Algebra 2

Building a polynomial from its zeros

You know where a polynomial is zero. How do you get the polynomial back, and when is there only one?

You need

Factored formAlgebra 1 Zeros and multiplicityAlgebra 2

Where you are 6 in Polynomial functions

Zeros and multiplicity Degree, end behavior and the shape of a polynomial graph Cubic, quartic and higher degree functions Graphing polynomial functions

Building a polynomial from its zeros

Modeling with polynomial functions

When you reach for it

Graphing a polynomial starts from the rule and draws the curve. This move runs the other way. You know where a polynomial is zero, how many times, and perhaps one other point on its graph, and you want the rule. That happens when you read a formula off a drawn graph, when you design a curve that must meet the axis at chosen places, and whenever a problem says “a cubic with zeros 2, 5 and 7” and the cubic has to be written down.

For a quadratic this is factored form read backwards. Here it works at any degree, with repeated zeros, fractional zeros and non-real zeros.

The procedure

Each zero gives a factor

By the factor theorem, c is a zero of p exactly when (x - c) is a factor of p. A zero of multiplicity m means (x - c)^m is a factor (zeros and multiplicity). Suppose p has degree n and zeros c_1, \dots, c_k with multiplicities m_1, \dots, m_k that add up to n. Then p contains the product

(x - c_1)^{m_1}(x - c_2)^{m_2}\cdots(x - c_k)^{m_k}

and that product already has degree n. Dividing p by it leaves a quotient of degree 0, a constant. So, for some number a \neq 0,

p(x) = a(x - c_1)^{m_1}\cdots(x - c_k)^{m_k}

Each factor carries the opposite sign of its zero: the zero -2 gives (x - (-2)) = (x + 2). The fundamental theorem of algebra says a polynomial of degree n has exactly n zeros counted with multiplicity, complex ones included, so a list whose multiplicities add to n is the whole list.

The leading coefficient is free

Every non-zero a gives a polynomial with the same zeros. Multiplying by a stretches the graph vertically, and flips it when a < 0, but a height of 0 stays 0, so the x-intercepts never move. The zeros alone describe a family of polynomials, one for each a.

Figure 1The family a\,x(x + 3)(x - 2) from Example 1. Every member has the same zeros; only a = -3 passes through (1, 12).

One more point fixes a

Suppose the graph also passes through (x_0, y_0), where x_0 is not a zero. Substituting,

y_0 = a(x_0 - c_1)^{m_1}\cdots(x_0 - c_k)^{m_k}

The product on the right is a number, and it is not 0, because no factor is 0. Dividing by it leaves exactly one value of a. The y-intercept (0, p(0)) is usually the easiest point, since every factor becomes a plain number.

Why the answer is unique. Any polynomial with the given degree and the given zeros and multiplicities has the form a times the product, and the point allows one value of a. So two polynomials that share the degree, the zeros with their multiplicities, and one point that is not a zero, are the same polynomial. Each piece is needed. Without the point, every a works. Without the degree, a factor such as (x^2 + 1) could be multiplied in and a chosen again, giving a second polynomial with the same real zeros through the same point.

Integer coefficients

A fractional zero such as \displaystyle \frac{5}{2} gives the factor \displaystyle \left(x - \frac{5}{2}\right), whose constant is a fraction. Since a is free, choose it to clear the fraction: \displaystyle 2\left(x - \frac{5}{2}\right) = 2x - 5. The zero is unchanged, because 2x - 5 = 0 exactly when \displaystyle x = \frac{5}{2}. In general a zero \displaystyle \frac{r}{s} gives the integer factor (sx - r).

