1. Polynomial Factor Theorem Polynomial Factor Theorem
Polynomial Factors
Polynomial Factor Theorem
This section explores some of the techniques employed to locate and evaluate the real zeros of polynomial functions. Discussion of complex zeros is left for later sessions on the Fundamental Theorem of Algebra.
This module should take at least four class sessions. Additional examples should be worked out with class participation. The step from concrete examples to the more abstract general case is often difficult for inexperienced student. Having them write down the general case steps and justify each can help to promote understanding of the basic concepts. Working out concrete examples enhances that understanding.
Polynomial Factors
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2. Polynomial Factor Theorem
The Remainder Theorem
Polynomial f(x) divided by x – k
yields a remainder of f(k)
Question:
What if the remainder is 0 ?
Then we know that
f(x) = (x – k)Q(x) + r(x)
The Remainder Theorem
This is the follow-up to the lead-in slides for discussion of the relationship between remainders and functional values. This in turn leads into the Factor Theorem to relate zero remainders and zeros of the polynomial.
When we divide a polynomial f(x) by a linear expression (x – k) the remainder is the value of f(x) at x = k, that is f(k). If f(k) = 0, then k is a zero for f.
The formal statement of this property is called the Factor Theorem.
= (x – k)Q(x)
and thus (x – k) is a factor of f(x)
… leading to …
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Polynomial Factors
Polynomial Factors
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3. Polynomial Factor Theorem
The Factor Theorem
Polynomial f(x) has factor x – k
if and only if
f(k) = 0
Factor Theorem Corollary
If (x – k) is a factor of
f(x) = anxn + an–1xn–1 + • • • + a1x + a0
then k is a factor of a0
The Remainder Theorem
This is the follow-up to the lead-in slides for discussion of the relationship between remainders and functional values. This in turn leads into the Factor Theorem to relate zero remainders and zeros of the polynomial.
When we divide a polynomial f(x) by a linear expression (x – k) the remainder is the value of f(x) at x = k, that is f(k). If f(k) = 0, then k is a zero for f.
The formal statement of this property is called the Factor Theorem.
Question:
What about the converse ?
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Polynomial Factors
Polynomial Factors
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4. Polynomial Factor Theorem
Example: f(x) = x4 + x3 – 19x2 + 11x + 30
Try factor k = 2
Try factor k = 6 2 – 6 – 2 6
–26
–30 6 42
138
894 1 3
–13
–15 1 7 23
149
924
≠ 0
(x – 2) is a factor of f(x)
(x – 6) NOT a factor of f(x)
Factor Theorem Corollary
This is a follow-up to the Factor Theorem and will help us to try only those values of k that have a chance of producing a factor of the polynomial f(x).
To see why this works, assume (x – k) is a factor of f(x) so that we can write
f(x) = (x – k)Q1(x) If f(x) = anxn + ∙∙∙ a1x + a0 and Q1(x) = bn–1 xn–1 + ∙∙∙ + b1 x + b0 then
f(x) = (x – k)Q1(x) = xQ1(x) – kQ1(x)
= [xQ1(x) – k(bn–1 xn–1 + ∙∙∙ + b1x)] – kb0
= [xQ1(x) – kQ2(x)] – kb0
= [anxn + ∙∙∙ a1x] + a0 = f(x)
Since polynomials are equal if and only if corresponding terms are equal, we must have a0 = – kb0 = k(– b0). Thus k is a factor of a0.
Thus, (x – k) being a factor of f(x) implies that k is a factor of the constant term of f(x). The example shows the converse is false, that is, if k is a factor of the constant term a0 then (x – k) is not necessarily a factor of f(x).
Checking out the possible factors of 30 in the example (all 14 of them), we see that k = –5, k = –1, k = 2, and k = 3 are all factors of 30 and corresponding linear factors (k + 5), (k + 1), (k – 2), and (k – 3) are factors of f(x). Note that k being a factor of the constant term of f(x) is necessary, but not sufficient, for (x – k) to be a factor of f(x). For example, in the illustration, k = 6 is a factor of 30 but (k – 6) is not a factor of f(x).
