x+3)(2x4−3x3+5x2−17 2x+
51 4 )+(−
Definition
A polynomial functionf is said to be a factor of pif p(x) = f(x)g(x) for some polynomial g(x).
Examples:
1. f(x) = 4 and g(x) = x−2 are factors of P(x) = 4x−8.
2. f(x) = x−
√ 2
2 and g(x) =x+ √
2
2 are factors of P(x) = x2−1
Definition
A polynomial functionf is said to be a factor of pif p(x) = f(x)g(x) for some polynomial g(x).
Examples:
1. f(x) = 4 and g(x) = x−2 are factors of P(x) = 4x−8.
2. f(x) =x−
√ 2
2 and g(x) =x+ √
2
2 are factors of P(x) = x2−1
Definition
Given a polynomial p, we say a complex number r is a zero of pif p(r) = 0.
Example: √2 and √−2 are zeros of p(x) = x4−4.
Definition
Given a polynomial p, we say a complex number r is a zero of pif p(r) = 0.
Example: √2 and √−2 are zeros of p(x) = x4−4.
Remainder Theorem
Letq be the quotient and R the remainder when p is divided by x−r,
p(x) =q(x)(x−r) +R.
Then,
p(r) =q(r)(r−r) +R=R.
Theorem (Remainder Theorem)
If p is a polynomial function, then the remainder when p(x)
Remainder Theorem
Letq be the quotient and R the remainder when p is divided by x−r,
p(x) =q(x)(x−r) +R.
Then,
p(r) =
q(r)(r−r) +R=R.
Theorem (Remainder Theorem)
If p is a polynomial function, then the remainder when p(x)
Remainder Theorem
Letq be the quotient and R the remainder when p is divided by x−r,
p(x) =q(x)(x−r) +R.
Then,
p(r) =q(r)(r−r) +R
=R.
Theorem (Remainder Theorem)
If p is a polynomial function, then the remainder when p(x)
Remainder Theorem
Letq be the quotient and R the remainder when p is divided by x−r,
p(x) =q(x)(x−r) +R.
Then,
p(r) =q(r)(r−r) +R=R.
Theorem (Remainder Theorem)
If p is a polynomial function, then the remainder when p(x)
Remainder Theorem
Letq be the quotient and R the remainder when p is divided by x−r,
p(x) =q(x)(x−r) +R.
Then,
p(r) =q(r)(r−r) +R=R.
Theorem (Remainder Theorem)
If p is a polynomial function, then the remainder when p(x)
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 =
16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 =
13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)?
By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Previous Example: (2x3−x2+x−1)÷(x−2) has remainder 13 .
Remainder Theorem: Evaluate the dividend at 2 to get the remainder
2(2)3−(2)2+ 2−1 = 16−4 + 2−1 = 13
Example: What is the remainder when
p(x) = 2x3+ 5x2−2x−1 is divided by (x+ 3)? By the Remainder Theorem, the remainder isp(−3).
p(−3) = 2(−3)3+ 5(−3)2−2(−3)−1 = −54 + 45 + 6−1
Example: Find the value of a so that the remainder is 2 when 5x100+ax80+ 2x25−5 is divided by (x+ 1). Solution:
Letp(x) = 5x100+ax80+ 2x25−5 andr =−1,
2 = 5(−1)100+a(−1)80+ 2(−1)25−5 2 = 5 +a−2−5
2 = −2 +a a = 4
Example: Find the value of a so that the remainder is 2 when 5x100+ax80+ 2x25−5 is divided by (x+ 1). Solution:
Letp(x) = 5x100+ax80+ 2x25−5 andr =−1,
2 = 5(−1)100+a(−1)80+ 2(−1)25−5
2 = 5 +a−2−5 2 = −2 +a a = 4
Example: Find the value of a so that the remainder is 2 when 5x100+ax80+ 2x25−5 is divided by (x+ 1). Solution:
Letp(x) = 5x100+ax80+ 2x25−5 andr =−1,
2 = 5(−1)100+a(−1)80+ 2(−1)25−5 2 = 5 +a−2−5
2 = −2 +a a = 4
Example: Find the value of a so that the remainder is 2 when 5x100+ax80+ 2x25−5 is divided by (x+ 1). Solution:
Letp(x) = 5x100+ax80+ 2x25−5 andr =−1,
2 = 5(−1)100+a(−1)80+ 2(−1)25−5 2 = 5 +a−2−5
2 = −2 +a
a = 4
Example: Find the value of a so that the remainder is 2 when 5x100+ax80+ 2x25−5 is divided by (x+ 1). Solution:
Letp(x) = 5x100+ax80+ 2x25−5 andr =−1,
2 = 5(−1)100+a(−1)80+ 2(−1)25−5 2 = 5 +a−2−5
2 = −2 +a a = 4
Factor Theorem
Supposep(r) = 0.
Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R
p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x),
then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Factor Theorem
Supposep(r) = 0. Then
p(x) = (x−r)q(x) +R p(x) = (x−r)q(x) +p(r)
p(x) = (x−r)q(x)
Therefore,x−r is a factor of p(x).
Conversely, ifx−r is a factor of p(x), then there is a polynomial g(x) such that
p(x) = (x−r)g(x)
⇒p(r) = (r−r)g(r)
⇒p(r) = 0
Theorem (Factor Theorem)
If p is a polynomial function, then
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3 −24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3 −24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30
= 48 + 16−96 + 2 + 30 = 0.
p(−3) = 3(−3)4+ 2(−3)3 −24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3 −24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3 −24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3−24(−3)2+ (−3) + 30
= 243−54−216−3 + 30 = 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3−24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3−24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Verify that 3x4+ 2x3−24x2+x+ 30 hasx−2 and x+ 3 as factors.
Solution:
Letp(x) = 3x4+ 2x3−24x2+x+ 30 and computep(2) and p(−3).
p(2) = 3(2)4+ 2(2)3−24(2)2+ (2) + 30 = 48 + 16−96 + 2 + 30
= 0.
p(−3) = 3(−3)4+ 2(−3)3−24(−3)2+ (−3) + 30 = 243−54−216−3 + 30
= 0.
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k 0 =−8 + 8k−6−k
14 = 7k
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k 0 =−8 + 8k−6−k
14 = 7k
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k
0 =−8 + 8k−6−k 14 = 7k
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k 0 =−8 + 8k−6−k
14 = 7k
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k 0 =−8 + 8k−6−k
14 = 7k
Ex. Find the value of k so that x+ 2 is a factor of x3 + 2kx2+ 3x−k.
Solution:
By the Factor Theorem,P(−2) = 0. Thus,
0 = (−2)3+ 2k(−2)2 + 3(−2)−k 0 =−8 + 8k−6−k
14 = 7k
GOAL:
Theorem (Fundamental Theorem of Algebra)
Every non-constant polynomial with complex coefficients has at least one complex zero.
Consequence:
Theorem (Factored Form of a Polynomial)
If a polynomialp(x) has degree n and leading coefficient an,
then
p(x) = an(x−r1)(x−r2)...(x−rn)
where r1, r2, ..., rn (not necessarily distinct) are the roots
Theorem (Fundamental Theorem of Algebra)
Every non-constant polynomial with complex coefficients has at least one complex zero.
Consequence:
Theorem (Factored Form of a Polynomial)
If a polynomialp(x) has degree n and leading coefficient an,
then
p(x) =an(x−r1)(x−r2)...(x−rn)
where r1, r2, ..., rn (not necessarily distinct) are the roots
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2.
We solve x2−3x−3 2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x= 3±
q
9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9
−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2)
2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3 2 =
x− 3+√15 2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Example: The only solutions to x2+ 1 = 0 are i and −i. Therefore,x2+ 1 = (x+i)(x−i).
Example: Find the factored form ofx2 −3x− 3 2. We solve x2−3x−3
2 = 0.
x=
3±q9−4(1)(−3 2) 2
= 3±
√
15 2
Thus,x2−3x−3
2 = x− 3+√15
2
x−3−√15 2
Multiplicity
Definition
Letp(x) be a polynomial of degreen
and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk. We say that r1 is a zero of p(x) with
multiplicitym1,r2 is a zero ofp(x) with multiplicitym2, ..., rk is a zero ofp(x) with multiplicity mk.
Example:
Multiplicity
Definition
Letp(x) be a polynomial of degreen and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk.
We say that r1 is a zero of p(x) with multiplicitym1,r2 is a zero ofp(x) with multiplicitym2, ..., rk is a zero ofp(x) with multiplicity mk.
Example:
Multiplicity
Definition
Letp(x) be a polynomial of degreen and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk. We say that r1 is a zero of p(x) with
multiplicitym1,
r2 is a zero ofp(x) with multiplicitym2, ..., rk is a zero ofp(x) with multiplicity mk.
