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Algebraic Function Structures · Tutorial 53 of 1495

Constructing Polynomials from Given Zeros

Translate each specified real zero and its multiplicity into linear factors, then choose the leading coefficient to obtain the required scale.

Beginner 9 min read

What You'll Learn

  • Convert a real zero \(r\) into the factor \(x-r\).
  • Use the multiplicity of each zero as the exponent on its factor.
  • Build a polynomial with a specified leading coefficient.
  • Check the degree and leading coefficient of a constructed polynomial.
  • Explain when the specified zeros and leading coefficient determine a unique polynomial.
  • Identify and correct a common sign error when writing linear factors.

From Zeros to Factors

A polynomial can be described by its coefficients in standard form, but that form does not always make its zeros easy to see. For example, the expression \( (x-2)(x+1) \) immediately identifies the zeros \(2\) and \(-1\). This tutorial uses that connection in reverse: starting with real zeros and their multiplicities, we will construct a polynomial that has those features.

The key translation is simple but important. A zero \(r\) corresponds to the factor \(x-r\), not \(x+r\) unless \(r\) is negative. If the zero has multiplicity greater than one, its factor is repeated by raising it to that multiplicity. A nonzero constant in front then sets the polynomial's leading coefficient.

Definition: A real number \(r\) is a zero of a polynomial \(P\) if \(P(r)=0\). If \(P\) is a nonzero polynomial and \(r\) is one of its zeros, the multiplicity of \(r\) is the positive integer \(m\) such that \((x-r)^m\) divides \(P(x)\), but \((x-r)^{m+1}\) does not.

The nonzero condition matters: every power of \(x-r\) divides the zero polynomial, so it does not have a finite multiplicity in this sense. In what follows, each constructed polynomial has a nonzero leading coefficient and is therefore nonzero.

The Construction Rule

Suppose distinct real numbers \(r_1,\ldots,r_k\) are specified as zeros, with respective positive integer multiplicities \(m_1,\ldots,m_k\). The basic polynomial built from exactly these linear factors is

$$ P(x)=a(x-r_1)^{m_1}(x-r_2)^{m_2}\cdots(x-r_k)^{m_k}, \qquad a\ne 0. $$

Here \(a\) is the leading coefficient. Each factor \(x-r_i\) is monic, meaning its leading coefficient is \(1\). Consequently, their product is also monic, and multiplying by \(a\) gives leading coefficient \(a\). The degree is the sum of the multiplicities:

$$ \deg(P)=m_1+m_2+\cdots+m_k. $$

This construction gives a polynomial with the specified zeros and multiplicities. It also uses no extra factors, so its real zeros are precisely the listed numbers. If a problem asks only for a polynomial that includes certain zeros, other choices may be possible: an additional polynomial factor can introduce more zeros or increase the degree. The product above is the direct construction when only the stated real zeros are wanted.

Why the Product Has the Required Properties

The construction works because a product is zero whenever one of its factors is zero. At \(x=r_i\), the factor \(x-r_i\) equals zero, so \(P(r_i)=0\). To check the multiplicity, separate that factor from all the others:

$$ P(x)=(x-r_i)^{m_i}Q_i(x), \qquad Q_i(x)=a\prod_{\substack{j=1\\j\ne i}}^k(x-r_j)^{m_j}. $$

All the listed roots are distinct. Thus, when \(x=r_i\), every factor in \(Q_i(r_i)\) is nonzero: \(a\ne0\), and \(r_i-r_j\ne0\) whenever \(j\ne i\). Therefore \(Q_i(r_i)\ne0\). The factor \((x-r_i)^{m_i}\) divides \(P\), while one more factor \(x-r_i\) cannot divide it, because the remaining factor \(Q_i\) does not vanish at \(r_i\). This proves that the multiplicity is exactly \(m_i\).

The leading coefficient and degree follow directly from multiplying the leading terms. Each \((x-r_i)^{m_i}\) has leading term \(x^{m_i}\), so the product of those terms is \(x^{m_1+\cdots+m_k}\). Multiplication by \(a\) makes the leading term \(a x^{m_1+\cdots+m_k}\).

Theorem: For distinct real numbers \(r_1,\ldots,r_k\), positive integer multiplicities \(m_1,\ldots,m_k\), and any nonzero real number \(a\), the polynomial \(a\prod_{i=1}^k(x-r_i)^{m_i}\) has each \(r_i\) as a zero of multiplicity exactly \(m_i\). Its degree is \(\sum_{i=1}^k m_i\), and its leading coefficient is \(a\).

Worked Examples

Worked Example: Zeros with Repeated Factors

Construct a polynomial with zeros \(-2\), \(1\), and \(3\) of multiplicities \(2\), \(1\), and \(2\), respectively, and leading coefficient \(2\).

For a zero \(r\), the factor is \(x-r\). Thus the zero \(-2\) gives \(x-(-2)=x+2\); the zero \(1\) gives \(x-1\); and the zero \(3\) gives \(x-3\). Apply the given multiplicities and leading coefficient:

$$ P(x)=2(x+2)^2(x-1)(x-3)^2. $$

The degree is \(2+1+2=5\). Each factor inside the product is monic, so the leading term is \(2x^5\), as required. To check the specified zeros, substitute each into its corresponding factor: at \(x=-2\), \(x+2=0\); at \(x=1\), \(x-1=0\); and at \(x=3\), \(x-3=0\). The exponents show the respective multiplicities are \(2\), \(1\), and \(2\); the other factors are nonzero at each of those values.