... and 2 is a factor of 30
... BUT 6 IS a factor of 30
Question:
What about other factors of 30 ?
What if
k = ±1, –2, ±3, ±5, –6, ±10, ±15 ?
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Polynomial Factors
Polynomial Factors
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5. Polynomial Factor Theorem
Factor Theorem Example
Given the graph of polynomial f(x)
Estimate the degree of f(x)
Even or odd degree ? x y
(2, 0)
(–7, 0)
(11, 0) ● ● ●
Odd degree !
WHY ?
Factor Theorem Example
Here we connect the graph of a polynomial with its zeros and factors. It is pointed out that simply knowing the zeros does not necessarily yield all there is to know about the function.
For example, what is the ultimate direction of the graph as x grows without bound in both directions? Where are the local extrema for the function? What about turning points? Is there a constant factor Q3(x) ?
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Polynomial Factors
Polynomial Factors
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6. Polynomial Factor Theorem
Factor Theorem Example
Given the graph of polynomial f(x)
Find the factors of f(x)
Since f(–7) = f(2) = f(11) = 0
… then factors are
(x + 7), (x – 2), (x – 11)
Note:
(x + 7)(x – 2)(x – 11) = x3 – 6x2 – 69x + 154 x y ●
(–7, 0)
(2, 0)
(11, 0)
Factor Theorem Example
Here we connect the graph of a polynomial with its zeros and factors. It is pointed out that simply knowing the zeros does not necessarily yield all there is to know about the function.
For example, what is the ultimate direction of the graph as x grows without bound in both directions? Where are the local extrema for the function? What about turning points? Is there a constant factor Q3(x) ?
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Polynomial Factors
Polynomial Factors
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7. Polynomial Factor Theorem
Factor Theorem Example
The graph of polynomial f(x)
Question:
Is f(x) = (x + 7)(x – 2)(x – 11) ? x y ●
(–7, 0)
(2, 0)
(11, 0)
Not necessarily !
Factor Theorem Example
Here we connect the graph of a polynomial with its zeros and factors. It is pointed out that simply knowing the zeros does not necessarily yield all there is to know about the function.
For example, what is the ultimate direction of the graph as x grows without bound in both directions? Where are the local extrema for the function? What about turning points? Is there a constant factor Q3(x) ?
f(x) = (x + 7)Q1(x)
= (x + 7)(x – 2)Q2(x)
= (x + 7)(x – 2)(x – 1)Q3(x)
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Polynomial Factors
Polynomial Factors
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8. Polynomial Factor Theorem
Factor Theorem Example
The graph of polynomial f(x)
Question: x y ●
(–7, 0)
(2, 0)
(11, 0)
f(x)
= (x + 7)(x – 2)(x – 1)Q3(x)
What is Q3(x) ?
Is Q3(x) constant ?
Factor Theorem Example
Here we connect the graph of a polynomial with its zeros and factors. It is pointed out that simply knowing the zeros does not necessarily yield all there is to know about the function.
For example, what is the ultimate direction of the graph as x grows without bound in both directions? Where are the local extrema for the function? What about turning points? Is there a constant factor Q3(x) ?
Does Q3(x) have factors x – k ?
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Polynomial Factors
Polynomial Factors
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9. Polynomial Factor Theorem
Factor Theorem Example
Given the graph of polynomial f(x)
estimate the degree of f(x)
Even or odd degree ?
Find the factors of f(x) x y
WHY ! ● ●
(–3, 0)
(5, 0)
Factor Theorem Example
In this example we amplify what we learned in the last example. In this case there are only two zeros but more factors. How can we tell? The graph gives us clues.
We know, for example, that the polynomial has even degree, since the end behavior of the graph shows both ends growing in the same direction (positive in this case). We can tell from the shape of the graph that the degree is not 2, that is, it is not a parabola. So it must 4, 6 or higher.
