The factor theorem has a number of fairly obvious but very useful consequences, which are presented here as six successive theorems.
A. Several Distinct Zeroes: Suppose that several distinct zeroes of a polynomial have been found, probably using test substitutions into the polynomial.
14 DISTINCT ZEROES: Suppose that α1, α2, . . . αs are distinct zeroes of a polyno-mialP (x). Then (x − α1)(x − α2)· · · (x − αs) is a factor ofP (x).
r
156 CHAPTER4: Polynomial Functions CAMBRIDGEMATHEMATICS3 UNITYEAR12r
Proof: Since α1 is a zero,x − α1 is a factor, andP (x) = (x − α1)p1(x).
SinceP (α2) = 0 but α2− α1 = 0, p1(α2) must be zero.
Hencex − α2 is a factor ofp1(x), and p1(x) = (x − α2)p2(x), and thusP (x) = (x − α1)(x − α1)p2(x).
Continuing similarly fors steps, (x − α1)(x − α2)· · · (x − αs) is a factor of P (x).
B. All Distinct Zeroes: If n distinct zeroes of a polynomial of degree n can be found, then the factorisation is complete, and the polynomial is the product of distinct linear factors.
15
ALL DISTINCT ZEROES: Suppose that α1,α2,. . . αn aren distinct zeroes of a poly-nomialP (x) of degree n. Then
P (x) = a(x − α1)(x − α2)· · · (x − αn), wherea is the leading coefficient of P (x).
Proof: By the previous theorem, (x − α1)(x − α2)· · · (x − αn) is a factor of P (x), soP (x) = (x − α1)(x − α2)· · · (x − αn)Q(x), for some polynomial Q(x).
ButP (x) and (x − α1)(x − α2)· · · (x − αn) both have degree n, so Q(x) is a constant.
Equating coefficients ofxn, the constantQ(x) must be the leading coefficient.
Factoring Polynomials — Finding Several Zeroes First: If we can find more than one zero of a polynomial, then we have found a quadratic or cubic factor, and the long divisions required can be reduced or even avoided completely.
16
FACTORING POLYNOMIALS — FINDING SEVERAL ZEROES FIRST:
• Use trial and error to find as many integer zeroes of P (x) as possible.
• Using long division, divide P (x) by the product of the known factors.
If the coefficients of P (x) are all integers, then any integer zero of P (x) must be one of the divisors of the constant term.
When this procedure is applied to the polynomial factored in the previous section, one rather than two long divisions is required.
WORKEDEXERCISE: FactorP (x) = x4+x3− 9x2+ 11x − 4 completely.
SOLUTION: As before, all the coefficients are integers, so the only integer zeroes are the divisors of the constant term−4, that is 1, 2, 4, −1, −2 and −4.
P (1) = 1 + 1 − 9 + 11 − 4 = 0, so x − 1 is a factor.
P (−4) = 256 − 64 − 144 − 44 − 4 = 0, so x + 4 is a factor.
After long division by (x − 1)(x + 4) = x2+ 3x − 4 (omitted), P (x) = (x2+ 3x − 4)(x2− 2x + 1).
Factoring the quadratic, P (x) = (x − 1)(x − 4) × (x − 1)2
= (x − 1)3(x + 4).
Note: The methods of the next section will allow this particular factoring to be done with no long divisions. The following worked exercise involves a polynomial that factors into distinct linear factors, so that nothing more than the factor theorem is required to complete the task.
WORKEDEXERCISE: FactorP (x) = x4− x3− 7x2+x + 6 completely.
r
CHAPTER4: Polynomial Functions 4E Consequences of the Factor Theorem 157r
SOLUTION: The divisors of the constant term 6 are 1, 2, 3, 6,−1, −2, −3 and −6.
P (1) = 1 − 1 − 7 + 1 + 6 = 0, so x − 1 is a factor.
P (−1) = 1 + 1 − 7 − 1 + 6 = 0, so x + 1 is a factor.
P (2) = 16 − 8 − 28 + 2 + 6 = −12 = 0, so x − 2 is not a factor.
P (−2) = 16 + 8 − 28 − 2 + 6 = 0, so x + 2 is a factor.
