evaluate a polynomial making use of the Remainder Theorem. Usage the aspect Theorem to settle a polynomial equation. Usage the reasonable Zero organize to find rational zeros. Discover zeros of a polynomial function. Use the linear Factorization theorem to uncover polynomials with provided zeros. Use Descartes’ dominance of Signs. Settle real-world applications that polynomial equations

A brand-new bakery supplies decorated paper cakes because that children’s birthday parties and other unique occasions. The bakery desires the volume that a tiny cake to be 351 cubic inches. The cake is in the form of a rectangular solid. They want the size of the cake come be 4 inches longer than the width of the cake and the elevation of the cake to it is in one-third that the width. What must the size of the cake pan be?

This problem can be fixed by creating a cubic function and solving a cubic equation for the volume of the cake. In this section, we will comment on a range of tools for writing polynomial functions and also solving polynomial equations.

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Evaluating a Polynomial making use of the Remainder Theorem

In the critical section, us learned how to division polynomials. We have the right to now usage polynomial department to evaluate polynomials making use of the Remainder Theorem. If the polynomial is divided by (x–k), the remainder might be uncovered quickly by evaluating the polynomial duty at (k), that is, (f(k)). Let’s walk with the evidence of the theorem.

Recall the the department Algorithm claims that, provided a polynomial dividend (f(x)) and a non-zero polynomial divisor (d(x)) where the level of (d(x)) is less than or equal to the degree of (f(x)),there exist distinctive polynomials (q(x)) and (r(x)) such that

If the divisor, (d(x)), is (x−k), this bring away the form

Since the divisor (x−k)

is linear, the remainder will be a constant, (r). And, if us evaluate this because that (x=k), we have

<eginalign* f(k)&=(k−k)q(k)+r \<4pt> &=0cdotq(k)+r \<4pt> &=r endalign*>

In other words, (f(k)) is the remainder derived by separating (f(x))by (x−k).

The Remainder Theorem

If a polynomial (f(x)) is divided by (x−k),then the remainder is the value (f(k)).

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Notice that 2 of the components of the constant term, 6, space the 2 numerators indigenous the original rational roots: 2 and also 3. Similarly, two of the determinants from the top coefficient, 20, room the two denominators indigenous the initial rational roots: 5 and 4.

We deserve to infer that the numerators of the rational roots will constantly be components of the constant term and the denominators will certainly be factors of the leading coefficient. This is the essence of the reasonable Zero Theorem; the is a method to offer us a swimming pool of possible rational zeros.

How to: offered a polynomial function (f(x)), use the reasonable Zero organize to find rational zeros.

recognize all components of the consistent term and also all components of the leading coefficient. Recognize all possible values the (dfracpq), whereby (p) is a factor of the constant term and also (q) is a factor of the leading coefficient. Be certain to incorporate both hopeful and an adverse candidates. Determine which feasible zeros space actual zeros by evaluating each case of (f(fracpq)).

Example (PageIndex3): Listing All possible Rational Zeros

List all possible rational zeros the (f(x)=2x^4−5x^3+x^2−4).


The only possible rational zeros of (f(x)) space the quotients the the determinants of the last term, –4, and also the determinants of the top coefficient, 2.

The consistent term is –4; the determinants of –4 room (p=±1,±2,±4).

The top coefficient is 2; the determinants of 2 space (q=±1,±2).

If any type of of the 4 real zeros space rational zeros, climate they will certainly be of one of the following factors of –4 divided by one of the factors of 2.

Note that (frac22=1) and also (frac42=2), i beg your pardon have already been listed. For this reason we have the right to shorten ours list.

Exercise (PageIndex3)

Use the rational Zero theorem to find the reasonable zeros that (f(x)=x^3−5x^2+2x+1).


There are no reasonable zeros.

Finding the Zeros the Polynomial Functions

The reasonable Zero organize helps united state to narrow down the list of feasible rational zeros because that a polynomial function. As soon as we have done this, we deserve to use synthetic division repetitively to determine all of the zeros that a polynomial function.