This means that the zero at –3 must produce at least two more factors. So, (x + 3)3 and (x – 5) are factors. If by some means we know that the degree of f(x) is 4, then we now have all the factors. The only thing missing is the lead coefficient.
f(–3) = f(5) = 0 so
(x + 3)
and
(x – 5)
are factors
Note:
(x + 3)(x – 5) = x2 – 2x2 – 15
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Polynomial Factors
Polynomial Factors
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10. Polynomial Factor Theoremx y
(–3, 0)
(5, 0) ●
Factor Theorem Example
Given the graph of f(x)
(x + 3)(x – 5) = x2 – 2x2 – 15
Question:
Is f(x) equal to x2 – 2x – 15 ?
Probably not !
What is the graph of x2 – 2x – 15 ?
Factor Theorem Example
In this example we amplify what we learned in the last example. In this case there are only two zeros but more factors. How can we tell? The graph gives us clues.
We know, for example, that the polynomial has even degree, since the end behavior of the graph shows both ends growing in the same direction (positive in this case). We can tell from the shape of the graph that the degree is not 2, that is, it is not a parabola. So it must 4, 6 or higher.
This means that the zero at –3 must produce at least two more factors. So, (x + 3)3 and (x – 5) are factors. If by some means we know that the degree of f(x) is 4, then we now have all the factors. The only thing missing is the lead coefficient.
Note:
f(x) = (x + 3)Q1(x) = (x + 3)(x – 5)Q2(x)
Does Q2(x) have factors (x – k) ?
Which ones and how many ?
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Polynomial Factors
Polynomial Factors
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11. Intercepts, Zeros and Factors
Polynomial Factor Theorem
These things are related
If f(k) = 0, then point (k, 0) on the graph of f is an x-intercept
The number k is a zero for f(x), i.e. f(k) = 0
(x – k) is a factor of f(x)
The number k is a factor of the constant term of f(x)
Intercepts, Zeros and Factors
An intercept is a point on the graph of the function f(x), which lies on an axis. So there are vertical intercepts (sometimes called the y-intercept) and horizontal intercepts (sometimes called x-intercepts). The point (0, y) = (0, f(0)) is the y-intercept, and clearly, there is only one such intercept for a function. An x-intercept is a point (k, 0) where the graph intersects the x-axis. Clearly there can be multiple x-intercepts.
The x-intercepts correspond exactly to the zeros of the function, that is, values such as x = k where f(k) = 0. The number k is the zero for f. If the polynomial function f(x) has a factor (x – k), then k is a zero of f(x), and conversely, if k is a zero of f(x) then (x – k) is a factor of f(x).
Thus, x-intercepts for the graph of f(x), zeros of f(x), and factors of f(x) are all related.
Recall from the Factor Theorem and the Factor Theorem Corollary that f(k) = 0 implies that (x – k) is a factor of f(x) and k is a factor of the constant term of f(x). To see this, consider the following, where Q1(x) and Q2(x) are derived by dividing f(x) by (x – k) and isolating the last term of –kQ1(x), respectively:
f(x) = [anxn + ∙∙∙ + a1x] + a0 = (x – k)Q1(x) = xQ1(x) – kQ1(x) = x(bn–1 xn–1 + ∙∙∙ + b1x + b0) – k(bn–1 xn–1 + ∙∙∙ + b1x) – k( b0)
= [xQ1(x) – kQ2(x)] – kb0
Since polynomials are equal if and only if corresponding terms are equal, we must have a0 = –kb0 = k(–b0). Thus k is a factor of a0.
WHY ?
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Polynomial Factors
Polynomial Factors
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12. Completely Factored Polynomials
Polynomial Factor Theorem
Example:
Completely Factor
f(x) = 2x4 + 14x3 + 18x2 – 54x – 108
Now use synthetic division to check out zeros
= 2(x4 + 7x3 + 9x2 – 27x – 54)
Completely Factored Polynomials
We show how the lead coefficient fits in with all of the factors to produce a completely factored polynomial.