P (3) = 81 − 27 − 63 + 3 + 6 = 0, so x − 3 is a factor.
We now have four distinct zeroes of a polynomial of degree 4.
Hence P (x) = (x − 1)(x + 1)(x + 2)(x − 3) (notice that P (x) is monic).
C. The Maximum Number of Zeroes: If a polynomial of degree n were to have n + 1 zeroes, then by the first theorem above, it would be divisible by a polynomial of degree n + 1, which is impossible.
17 MAXIMUM NUMBER OF ZEROES: A polynomial of degree n has at most n zeroes.
D. A Vanishing Condition: The previous theorem translates easily into a condition for a polynomial to be the zero polynomial.
18
AVANISHING CONDITION: Suppose thatP (x) is a polynomial in which no terms have degree more than n, yet which is zero for at least n + 1 distinct values of x.
Then P (x) is the zero polynomial.
Proof: Suppose thatP (x) had a degree. This degree must be at most n since there is no term of degree more thann. But the degree must also be at least n+1 since there aren+1 distinct zeroes. This is a contradiction, so P (x) has no degree, and is therefore the zero polynomial.
Note: Once again, the zero polynomialZ(x) = 0 is seen to be quite different in nature from all other polynomials. It is the only polynomial with an infinite number of zeroes; in fact every real number is a zero of Z(x). Associated with this is the fact that x − α is a factor of Z(x) for all real values of α, since Z(x) = (x−α)Z(x) (which is trivially true, because both sides are zero for all x).
It is no wonder then that the zero polynomial does not have a degree.
E. A Condition for Two Polynomials to be Identically Equal: A most important conse-quence of this last theorem is a condition for two polynomialsP (x) and Q(x) to be identically equal — written asP (x) ≡ Q(x), and meaning that P (x) = Q(x) for all values ofx.
19
AN IDENTICALLY EQUAL CONDITION: Suppose that P (x) and Q(x) are polynomials of degree n which have the same values for at least n + 1 values of x. Then the polynomials P (x) and Q(x) are identically equal, written as P (x) ≡ Q(x), that is, they are equal for all values of x.
Proof: Let F (x) = P (x) − Q(x).
Since F (x) is zero whenever P (x) and Q(x) have the same value, it follows thatF (x) is zero for at least n + 1 values of x,
so by the previous theorem,F (x) is the zero polynomial, and P (x) ≡ Q(x).
r
158 CHAPTER4: Polynomial Functions CAMBRIDGEMATHEMATICS3 UNITYEAR12r
WORKEDEXERCISE: Finda, b, c and d, if x3−x = a(x−2)3+b(x−2)2+c(x−2)+d for at least four values ofx.
SOLUTION: Since they are equal for four values ofx, they are identically equal.
Substitutingx = 2, 6 =d.
Equating coefficients ofx3, 1 =a.
Substitutingx = 0, 0 =−8 + 4b − 2c + 6 2b − c = 1.
Substitutingx = 1, 0 =−1 + b − c + 6 b − c = −5.
Henceb = 6 and c = 11.
F. Geometrical Implications of the Factor Theorem: Here are some of the geometrical versions of the factor theorem — they are translations of the consequences given above into the language of coordinate geometry. They are simply generalisations of the similar remarks about the graphs of quadratics in Box 25 of Section 8I of the Year 11 volume.
20
GEOMETRICAL IMPLICATIONS OF THE FACTOR THEOREM:
1. The graph of a polynomial function of degree n is completely determined by any n + 1 points on the curve.
2. The graphs of two distinct polynomial functions cannot intersect in more points than the maximum of the two degrees.
3. A line cannot intersect the graph of a polynomial of degree n in more than n points.
In parts (2) and (3), points where the two curves are tangent to each other count according to their multiplicity.
WORKEDEXERCISE: By factoring the differenceF (x) = P (x) − Q(x), describe the intersections between the curvesP (x) = x4+ 4x3+ 2 andQ(x) = x4+ 3x3+ 3x, and find whereP (x) is above Q(x).
SOLUTION: Subtracting, F (x) = x3− 3x + 2.