How to: given a polynomial role (f), usage synthetic division to discover its zeros.

use the rational Zero theorem to list all possible rational zeros that the function. Usage synthetic department to evaluate a given possible zero by synthetically separating the candidate into the polynomial. If the remainder is 0, the candidate is a zero. If the remainder is no zero, discard the candidate. Repeat step two using the quotient discovered with man-made division. If possible, continue until the quotient is a quadratic. Uncover the zeros the the quadratic function. Two feasible methods for fixing quadratics are factoring and also using the quadratic formula.

Example (PageIndex5): recognize the Zeros of a Polynomial role with Repeated real Zeros

Find the zeros of (f(x)=4x^3−3x−1).


The rational Zero organize tells united state that if (dfracpq) is a zero of (f(x)), climate (p) is a variable of –1 and (q) is a variable of 4.

<eginalign*dfracpq=dfracfactorspace ofspace constantspace termfactorspace ofspace leadingspace coefficient \<4pt> =dfracfactorspace ofspace -1factorspace ofspace 4 endalign*>

The components of –1 space ±1 and also the factors of 4 room ±1,±2, and also ±4. The feasible values because that (dfracpq) are (±1),(±dfrac12), and also (±dfrac14). These room the feasible rational zeros for the function. We will use synthetic division to evaluate each possible zero till we uncover one that offers a remainder that 0. Let’s begin with 1.

Dividing by ((x−1)) provides a remainder that 0, for this reason 1 is a zero that the function. The polynomial have the right to be written as

<(x−1)(4x^2+4x+1) onumber>

The quadratic is a perfect square. (f(x)) have the right to be written as

<(x−1)(2x+1)^2 onumber>

We currently know the 1 is a zero. The other zero will have actually a multiplicity of 2 since the aspect is squared. To uncover the other zero, we can set the variable equal to 0.

< eginalign* 2x+1=0 \<4pt> x &=−dfrac12 endalign*>

The zeros of the function are 1 and also (−frac12) with multiplicity 2.


Look at the graph of the duty (f) in number (PageIndex1). Notice, in ~ (x =−0.5), the graph bounces turn off the x-axis, denote the also multiplicity (2,4,6…) for the zero −0.5. In ~ (x=1), the graph the cross the x-axis, denote the weird multiplicity (1,3,5…) because that the zero (x=1).

Figure (PageIndex1).

THE an essential THEOREM that ALGEBRA

The an essential Theorem the Algebra states that, if (f(x)) is a polynomial of degree (n > 0), then (f(x)) has at least one complex zero.

We have the right to use this theorem to argue that, if (f(x)) is a polynomial of degree (n >0), and also a is a non-zero actual number, climate (f(x)) has specifically (n) direct factors

where (c_1,c_2),...,(c_n) are facility numbers. Therefore, (f(x)) has actually (n) root if we allow for multiplicities.

Q&A: does every polynomial have actually at least one imaginary zero?

No. Genuine numbers room a subset of complex numbers, but not the other method around. A complicated number is not necessarily imaginary. Genuine numbers space also complicated numbers.

We deserve to then set the quadratic same to 0 and solve to discover the other zeros the the function.

< eginalign* 3x^2+1=0 \<4pt> x^2 &=−dfrac13 \<4pt> x&=±−sqrtdfrac13 \<4pt> &=±dfracisqrt33 endalign*>

The zeros of (f(x)) are (–3) and (±dfracisqrt33).


Look at the graph of the role (f) in number (PageIndex2). Notice that, in ~ (x =−3), the graph the cross the x-axis, describe an weird multiplicity (1) for the zero (x=–3). Likewise note the presence of the two turning points. This means that, since there is a (3^rd) degree polynomial, we are looking at the maximum number of turning points. So, the end actions of boosting without bound to the right and decreasing without bound to the left will continue. Thus, every the x-intercepts because that the function are shown. So one of two people the multiplicity the (x=−3) is 1 and also there room two facility solutions, i m sorry is what us found, or the multiplicity in ~ (x =−3) is three. One of two people way, our an outcome is correct.

Figure (PageIndex2).

Using the direct Factorization theorem to discover Polynomials with given Zeros

A critical implication of the Fundamental to organize of Algebra, together we declared above, is the a polynomial role of degree n will have (n) zeros in the collection of complicated numbers, if we allow for multiplicities. This method that we can factor the polynomial role into (n) factors. The Linear Factorization theorem tells us that a polynomial role will have actually the same number of factors as its degree, and also that each aspect will be in the kind ((x−c)), whereby c is a facility number.