A polynomial is completely factored if and only if it can be written as the product of a constant and one or more linear factors. This includes the possibility that two or more of zeros of the polynomial are complex numbers (which occur in complex conjugate pairs). If we insist that only real numbers be considered, then the pairs of complex conjugates form irreducible quadratic factors. By irreducible quadratics we mean quadratic factors that cannot be factored using only real numbers, e.g. x2 + x + 1.
We first factor out the constant 2 from every term, leaving the lead coefficient as 1. Then, in the illustration, synthetic division is used to show that (x + 3) is a factor of the polynomial f(x), that is, that -3 is a zero for f(x), leaving a lower degree polynomial Q1. The Q1 is then divided by (x + 3) again, showing the there are two factors of (x + 3) with a remaining polynomial Q2 of lower degree than Q1. The process is repeated leaving finally a linear polynomial (x – 2) as the last factor.
We note that (x + 3) appears as a factor three times, i.e. (x + 3)3 is a factor of f(x). This is called a zero of multiplicity 3. Having already factored the constant, we can then write down the polynomial in completely factored form.
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Polynomial Factors
Polynomial Factors
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13. Completely Factored Polynomials
Polynomial Factor Theorem
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)
x – k = x – (–3) –3
–27 –54 –3
–12 9 54 1 4 –3
–18
(x + 3) is a factor
Completely Factored Polynomials
We show how the lead coefficient fits in with all of the factors to produce a completely factored polynomial.
A polynomial is completely factored if and only if it can be written as the product of a constant and one or more linear factors. This includes the possibility that two or more of zeros of the polynomial are complex numbers (which occur in complex conjugate pairs). If we insist that only real numbers be considered, then the pairs of complex conjugates form irreducible quadratic factors. By irreducible quadratics we mean quadratic factors that cannot be factored using only real numbers, e.g. x2 + x + 1.
We first factor out the constant 2 from every term, leaving the lead coefficient as 1. Then, in the illustration, synthetic division is used to show that (x + 3) is a factor of the polynomial f(x), that is, that -3 is a zero for f(x), leaving a lower degree polynomial Q1. The Q1 is then divided by (x + 3) again, showing the there are two factors of (x + 3) with a remaining polynomial Q2 of lower degree than Q1. The process is repeated leaving finally a linear polynomial (x – 2) as the last factor.
We note that (x + 3) appears as a factor three times, i.e. (x + 3)3 is a factor of f(x). This is called a zero of multiplicity 3. Having already factored the constant, we can then write down the polynomial in completely factored form. –3
–3 –18 Q1 –3 –3 18
(x + 3) is a second factor 1 1 –6
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Polynomial Factors
Polynomial Factors
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14. Completely Factored Polynomials
Polynomial Factor Theorem
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)
–27 –54 1 4
–18 –3
(x + 3) is a factor 1 –6
(x + 3) is a second factor
2(x + 3)3(x – 2) –3 –6 Q2 –3 6
(x + 3) is a third factor 1 –2
Completely Factored Polynomials
We show how the lead coefficient fits in with all of the factors to produce a completely factored polynomial.
A polynomial is completely factored if and only if it can be written as the product of a constant and one or more linear factors. This includes the possibility that two or more of zeros of the polynomial are complex numbers (which occur in complex conjugate pairs). If we insist that only real numbers be considered, then the pairs of complex conjugates form irreducible quadratic factors. By irreducible quadratics we mean quadratic factors that cannot be factored using only real numbers, e.g. x2 + x + 1.
We first factor out the constant 2 from every term, leaving the lead coefficient as 1. Then, in the illustration, synthetic division is used to show that (x + 3) is a factor of the polynomial f(x), that is, that -3 is a zero for f(x), leaving a lower degree polynomial Q1. The Q1 is then divided by (x + 3) again, showing the there are two factors of (x + 3) with a remaining polynomial Q2 of lower degree than Q1. The process is repeated leaving finally a linear polynomial (x – 2) as the last factor.