Real coefficients and non-real zeros

If p has real coefficients and z is a non-real zero, then its conjugate \bar{z} is a zero too (conjugate pairs). So when real coefficients are wanted and a zero h + ki is given, h - ki comes with it, unstated. Their two factors multiply to a real quadratic, by the difference of squares and i^2 = -1:

\begin{aligned} &(x - (h + ki))(x - (h - ki)) \\ &\quad = (x - h)^2 - (ki)^2 = (x - h)^2 + k^2 \end{aligned}

Reading zeros off a graph

A graph shows the real zeros as its x-intercepts, and the parity of each multiplicity in what the graph does there: a crossing at a slant suggests multiplicity 1, a touch that turns back 2, a flat crossing 3. Take the smallest multiplicities that fit. They give the polynomial of lowest degree the graph allows, which is the usual question, since a picture cannot tell 2 from 4. Fix a with one more point the graph shows, then check its sign against the end behavior. If the ends of the graph disagree with the sign of a and the degree, a zero or a multiplicity was misread.

Worked examples

Example 1

A family, then one member

Find every polynomial of degree 3 with zeros -3, 0 and 2. Then find the one whose graph passes through (1, 12).

Solution

The zero -3 gives (x + 3), the zero 0 gives (x - 0) = x, and the zero 2 gives (x - 2). Three linear factors already make degree 3, so the only freedom left is a constant a \neq 0:

p(x) = a\,x(x + 3)(x - 2)

Since 0 is a zero, every member passes through (0, 0), so the y-intercept cannot pick one out. Use (1, 12) instead:

12 = a(1)(1 + 3)(1 - 2) = -4a

so a = -3. Expanding, x(x + 3)(x - 2) = x(x^2 + x - 6) = x^3 + x^2 - 6x, and

p(x) = -3x^3 - 3x^2 + 18x

Check. p(1) = -3 - 3 + 18 = 12.

Example 2

A repeated zero and the y-intercept

Find the polynomial of degree 4 with zeros -2, 1 (multiplicity 2) and 4, whose graph passes through (0, 16).

Solution

The double zero gives (x - 1)^2. The multiplicities add to 1 + 2 + 1 = 4, the degree, so

p(x) = a(x + 2)(x - 1)^2(x - 4)

At x = 0 every factor is a number:

16 = a(2)(-1)^2(-4) = -8a

so a = -2 and p(x) = -2(x + 2)(x - 1)^2(x - 4). To expand, pair the factors so they share a piece: (x + 2)(x - 4) = x^2 - 2x - 8 and (x - 1)^2 = x^2 - 2x + 1. With u = x^2 - 2x their product is (u - 8)(u + 1) = u^2 - 7u - 8, so

\begin{aligned} p(x) &= -2(u^2 - 7u - 8) \\ &= -2(x^4 - 4x^3 - 3x^2 + 14x - 8) \\ &= -2x^4 + 8x^3 + 6x^2 - 28x + 16 \end{aligned}

The constant term is 16, as the y-intercept requires. The factored form is usually the more useful answer, because it shows the zeros at a glance.

Example 3

A fractional zero and integer coefficients

Find a polynomial of degree 3 with integer coefficients whose zeros are 3, -1 and \displaystyle \frac{5}{2}.

Solution

The family is \displaystyle a(x - 3)(x + 1)\left(x - \frac{5}{2}\right). With a = 1 the constant term would be \displaystyle (-3)(1)\left(-\frac{5}{2}\right) = \frac{15}{2}, not an integer. Take a = 2 and give the 2 to the fractional factor, \displaystyle 2\left(x - \frac{5}{2}\right) = 2x - 5, which is still zero exactly at \displaystyle x = \frac{5}{2}:

\begin{aligned} p(x) &= (x - 3)(x + 1)(2x - 5) \\ &= (x^2 - 2x - 3)(2x - 5) \\ &= 2x^3 - 9x^2 + 4x + 15 \end{aligned}

Check at \displaystyle \frac{5}{2}: \displaystyle 2 \cdot \frac{125}{8} - 9 \cdot \frac{25}{4} + 10 + 15 is \displaystyle \frac{125}{4} - \frac{225}{4} + 25 = 0.

The question says “a polynomial” because there are more. Every answer of degree 3 is k times this one, and its coefficients are integers exactly when 2k and 9k are, which forces k = 9k - 4(2k) to be an integer.