Let (f) be a polynomial function with real coefficients, and also suppose (a +bi), (b≠0), is a zero of (f(x)). Then, by the variable Theorem, (x−(a+bi)) is a aspect of (f(x)). For (f) to have real coefficients, (x−(a−bi)) must likewise be a element of (f(x)). This is true because any kind of factor various other than (x−(a−bi)), as soon as multiplied by (x−(a+bi)), will leave imaginary materials in the product. Only multiplication through conjugate bag will remove the imaginary components and an outcome in real coefficients. In other words, if a polynomial duty (f) with actual coefficients has a complex zero (a +bi), then the facility conjugate (a−bi) must also be a zero the (f(x)). This is referred to as the Complex Conjugate Theorem.


According come the direct Factorization Theorem, a polynomial duty will have the same number of factors as its degree, and each factor will be in the form ((x−c)), wherein (c) is a facility number.

If the polynomial duty (f) has actually real coefficients and a facility zero in the kind (a+bi), then the complicated conjugate the the zero, (a−bi), is likewise a zero.


If (2+3i) were given as a zero the a polynomial with genuine coefficients, would (2−3i) likewise need to be a zero?

Yes. When any facility number with an imaginary component is offered as a zero the a polynomial with real coefficients, the conjugate must also be a zero the the polynomial.

Using Descartes’ dominance of Signs

There is a straightforward method to identify the feasible numbers of hopeful and negative real zeros for any type of polynomial function. If the polynomial is composed in to decrease order, Descartes’ dominance of Signs tells united state of a relationship between the variety of sign changes in (f(x)) and also the number of positive real zeros. For example, the polynomial role below has actually one sign change.


This tells united state that the role must have actually 1 confident real zero.

There is a similar relationship in between the number of sign alters in (f(−x)) and also the number of an unfavorable real zeros.

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Again, there room two sign changes, therefore there are either 2 or 0 an adverse real roots.

There are four possibilities, as we can see in Table (PageIndex1).

Table (PageIndex1) confident Real Zeros an adverse Real Zeros facility Zeros total Zeros
2 2 0 4
2 0 2 4
0 2 2 4
0 0 4 4

We can confirm the number of positive and an adverse real roots by evaluating a graph of the function. See figure (PageIndex3). We deserve to see native the graph that the role has 0 hopeful real roots and also 2 an adverse real roots.

Figure (PageIndex3).

Solving Real-World Applications

We have now presented a selection of devices for solving polynomial equations. Let’s usage these devices to deal with the bakery problem from the beginning of the section.

Solving Polynomial Equations

A brand-new bakery offers decorated sheet cakes because that children’s birthday parties and other special occasions. The bakery desires the volume of a small cake to it is in 351 cubic inches. The cake is in the shape of a rectangle-shaped solid. They desire the length of the cake to be four inches much longer than the width of the cake and also the elevation of the cake to it is in one-third of the width. What must the size of the cake pan be?


Begin by creating an equation because that the volume that the cake. The volume of a rectangle-shaped solid is offered by (V=lwh). We were provided that the length must be four inches much longer than the width, for this reason we have the right to express the length of the cake as (l=w+4). Us were provided that the height of the cake is one-third the the width, therefore we can express the elevation of the cake as (h=dfrac13w). Let’s write the volume that the cake in terms of width of the cake.

Substitute the offered volume into this equation.

(351=13w^3+43w^2) instead of 351 for V.(1053=w^3+4w^2) main point both sides by 3.(0=w^3+7w^2−1053) Subtract 1053 from both sides.

Descartes" dominion of indicators tells united state there is one hopeful solution. The rational Zero to organize tells united state that the feasible rational zeros space (pm 1,±3,±9,±13,±27,±39,±81,±117,±351,) and also (±1053). We can use synthetic department to check these feasible zeros. Just positive number make feeling as dimensions because that a cake, therefore we require not test any negative values. Let’s start by testing values that make the most sense as dimensions for a little sheet cake. Usage synthetic division to examine (x=1).

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Since 1 is no a solution, us will check (x=3).