We note that (x + 3) appears as a factor three times, i.e. (x + 3)3 is a factor of f(x). This is called a zero of multiplicity 3. Having already factored the constant, we can then write down the polynomial in completely factored form.
(x – 2) is a fourth factor
Note:
x = –3 is a repeated zero of multiplicity 3
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Polynomial Factors
Polynomial Factors
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15. Completely Factored Polynomials
Polynomial Factor Theorem
f(x) = 2(x4 + 7x3 + 9x2 – 27x – 54)
x – k = x – (–3) –3
–27 –54 –3
–12 9 54 1 4 –3
–18
(x + 3) is a factor –3
–3 –18 Q1 –3 –3 18
2(x + 3)3(x – 2)
(x + 3) is a second factor 1 1 –6
Completely Factored Polynomials
We show how the lead coefficient fits in with all of the factors to produce a completely factored polynomial.
A polynomial is completely factored if and only if it can be written as the product of a constant and one or more linear factors. This includes the possibility that two or more of zeros of the polynomial are complex numbers (which occur in complex conjugate pairs). If we insist that only real numbers be considered, then the pairs of complex conjugates form irreducible quadratic factors. By irreducible quadratics we mean quadratic factors that cannot be factored using only real numbers, e.g. x2 + x + 1.
We first factor out the constant 2 from every term, leaving the lead coefficient as 1. Then, in the illustration, synthetic division is used to show that (x + 3) is a factor of the polynomial f(x), that is, that -3 is a zero for f(x), leaving a lower degree polynomial Q1. The Q1 is then divided by (x + 3) again, showing the there are two factors of (x + 3) with a remaining polynomial Q2 of lower degree than Q1. The process is repeated leaving finally a linear polynomial (x – 2) as the last factor.
We note that (x + 3) appears as a factor three times, i.e. (x + 3)3 is a factor of f(x). This is called a zero of multiplicity 3. Having already factored the constant, we can then write down the polynomial in completely factored form.
Note: –3 –6 Q2
x = –3 is a repeated zero of multiplicity 3 –3 6
(x + 3) is a third factor 1 –2
(x – 2) is a fourth factor
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Polynomial Factors
Polynomial Factors
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16. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring with Multiple Zeros
General Form
f(x) = anxn + an–1 xn– a1x + a0
= an(x – kn)(x – kn–1 ) ... (x – k1)
Some of the zeros ki may be repeated
Complete Factoring and Multiple Zeros
Here we connect all the factors, including those with multiplicity greater than 1, with a lead coefficient to produce a polynomial in completely factored form.
As the illustration shows, the numbers of zeros for f(x), counting zeros with multiplicity greater than 1, is always n, the degree of the polynomial. The number of real zeros, that is zeros that are real numbers, is always less than or equal to n.
In the example, there is one real zero at x = 2 (with multiplicity 1) and one real zero at x = -3, with multiplicity 3. The total of multiplicities is 4, exactly matching the degree of the polynomial.
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Polynomial Factors
Polynomial Factors
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17. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring with Multiple Zeros
f(x) = an(x – kn)(x – kn–1 ) ... (x – k1)
Degree of f(x) = n
Number of zeros is m, m ≤ n
Real zeros occur at x-intercepts
Counting multiplicities, total number of zeros is n
Complete Factoring and Multiple Zeros
Here we connect all the factors, including those with multiplicity greater than 1, with a lead coefficient to produce a polynomial in completely factored form.
As the illustration shows, the numbers of zeros for f(x), counting zeros with multiplicity greater than 1, is always n, the degree of the polynomial. The number of real zeros, that is zeros that are real numbers, is always less than or equal to n.
In the example, there is one real zero at x = 2 (with multiplicity 1) and one real zero at x = -3, with multiplicity 3. The total of multiplicities is 4, exactly matching the degree of the polynomial.