For an additional check, expand. First,

$$ (x+2)^2(x-3)^2 =(x^2+4x+4)(x^2-6x+9) =x^4-2x^3-11x^2+12x+36. $$

Multiplying by \(x-1\) gives \(x^5-3x^4-9x^3+23x^2+24x-36\). Multiplying by \(2\) gives

$$ P(x)=2x^5-6x^4-18x^3+46x^2+48x-72. $$

The expanded form confirms that the degree is \(5\) and the leading coefficient is \(2\). The factored form makes the zeros and multiplicities clear.

Worked Example: A Zero at the Origin and a Negative Scale

Construct a polynomial with zero \(0\) of multiplicity \(3\), zero \(2\) of multiplicity \(2\), and leading coefficient \(-3\).

The factors are \(x-0=x\) and \(x-2\). Applying the multiplicities and the leading coefficient gives

$$ P(x)=-3x^3(x-2)^2. $$

The degree is \(3+2=5\). The factors have leading terms \(x^3\) and \(x^2\), so the leading term of the product is \(-3x^5\). At \(x=0\), the factor \(x^3\) vanishes and \((0-2)^2=4\ne0\); at \(x=2\), the factor \((x-2)^2\) vanishes and \(2^3=8\ne0\). Thus the multiplicities are exactly \(3\) and \(2\).

Expanding provides a further check:

$$ \begin{aligned} P(x) &=-3x^3(x^2-4x+4)\\ &=-3x^5+12x^4-12x^3. \end{aligned} $$

The expanded polynomial has leading coefficient \(-3\), degree \(5\), and vanishes at \(0\) and \(2\), as required. The factorization also confirms the exact multiplicities.

Worked Example: Matching a Specified Leading Term

Construct a polynomial whose zeros are \(1\) of multiplicity \(2\) and \(-2\) of multiplicity \(4\), and whose leading term is \(7x^6\).

The multiplicities add to \(2+4=6\), so the product of the monic factors has leading term \(x^6\). To change that leading term to \(7x^6\), multiply the product by \(7\). The factors are \(x-1\) and \(x-(-2)=x+2\), so

$$ P(x)=7(x-1)^2(x+2)^4. $$

The leading term is \(7x^2x^4=7x^6\). At \(x=1\), the factor \((x-1)^2\) is zero, while \((1+2)^4=81\ne0\). At \(x=-2\), the factor \((x+2)^4\) is zero, while \((-2-1)^2=9\ne0\). Hence the multiplicities are exactly \(2\) and \(4\). The polynomial has degree \(6\) and the requested leading term.

What the Data Determine

Specified zeros and multiplicities determine the factors, but not by themselves the leading coefficient. For instance, multiplying the same product by \(2\) or by \(-5\) preserves its zeros and their multiplicities while changing its leading coefficient. The zero polynomial is not an option when exact multiplicities are specified, because multiplicity here is defined for nonzero polynomials.

When the zeros are stated to be all the zeros and the degree equals the sum of their multiplicities, the leading coefficient determines the polynomial uniquely. To see why, repeatedly use the Factor Theorem: if \(r\) is a zero of a polynomial, then \(x-r\) is a factor. Applying this for the specified multiplicities shows that the product of all the stated factors divides the polynomial. If the polynomial has degree equal to the degree of that product, the quotient must have degree zero, so it is a constant. The leading coefficient fixes that constant.

If only some zeros are specified and the degree is not fixed, there is no such uniqueness: a polynomial can contain those factors and also have additional factors. The direct construction is still valid, but it is one particular choice.

A Common Sign Error

A frequent mistake is to use \(x+r\) for a zero \(r\). If the desired zero is \(-2\), the incorrect factor \(x-2\) does not vanish there:

$$ (-2)-2=-4\ne0. $$

The correct factor is \(x-r\). With \(r=-2\), this becomes

$$ x-r=x-(-2)=x+2, \qquad (-2)+2=0. $$

A reliable check is to substitute the proposed zero into its factor. The factor must equal zero. After building the product, also check the exponents, the sum of multiplicities for the degree, and the constant multiplier for the leading coefficient.

1
Write a factor for each distinct zero.
For a zero \(r\), use \(x-r\), including the sign inside the subtraction.
2
Apply the multiplicities.
Raise each factor to its specified positive integer power.
3
Set the leading coefficient.
Multiply the product by the required nonzero leading coefficient.
4
Verify the construction.
Check zeros, exact multiplicities, degree, and leading coefficient.

Why This Representation Is Useful

The factored representation stores several kinds of information at once. It makes the specified zeros visible, records how many times each corresponding factor occurs, and gives the degree by adding the exponents. It also allows the leading coefficient to be read without expanding. Expansion can be useful when standard form is requested, but it is not necessary to establish these features.

When constructing a polynomial, begin with its required features rather than expanding prematurely. Build the factors first, attach the scale last, and verify each condition in the representation where it is easiest to see. This approach reduces sign errors and keeps the relationship between zeros, multiplicities, degree, and leading coefficient explicit.

Check Your Understanding

For each question, use the factor form to organize your reasoning before expanding, if expansion is requested.

  1. Write a polynomial with zeros \(-3\) and \(4\), each of multiplicity \(1\), and leading coefficient \(5\).
  2. What degree does a polynomial have if its specified real zeros have multiplicities \(2\), \(3\), and \(1\), and these are all its zeros?
  3. What factor corresponds to the zero \(-5\), and how can substitution check its sign?
  4. Construct a polynomial with zero \(0\) of multiplicity \(2\), zero \(3\) of multiplicity \(1\), and leading coefficient \(-2\).
  5. Why do specified zeros and multiplicities alone not determine the leading coefficient?

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