Example 4

A non-real zero and real coefficients

Find the polynomial of degree 3 with real coefficients, zeros 2 and 4 + i, and y-intercept 68.

Solution

Real coefficients force the conjugate 4 - i to be a zero as well, which makes three zeros for degree 3. Leaving it out would not even give real coefficients: (x - 2)(x - (4 + i)) expands to x^2 - (6 + i)x + 8 + 2i.

Pair the conjugate factors first:

\begin{aligned} (x - 4 - i)(x - 4 + i) &= (x - 4)^2 - i^2 \\ &= x^2 - 8x + 17 \end{aligned}

So p(x) = a(x - 2)(x^2 - 8x + 17), and at x = 0, 68 = a(-2)(17) = -34a, so a = -2:

\begin{aligned} p(x) &= -2(x - 2)(x^2 - 8x + 17) \\ &= -2(x^3 - 10x^2 + 33x - 34) \\ &= -2x^3 + 20x^2 - 66x + 68 \end{aligned}

The graph crosses the x-axis only once, at x = 2. The pair 4 \pm i leaves no intercept.

Example 5

Reading a polynomial off its graph

The graph below touches the x-axis at -3, crosses it at 2, and passes through (0, -18). Find the polynomial of lowest degree it could be.

Figure 2The graph in Example 5. It touches the axis at -3, crosses at 2, and passes through (0, -18).

Solution

At -3 the graph touches and turns back, so that zero has even multiplicity; the smallest is 2, giving (x + 3)^2. At 2 it crosses at a slant, not flat, so multiplicity 1, giving (x - 2). The lowest degree is 2 + 1 = 3:

p(x) = a(x + 3)^2(x - 2)

The y-intercept gives -18 = a(3)^2(-2) = -18a, so a = 1 and

\begin{aligned} p(x) &= (x + 3)^2(x - 2) \\ &= x^3 + 4x^2 - 3x - 18 \end{aligned}

Check against the ends. Odd degree with a > 0 falls on the left and rises on the right, which is what the graph does. A rule such as a(x + 3)^4(x - 2) also touches at -3 and crosses at 2; only the words “lowest degree” rule it out.

Example 6

A cubic that takes one value three times

Find the polynomial p of degree 3 with p(-1) = p(1) = p(4) = 6 and p(0) = 2.

Solution

p is not zero at those places, but q(x) = p(x) - 6 is: q(-1) = q(1) = q(4) = 0. Subtracting a constant does not change the degree, so q is a cubic with three known zeros,

q(x) = a(x + 1)(x - 1)(x - 4)

and q(0) = p(0) - 6 = -4. So -4 = a(1)(-1)(-4) = 4a, and a = -1. Then

\begin{aligned} p(x) &= -(x + 1)(x - 1)(x - 4) + 6 \\ &= -(x^3 - 4x^2 - x + 4) + 6 \\ &= -x^3 + 4x^2 + x + 2 \end{aligned}

Check. p(4) = -64 + 64 + 4 + 2 = 6 and p(0) = 2. On the graph, the line y = 6 meets the curve at x = -1, 1 and 4, and those are the zeros of the curve moved down by 6.

Boundaries

The extra point must not be a zero. At a zero the equation reads 0 = a \cdot 0, true for every a. When 0 is itself a zero, the y-intercept is such a point, as in Example 1.

The multiplicities must add to the degree. If they add to less, what remains of p is not a constant, and one number a cannot describe it. For a real polynomial the gap may be a pair of non-real zeros, which a graph never shows. If they add to more, no polynomial of that degree has those zeros.

Conjugates come from asking for real coefficients. With complex coefficients allowed, (x - 2)(x - (4 + i)) is a perfectly good polynomial of degree 2. Asking for real coefficients adds the conjugate, and each conjugate costs one degree.