Substituting, F (1) = 1 − 3 + 2 = 0, so x − 1 is a factor.
F (−2) = −8 + 6 + 2 = 0, so x + 2 is a factor.
After long division by (x − 1)(x + 2) = x2+x − 2, F (x) = (x − 1)2(x + 2).
Hencey = P (x) and y = Q(x) are tangent at x = 1, but do not cross there, and intersect also atx = −2, where they cross at an angle.
SinceF (x) is positive for −2 < x < 1 or x > 1, and negative for x < −2, P (x) is above Q(x) for −2 < x < 1 or x > 1, and below it for x < −2.
A note for 4 Unit students: The fundamental theorem of algebra is stated, but cannot be proven, in the 4 Unit course. It tells us that the graph of a polynomial of degreen intersects every line in exactly n points, provided first that points where the curves are tangent are counted according to their multiplicity, and secondly that complex points of intersection are also counted. As its name implies, this most important theorem provides the fundamental link between the algebra of polynomials and the geometry of their graphs, and allows the
r
CHAPTER4: Polynomial Functions 4E Consequences of the Factor Theorem 159r
degree of a polynomial to be defined either algebraically, as the highest index, or geometrically, as the number of times every line crosses it.
G. Behaviour at Simple and Multiple Zeroes — A Proof: We can now give a satisfactory proof of the theorem stated in Box 7 of Section 4B:
‘Suppose thatx = α is a zero of multiplicity m ≥ 1 of a polynomial P (x).
• If x = α has even multiplicity, the curve is tangent to the x-axis at x = α, and does not cross the x-axis there.
• If x = α has odd multiplicity at least 3, the curve is tangent to the x-axis at x = α, and crosses the x-axis there at a point of inflexion.
• If x = α is a simple zero, then the curve crosses the x-axis at x = α and is not tangent to thex-axis there.’
Proof: [The proof here is more suited to those taking the 4 Unit course.]
A. Differentiation is required since tangents are involved.
Let P (x) = (x − α)mQ(x), where Q(α) = 0, that is, where (x − α) is not a factor of Q(x).
Using the product rule, P(x) = m(x − α)m −1Q(x) + (x − α)mQ(x)
= (x − α)m −1
mQ(x) + (x − α)Q(x) . Whenm = 1, P(α) = Q(α), which is not zero since Q(α) = 0, but whenm ≥ 2, P(α) = 0.
Hencex = α is a zero of P(x) if and only if m ≥ 2,
That is, the curve is tangent to thex-axis at x = α if and only if m ≥ 2.
B. Since Q(α) = 0, P (x) = (x − α)mQ(x) will change sign around x = α when m is odd, and will not change sign around x = α when m is even. This completes the proof.
Exercise 4E
1. Use the factor theorem to write down in factored form:
(a) a monic cubic polynomial with zeroes−1, 3 and 4.
(b) a monic quartic polynomial with zeroes 0,−2, 3 and 1.
(c) a cubic polynomial with leading coefficient 6 and zeroes at 13,−12 and 1.
2. (a) Show that 2 and 5 are zeroes ofP (x) = x4− 3x3− 15x2+ 19x + 30.
(b) Hence explain why (x − 2)(x − 5) is a factor of P (x).
(c) DivideP (x) by (x − 2)(x − 5) and hence express P (x) as the product of four linear factors.
3. Use trial and error to find as many integer zeroes ofP (x) as possible. Use long division to divide P (x) by the product of the known factors and hence express P (x) in factored form.
(a) P (x) = 2x4− 5x3− 5x2+ 5x + 3 (b) P (x) = 2x4− 5x3− 5x2+ 20x − 12
(c) P (x) = 6x4− 25x3+ 17x2+ 28x − 20 (d) P (x) = 9x4− 51x3+ 85x2− 41x + 6 4. (a) The polynomial (a − 2)x2+ (1− 3b)x + (5 − 2c) has three zeroes. What are the values
ofa, b and c?
