Title: Default for 3.3 Solving Polynomial Equations
Course: HELM Workbooks 2023-2024

Multiplying polynomials together

Key Point

A polynomial expression is one of the form

\[a_nx^n + a_{n-1}x^{n-1} + \ldots +a_2x^2+a_1x+a_0\]

where \(a_0\), \(a_1\),\(\ldots\), \(a_n\) are known coefficients (numbers), \(a_n \ne 0\), and \(x\) is a variable.

\(n\) must be a positive integer.

For example \(x^3-17x^2+54x-8\) is a polynomial expression in \(x\). The polynomial may be expressed in terms of a variable other than \(x\). So, the following are also polynomial expressions:

\(\qquad t^3-t^2+t-3\qquad z^5-1\qquad w^4+10w^2-12\qquad s+1. \)

Note that only non-negative whole number powers of the variable (usually \(x\)) are allowed in a polynomial expression. In this Section you will learn how to factorise simple polynomial expressions and how to solve some polynomial equations. You will also learn the technique of equating coefficients. This process is very important when we need to perform calculations involving partial fractions which will be considered in Section 6.

The degree of a polynomial is the highest power to which the variable is raised. Thus \(x^3+6x+2\) has degree \(3\), \(t^6-6t^4+2t\) has degree \(6\), and \(5x+2\) has degree \(1\).

Let us consider what happens when two polynomials are multiplied together. For example

\(\qquad (x+1)(3x-2) \)

is the product of two first degree polynomials. Expanding the brackets we obtain

\(\qquad (x+1)(3x-2)= 3x^2+x-2 \)

which is a second degree polynomial.

In general we can regard a second degree polynomial, or quadratic, as the product of two first degree polynomials, provided that the quadratic can be factorised. Similarly

\(\qquad (x-1)(x^2+3x-7) = x^3+2x^2-10x+7 \)

is a third degree, or cubic, polynomial which is thus the product of a linear polynomial and a quadratic polynomial.

In general we can regard a cubic polynomial as the product of a linear polynomial and a quadratic polynomial or the product of three linear polynomials. This fact will be important in the following Section when we come to factorise cubics.

Key Point

A cubic expression can always be formulated as a linear expression times a quadratic expression.

Exercise

If \[[value] = (x - [value]) \times \hbox{ (a polynomial)},\] the degree of the undefined polynomial is .
Feedback: The previous key point tells us that a cubic is the product of a linear factor and a quadratic polynomial.

Exercise

1. If \[[value] x^2 + [value] x + [value] = (x + [value]) \times p(x),\] the degree of the polynomial \(p(x)\) is .
2. What is the coefficient of \(x\) in the polynomial \(p(x)\)?
Feedback: 1. A quadratic is the product of two linear polynomials. 2. The coefficient must be \([value]\) in order to generate the term \([value]x^2\) when the linear factors are multiplied together.

Exercise

If \([value] x^2 + [value] x + [value] = (x + [value]) \times \hbox{ (a polynomial)}\), what must be the coefficient of \(x\) in this unknown polynomial?
Feedback: We are given \([value] x^2 + [value] x + [value] = (x + [value]) \times \hbox{ (a polynomial)}\). We know that the polynomial multiplied by a linear function, \(x + [value]\), must give a quadratic. Therefore the polynomial must also be a linear function, i.e. the polynomial is of the form \(ax + b\). Therefore \[(x + [value])(ax + b) = ax^2 + (a\cdot[value] + b)x + b \cdot [value]\]. Comparing coefficients of the \(x^2\) terms we see that \(a=[value]\).

Exercise

Two quadratic polynomials are multiplied together. What is the degree of the resulting polynomial?
Feedback: The degree of the product of two polynomials is the sum of the degrees of these polynomials. A quadratic has degree \(2\) so the product of two quadratics is \(4.\)

Factorising polynomials and equating coefficients

We will consider how we might find the solution to some simple polynomial equations. An important part of this process is being able to express a complicated polynomial into a product of simpler polynomials. This involves factorisation.

Factorisation of polynomial expressions can be achieved more easily if one or more of the factors is already known. This requires a knowledge of the technique of 'equating coefficients' which is illustrated in the following example.

Example

Factorise the expression \(x^3-17x^2+54x-8\) given that one of the factors is \((x-4)\).

Solution

Given that \(x-4\) is a factor we can write

\(\qquad x^3-17x^2+54x-8 = (x-4) \times \hbox{(a quadratic polynomial)} \).

The polynomial must be quadratic because the expression on the left is cubic and \(x-4\) is linear. Suppose we write this quadratic as \(a x^2+b x+c\) where \(a\), \(b\), and \(c\) are unknown numbers which we need to find. Then

\(\qquad x^3-17x^2+54x-8 = (x-4)(a x^2+b x+c) \).

Removing the brackets on the right and collecting like terms together we have

\( \qquad x^3-17x^2+54x-8 = a x^3 + (b-4a)x^2 +(c-4b)x -4c \).

Like terms are those which involve the same power of the variable (\(x\)).