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Polynomial Factors
Polynomial Factors
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18. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring with Multiple Zeros
Example
f(x) = 2(x + 3)3(x – 2)
deg f(x) = 4
One zero at 2, one at –3 of multiplicity 3
Total zeros, counting multiplicities,
1 + 3 = 4 = deg f(x)
Complete Factoring and Multiple Zeros
Here we connect all the factors, including those with multiplicity greater than 1, with a lead coefficient to produce a polynomial in completely factored form.
As the illustration shows, the numbers of zeros for f(x), counting zeros with multiplicity greater than 1, is always n, the degree of the polynomial. The number of real zeros, that is zeros that are real numbers, is always less than or equal to n.
In the example, there is one real zero at x = 2 (with multiplicity 1) and one real zero at x = -3, with multiplicity 3. The total of multiplicities is 4, exactly matching the degree of the polynomial.
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Polynomial Factors
Polynomial Factors
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19. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring Examples
1. f(x) = 5x3 – 10x2 – 15x
= 5x(x2 – 2x – 3)
= 5x(x – 3)(x + 1)
= 5(x + 1)(x – 0)(x – 3)
Zeros of f(x) are –1 , 0 , 3
Complete Factoring Examples
The first example relies merely on the appearance of the polynomial. The first step is factoring the constant 5 from every term. The cubic polynomial that is left has a factor of x in every term, allowing us to factor x from every term and reducing the problem to the simpler one of factoring a quadratic. The quadratic, in turn, is factorable into two distinct linear factors. Thus, the given polynomial has four factors: the constant 5 and the three linear factors (x – 3), (x + 1), and x. Note that x can be written as (x – 0) in the same form as the other linear factors.
The second problem relies on indirect knowledge of the polynomial, that is, the lead coefficient and two zeros, to construct the desired polynomial. From the factor theorem we know that the zeros at -3 and 2 correspond to linear factors (x – (-3)) = (x + 3), and (x – 2). The product of these two factors is a quadratic with lead coefficient 1, that is, x2 + x – 6. Since the lead coefficient is given as the constant 7, we add this as a factor, giving 7x2 + 7x – 42 as the polynomial. This is shown in the illustration in factored form.
Note that in the above, there are two types of negative sign: “–” and “-”. This is intentional, since they have different meanings. We use “–” for the operation of subtraction, and we use “-” to identify the negative of the positive integer 3. Since subtraction is defined as the addition of a negative, we define (x – 3) as (x + (-3)) to simplify the symbols. Technically, it would not be correct to write –3 to represent the negative of 3, since there is not operation intended. In practice, most people ignore this difference, sometimes leading to confusion.
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Polynomial Factors
Polynomial Factors
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20. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring Examples
2. Given:
f(x) is a quadratic polynomial
lead coefficient is 7
f(–3) = 0 and f(2) = 0
Write f(x) in completely factored form
Complete Factoring Examples
The first example relies merely on the appearance of the polynomial. The first step is factoring the constant 5 from every term. The cubic polynomial that is left has a factor of x in every term, allowing us to factor x from every term and reducing the problem to the simpler one of factoring a quadratic. The quadratic, in turn, is factorable into two distinct linear factors. Thus, the given polynomial has four factors: the constant 5 and the three linear factors (x – 3), (x + 1), and x. Note that x can be written as (x – 0) in the same form as the other linear factors.
The second problem relies on indirect knowledge of the polynomial, that is, the lead coefficient and two zeros, to construct the desired polynomial. From the factor theorem we know that the zeros at -3 and 2 correspond to linear factors (x – (-3)) = (x + 3), and (x – 2). The product of these two factors is a quadratic with lead coefficient 1, that is, x2 + x – 6. Since the lead coefficient is given as the constant 7, we add this as a factor, giving 7x2 + 7x – 42 as the polynomial. This is shown in the illustration in factored form.