Clearing denominators works for rational zeros. A zero \displaystyle \frac{r}{s} becomes (sx - r). An irrational zero such as \displaystyle \sqrt{2} has no linear factor with integer coefficients; it needs the partner \displaystyle -\sqrt{2} as well, and \displaystyle (x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2.

A graph does not show the exact multiplicity. A touch could be 2 or 4, a flat crossing 3 or 5. “Lowest degree” picks the smallest multiplicity that fits each intercept. That is a choice the question makes, not a fact the picture proves.

a is never 0. a = 0 gives the zero polynomial, which is zero everywhere and has no degree.

Where this goes next

This is the step that turns a picture or a list of zeros into a formula. Modeling with polynomials uses it when the zeros of a quantity are known from the situation, such as the cut size at which a folded box has no volume. The fundamental theorem of algebra guarantees that a polynomial of degree n always has n zeros to build from, so the factored form never misses a piece. Example 6 is a first taste of a larger idea: a polynomial of degree n is fixed by its values at n + 1 points, which is how a curve is threaded through measured data.

Exercises

Zeros and one point

Write every polynomial of degree 3 with zeros -4, 1 and 3. Then expand the one with leading coefficient 1.
Find the polynomial of degree 3 with zeros -2, 3 and 5 whose graph passes through (4, 12).
Find the polynomial of degree 4 with zeros 3 (multiplicity 2), -1 and -4, and y-intercept -18.
Find the polynomial of degree 4 with zeros -1 (multiplicity 3) and 2, and p(1) = 24.
Find the polynomial of degree 3 with zeros 0, 4 and -4 whose graph passes through (2, 6). Expand it.

Integer coefficients

Find a polynomial of degree 3 with integer coefficients and zeros \displaystyle \frac{3}{2}, -2 and 1.
Find a polynomial of degree 3 with integer coefficients and zeros \displaystyle \frac{2}{3}, \displaystyle -\frac{1}{4} and 1.

Non-real zeros

Find the polynomial of degree 3 with real coefficients, leading coefficient 1, and zeros -1 and 3 - 2i.
Find the polynomial of degree 4 with real coefficients, zeros 2i and -3 (multiplicity 2), and p(0) = 72.
Find the polynomial of degree 4 with real coefficients and leading coefficient 1 whose zeros include 1 + i and 2 - i.
What is the lowest degree of a polynomial with real coefficients whose zeros include 1, i and 2i? Write one.

From a graph

Figure 3Graphs A and B for exercises 12 and 13. Each is a polynomial of lowest degree.

Find the polynomial of lowest degree for graph A, which passes through (0, 8).
Find the polynomial of lowest degree for graph B, which passes through (-1, 4).
The graph of a cubic touches the x-axis at 4, crosses it at -1, and passes through (0, 32). Find the cubic, and say which way its ends point.

Using the idea

Find the polynomial p of degree 3 with p(-1) = p(2) = p(5) = 4 and p(0) = 14.
A polynomial p of degree 3 has zeros -2, 1 and 3, and p(2) = 8. Find p(4).
The polynomial p(x) = (x - 1)(x - 2)(x - k) has p(0) = 6. Find k.

Reasoning

Tullia and Uchenna each write a polynomial of degree 3 with zeros -1, 2 and 6, and their polynomials are different. Explain how both can be right, and what one extra fact would make them agree.
Rozenn knows p has degree 3 and zeros -1, 2 and 6, and tries to find a from the point (2, 0). Explain why this fails.
Prove that if p and q both have degree n, the same zeros with multiplicities adding to n, and p(x_0) = q(x_0) \neq 0 for one number x_0, then p = q.
Explain why no quadratic with real coefficients has zeros 3 and 1 - 2i, and give the lowest degree a polynomial with real coefficients and both zeros can have.