Example:
Multiplicity
Definition
Letp(x) be a polynomial of degreen and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk. We say that r1 is a zero of p(x) with
multiplicitym1,r2 is a zero ofp(x) with multiplicity m2,
..., rk is a zero ofp(x) with multiplicity mk.
Example:
Multiplicity
Definition
Letp(x) be a polynomial of degreen and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk. We say that r1 is a zero of p(x) with
multiplicitym1,r2 is a zero ofp(x) with multiplicity m2, ..., rk is a zero ofp(x) with multiplicity mk.
Example:
Multiplicity
Definition
Letp(x) be a polynomial of degreen and suppose we can factor p(x) as
p(x) =an(x−r1)m1(x−r2)m2...(x−rk)mk
for some positive integerk ≤n and distinct complex numbersr1, . . . , rk. We say that r1 is a zero of p(x) with
multiplicitym1,r2 is a zero ofp(x) with multiplicity m2, ..., rk is a zero ofp(x) with multiplicity mk.
Example:
Note:
1. The degree ofp(x) is m1+m2+...+mk.
2. If a root ofp(x) has multiplicity one, we say it is a
simple root. If it has multiplicity two, it is a double root. Triple roots have multiplicity three, and so on. 3. A polynomial can be expressed as a product of linear
factors with complex coefficients.
Theorem
A polynomial of degree n has exactly n complex zeros, counting multiplicities.
Note:
1. The degree ofp(x) is m1+m2+...+mk.
2. If a root ofp(x) has multiplicity one, we say it is a
simple root. If it has multiplicity two, it is a double root. Triple roots have multiplicity three, and so on.
3. A polynomial can be expressed as a product of linear factors with complex coefficients.
Theorem
A polynomial of degree n has exactly n complex zeros, counting multiplicities.
Note:
1. The degree ofp(x) is m1+m2+...+mk.
2. If a root ofp(x) has multiplicity one, we say it is a
simple root. If it has multiplicity two, it is a double root. Triple roots have multiplicity three, and so on. 3. A polynomial can be expressed as a product of linear
factors with complex coefficients.
Theorem
A polynomial of degree n has exactly n complex zeros, counting multiplicities.
Note:
1. The degree ofp(x) is m1+m2+...+mk.
2. If a root ofp(x) has multiplicity one, we say it is a
simple root. If it has multiplicity two, it is a double root. Triple roots have multiplicity three, and so on. 3. A polynomial can be expressed as a product of linear
factors with complex coefficients.
Theorem
A polynomial of degree n has exactly n complex zeros, counting multiplicities.
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x):
x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2,
x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4,
(x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Example: Find a polynomial functionp(x) of least degree having 2 and 4 as simple roots, with−1 being a zero of multiplicity two, and p(1) = 3.
Solution:
Factors ofp(x): x−2, x−4, (x+ 1)2
⇒p(x) = an(x−2)(x−4)(x+1)2
an(1−2)(1−4)(1 + 1)2 = 3
an(−1)(−3)(2)2 = 3
12an= 3
an= 14
p(x) = 1
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4
= −8i+ 4 + 8i−4 = 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp,
−2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p.
Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Complex Zeros
Theorem
If z is a zero of the polynomial function p having real coefficients, then its complex conjugate z is also a zero of p.
Example: Consider p(x) = x3−x2+ 4x−4.
p(2i) = (2i)3−(2i)2+ 4(2i)−4 = −8i+ 4 + 8i−4
= 0.
Since 2i is a zero ofp, −2iis also a zero of p. Indeed,
p(−2i) = (−2i)3−(−2i)2+ 4(−2i)−4 = 8i+ 4−8i−4
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution:
Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and1 +i as a zero of multiplicity 1such that P(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)]
= k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i]
= k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficientshaving 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2]
= k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1))
= k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36 k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1−i)] = k(x−3)2[x−1−i][x−1 +i] = k(x−3)2[(x−1)2−(i)2] = k(x−3)2(x2−2x+ 1−(−1)) = k(x−3)2(x2−2x+ 2)
P(0) = 36
k(−3)2(2) = 36
18k = 36 k = 2
Thus the polynomial function is
Example: Find a polynomial functionP with real
coefficients having 3 as a zero of multipicity 2 and 1 +i as a zero of multiplicity 1 such thatP(0) = 36.
Solution: Since 1 +i is a zero ofP, then so is 1−i.
P(x) = k(x−3)2[x−(1 +i)][x−(1