Note that in the above, there are two types of negative sign: “–” and “-”. This is intentional, since they have different meanings. We use “–” for the operation of subtraction, and we use “-” to identify the negative of the positive integer 3. Since subtraction is defined as the addition of a negative, we define (x – 3) as (x + (-3)) to simplify the symbols. Technically, it would not be correct to write –3 to represent the negative of 3, since there is not operation intended. In practice, most people ignore this difference, sometimes leading to confusion.
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Polynomial Factors
Polynomial Factors
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21. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Complete Factoring Examples
Write f(x) in completely factored form
Note that –3 and 2 are zeros of f(x)
From the Factor Theorem
x –(–3) and x – 2 are factors of f(x)
Thus
f(x) = 7(x + 3)(x – 2)
Complete Factoring Examples
The first example relies merely on the appearance of the polynomial. The first step is factoring the constant 5 from every term. The cubic polynomial that is left has a factor of x in every term, allowing us to factor x from every term and reducing the problem to the simpler one of factoring a quadratic. The quadratic, in turn, is factorable into two distinct linear factors. Thus, the given polynomial has four factors: the constant 5 and the three linear factors (x – 3), (x + 1), and x. Note that x can be written as (x – 0) in the same form as the other linear factors.
The second problem relies on indirect knowledge of the polynomial, that is, the lead coefficient and two zeros, to construct the desired polynomial. From the factor theorem we know that the zeros at -3 and 2 correspond to linear factors (x – (-3)) = (x + 3), and (x – 2). The product of these two factors is a quadratic with lead coefficient 1, that is, x2 + x – 6. Since the lead coefficient is given as the constant 7, we add this as a factor, giving 7x2 + 7x – 42 as the polynomial. This is shown in the illustration in factored form.
Note that in the above, there are two types of negative sign: “–” and “-”. This is intentional, since they have different meanings. We use “–” for the operation of subtraction, and we use “-” to identify the negative of the positive integer 3. Since subtraction is defined as the addition of a negative, we define (x – 3) as (x + (-3)) to simplify the symbols. Technically, it would not be correct to write –3 to represent the negative of 3, since there is not operation intended. In practice, most people ignore this difference, sometimes leading to confusion.
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Polynomial Factors
Polynomial Factors
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22. Polynomial Factor Theorem
Polynomial Functions
Polynomial Factor Theorem
Even/Odd Multiplicity Examples x
y(x) x
y(x) x
y(x)
y = x + 3 ● ● ● ●
y = (x – 3)2
y = (x – 3)3
y = (x – 3)4 x
y(x) x
y(x) x
y(x)
Even/Odd Multiplicities
Multiplicities of zeros for polynomials of even or odd degree give the graphs different behaviors in a small neighborhood of the zero.
In the examples shown from left to right, the zeros of the polynomials have multiplicities of 1, 2, 3, 4, and 5. The last two polynomials each have two distinct zeros, -3 and +3, with multiplicity 3 and 1, respectively, and -2 and +3 with multiplicities 3 and 2, respectively. We can use the multiplicity of each zero to tell something about the behavior of the graph near the zero.
In the cases of odd multiplicity (zeros 1, 3, 5, -2 and -3), as we move away from the zero in both positive and negative directions, the graph rises on one side of the zero and falls on the other. In other words, in any small interval containing the zero, the graph is either increasing or decreasing throughout the interval.
In the cases of even multiplicity (zero at x = 3, multiplicities 2 and 4), as we move away from the zero in both directions, the graph either rises on both sides or falls on both sides. In other words, the zero is a turning point for the graph, so that it changes from increasing to decreasing or from decreasing to increasing at the zero.
These facts are very helpful in sketching the graphs of polynomials to give a general impression of its overall behavior. This is quite often all that is needed in understanding the behavior of the function. ● ● ● ● ●
y = (x – 3)5
y = (x + 3)3(x – 3)
y = (x + 2)3(x – 3)2
7/23/2013
Polynomial Factors
Polynomial Factors
7/23/2013

23. Polynomial Factor Theorem
Think about it !
7/23/2013
Polynomial Factors
Polynomial Factors
7/23/2013