Answers

a(x + 4)(x - 1)(x - 3) with a \neq 0; for a = 1, (x + 4)(x^2 - 4x + 3), which is x^3 - 13x + 12.
p(x) = -2(x + 2)(x - 3)(x - 5). 12 = a(6)(1)(-1) = -6a, so a = -2.
\displaystyle p(x) = -\frac{1}{2}(x - 3)^2(x + 1)(x + 4). -18 = a(9)(1)(4) = 36a.
p(x) = -3(x + 1)^3(x - 2). 24 = a(2)^3(-1) = -8a.
\displaystyle p(x) = -\frac{1}{4}x(x - 4)(x + 4), which expands to \displaystyle -\frac{1}{4}x^3 + 4x. Since 0 is a zero, the y-intercept cannot fix a, so use (2, 6): 6 = a(2)(-2)(6) = -24a.
(2x - 3)(x + 2)(x - 1), which expands to 2x^3 - x^2 - 7x + 6, or any non-zero integer multiple. The fraction clears as \displaystyle 2\left(x - \frac{3}{2}\right) = 2x - 3.
(3x - 2)(4x + 1)(x - 1), which expands to 12x^3 - 17x^2 + 3x + 2, or any non-zero integer multiple. Each fraction clears in its own factor.
(x + 1)(x^2 - 6x + 13), which expands to x^3 - 5x^2 + 7x + 13. The conjugate 3 + 2i is also a zero, and the two conjugate factors multiply to (x - 3)^2 + 4.
p(x) = 2(x^2 + 4)(x + 3)^2. The conjugate -2i joins 2i, and (x - 2i)(x + 2i) = x^2 + 4. Then 72 = a(4)(9) = 36a.
(x^2 - 2x + 2)(x^2 - 4x + 5), which expands to x^4 - 6x^3 + 15x^2 - 18x + 10. The conjugates 1 - i and 2 + i are forced, which makes four zeros for degree 4.
Degree 5, for example (x - 1)(x^2 + 1)(x^2 + 4). The conjugates -i and -2i are forced, so there are at least five zeros.
\displaystyle p(x) = -\frac{1}{2}(x + 1)(x - 2)^2(x - 4). Slanted crossings at -1 and 4, a touch at 2; then 8 = a(1)(4)(-4) = -16a. Even degree with a < 0 matches both ends pointing down.
\displaystyle p(x) = -\frac{1}{2}(x + 2)(x - 1)^3. A slanted crossing at -2, a flat crossing at 1; then 4 = a(1)(-2)^3 = -8a. Even degree with a < 0 matches both ends pointing down.
p(x) = 2(x - 4)^2(x + 1); it falls on the left and rises on the right. 32 = a(16)(1), so a = 2 > 0, with odd degree.
p(x) = (x + 1)(x - 2)(x - 5) + 4, which expands to x^3 - 6x^2 + 3x + 14. p(x) - 4 has zeros -1, 2 and 5, and p(0) - 4 = 10 = a(1)(-2)(-5) = 10a, so a = 1.
p(4) = -36. 8 = a(4)(1)(-1) gives a = -2, and p(4) = -2(6)(3)(1).
k = -3. p(0) = (-1)(-2)(-k) = -2k = 6.
The zeros fix the polynomial only up to the constant factor: both are a(x + 1)(x - 2)(x - 6), with different values of a. Agreeing on one point (x_0, y_0) with x_0 not a zero, or on the leading coefficient, fixes a and makes them the same.
2 is a zero, so substituting gives 0 = a(3)(0)(-4), which is 0 = 0 for every a. Every member of the family passes through (2, 0). She needs a point whose x is not -1, 2 or 6.
By the factor theorem and the meaning of multiplicity, both contain the product f(x) = (x - c_1)^{m_1}\cdots(x - c_k)^{m_k}, which has degree n, so each quotient has degree 0: p = af and q = bf for constants a and b. Since p(x_0) \neq 0, f(x_0) \neq 0, and af(x_0) = bf(x_0) gives a = b. So p = q.
A polynomial with real coefficients and the zero 1 - 2i also has the zero 1 + 2i. With 3 that is three zeros, and a quadratic has at most two. The lowest degree is 3, for example (x - 3)(x^2 - 2x + 5).