(b) The polynomial (a + 1)x3+ (b − 3)x2 + (2c − 1)x + (5 − 4d) has four zeroes. What are the values of a, b, c and d?
r
160 CHAPTER4: Polynomial Functions CAMBRIDGEMATHEMATICS3 UNITYEAR12r
D E V E L O P M E N T
5. (a) If 3x2− 4x + 7 ≡ a(x + 2)2+b(x + 2) + c, find a, b and c.
(b) If 2x3− 8x2+ 3x − 4 ≡ a(x − 1)3+b(x − 1)2+c(x − 1) + d, find a, b, c and d.
(c) Use similar methods to express x3+ 2x2− 3x + 1 as a polynomial in (x + 1).
(d) If the polynomials 2x2+ 4x + 4 and a(x + 1)2+b(x + 2)2 +c(x + 3)2 are equal for three values of x, find a, b and c.
6. (a) A polynomial of degree 3 has a double zero at 2. Whenx = 1 it takes the value 6 and when x = 3 it takes the value 8. Find the polynomial.
(b) Two zeroes of a polynomial of degree 3 are 1 and −3. When x = 2 it takes the value
−15 and when x = −1 it takes the value 36. Find the polynomial.
7. Show thatx2− 3x + 2 is a factor of P (x) = xn(2m − 1) + xm(1− 2n) + (2n− 2m), where m and n are positive integers.
8. If two polynomials have degrees m and n respectively, what is the maximum number of points that their graphs can have in common?
9. Explain why a cubic with three distinct zeroes must have two turning points.
10. The line y = k meets the curve y = ax3+bx2+cx + d four times. Find the values of a, b, c, d and k.
11. Find the turning points of the following polynomials and hence state how many zeroes they have.
(a) 12x − x3+ 3 (b) 7 + 5x − x2− x3
(c) 4x3− x4− 1 (d) 3x4− 4x3+ 2
12. By factoring the difference F (x) = P (x) − Q(x), describe the intersections between the curvesP (x) and Q(x).
(a) P (x) = 2x3− 4x2+ 3x + 1, Q(x) = x3+x2− 8 (b) P (x) = x4+x3+ 10x − 4, Q(x) = x4+ 7x2− 6x + 8
(c) P (x) = −2x3+ 3x2− 25, Q(x) = −3x3− x2+ 11x + 5 (d) P (x) = x4− 3x2− 2, Q(x) = x3− 5x
(e) P (x) = x4+ 4x3− x + 5, Q(x) = x3− 3x2− 2x + 5
13. Suppose that the polynomial equation P (x) = 0 has a double root at x = α. That is, P (x) = (x − α)2Q(x), for some polynomial Q(x), where Q(α) = 0.
(a) Find P(x) and show that P(α) = 0.
(b) If x = 1 is a double root of x4+ax3+bx2− 5x + 1 = 0, find a and b.
(c) Given that P (x) = 2x4− 20x3+ 74x2− 120x + 72, find P(x) and hence show that 2 and 3 are double roots of P (x). Factor P (x) completely.
E X T E N S I O N
14. Show that if the polynomialsx3+ax2− x + b and x3+bx2− x + a have a common factor of degree 2, thena + b = 0.
15. [Wallis’ product and sine as an infinite product] This extraordinary expression of sinπx as an infinite product result is not possible for us to prove rigorously:
sinπx = πx
1−x2
12
1−x2
22
1− x2
32
1−x2
42
· · · .
r
CHAPTER4: Polynomial Functions 4F The Zeroes and the Coefficients 161r
(a) Here is a wildly invalid, but still interesting, justification inspired by the factor theorem for polynomials. First, the function sinπx is zero at every integer value of x, so regarding sinπx as a sort of polynomial of infinite degree, (x − n) must be a factor for all n. Writing this another way, x is a factor, and
1− x2
n2
is a factor for all n. Secondly, the constant multiple π can be justified (invalidly again) because sinπx → πx as x → 0, and each of the other factors on the RHS has limit 1 as x → 0.
(b) By substitutingx = 12 into the expression in part (a), derive from it the identity called Wallis’ product (which is, in contrast, accessible by methods of the 4 Unit course):
π and use a calculator or computer to investigate the speed of convergence.
(c) By other substitutions into part (a), and using part (b), prove that
√2 = 22