Equating coefficients means that we compare the coefficients of each term on the left with the corresponding term on the right. Thus if we look at the \(x^3\) terms on each side we see that \(x^3 = a x^3\) which implies \(a\) must equal \(1\). Similarly by equating coefficients of \(x^2\) we find \(-17 = b-4a\). With \(a=1\) we have \(-17=b-4\) so \(b\) must equal to \(-13\). Finally, equating constant terms we find \(-8 = -4c\) so that \(c=2\).

As a check we look at the coefficient of \(x\) to ensure it is the same on both sides. Now that we know \(a=1\), \(b=-13\), \(c=2\) we can write the polynomial expression as

\(\qquad x^3-17x^2+54x-8=(x-4)(x^2-13x+2)\).

Exercise

Factorise [value] into a linear and quadratic product given that [value] is a factor:

\( [value] = ( [value]) ( \)  \( x^2 + \)  \( x + \)  \( ) \)
Feedback: We can write \( [value] = ( [value]) (ax^2 + bx + c) \). Try expanding the right hand side and equating the coefficients to find \(a\), \(b\), and \(c\). Be careful not to make a slip in your algebra! We obtain \( [value] = ( [value]) ( x^2 + ([value]) x + [value] ) \)

Exercise

Factorise [value] into a linear and quadratic product given that [value] is a factor:

\( [value] = ( [value]) ( \)  \( x^2 + \)  \( x + \)  \( ) \)
Feedback: We can write \( [value] = ( [value]) (ax^2 + bx + c) \). Try expanding the right hand side and equating the coefficients to find \(a\), \(b\), and \(c\). Be careful not to make a slip in your algebra! We obtain \( [value] = ( [value]) ( x^2 + ([value]) x [value] ) \)

Exercise

Factorise [value] into a linear and quadratic product given that [value] is a factor:

\( [value] = ( [value]) ( \)  \( x^2 + \)  \( x + \)  \( ) \)
Feedback: We can write \( [value] = ( [value]) ( a x^2 + b x + c ) \). Try expanding the right hand side and equating the coefficients to find \(a\), \(b\), and \(c\). Be careful not to make a slip in your algebra! We obtain \( [value] = ( [value]) ( [value] x^2 [value] x [value] ) \)

Polynomial equations

When a polynomial expression is equated to zero, a polynomial equation is obtained. Linear and quadratic equations, which you have already met, are particular types of polynomial equation.

Key Point

A polynomial equation has the form

\[a_nx^n+a_{n-1}x^{n-1} + \ldots a_2x^2+a_1x+a_0 = 0\]

where \(a_0\), \(a_1\), \(\dots\), \(a_n\) are known coefficients, \(a_n \ne 0\), and \(x\) represents an unknown whose value(s) are to be found.

Polynomial equations of low degree have special names. A polynomial equation of degree 1 is a linear equation and the solution of such equations have already been shown. Degree 2 polynomials are called quadratics; degree 3 polynomials are called cubics; degree 4 equations are called quartics and so on. The following are examples of polynomial equations:

\(\qquad 5x^6-3x^4+x^2+7=0, \qquad -7x^4+x^2+9=0,\qquad t^3-t+5=0,\qquad w^7-3w-1=0 \)

Recall that the degree of the equation is the highest power of \(x\) occurring. The solutions or roots of the equation are those values of \(x\) which satisfy the equation.

Key Point

A polynomial equation of degree \(n\) has \(n\) roots.

Some (possibly all) of the roots may be repeated.

Some (possibly all) of the roots may be complex.

Example

Verify that \(x=-1\), \(x=1\), and \(x=0\) are solutions (roots) of the equation

\( \qquad x^3-x=0 \).

Solution

We substitute each value in turn into \(x^3-x\).

\( \qquad (-1)^3 - (-1) = -1 + 1 = 0 \)

so \(x=-1\) is clearly a root.

It is easy to verify similarly that \(x=1\) and \(x=0\) are also solutions.

In the next subsection we will consider ways in which polynomial equations of higher degree than quadratic can be solved.

Q: Exercise

Verify that the given values are solutions of the given equations.

1. \(x^2-5x+6=0, \qquad x=3, x=2 \qquad \qquad \)
[...]
2. \(2t^3+t^2-t=0, \qquad t=0, t=-1, t=\dfrac{1}{2} \qquad\)
[...]
Also offered: I have not verified the given values
A: I have verified the given values; I have verified the given values
If partly right: Your answer is partially correct. That's a shame, as this was an incredibly easy mark.
If wrong: Your answer is incorrect. That's a shame, as these were incredibly easy marks.

Solving polynomial equations when one solution is known

We have already seen a formula which can be used to solve quadratic equations. Unfortunately when dealing with equations of higher degree no simple formulae exist. If one of the roots can be spotted or is known we can sometimes find the others by the method shown in the next example.

Example

Let the polynomial expression \(x^3-17x^2+54x-8\) be denoted by \(P(x)\). Verify that \(x=4\) is a solution of the equation \(P(x)=0\). Hence find the other solutions.

Solution

We substitute \(x=4\) into the polynomial expression \(P(x)\):

\(\qquad P(4)=4^3 - 17(4^2) + 54(4) - 8 = 64 -272 + 216 - 8 = 0 \).

So, when \(x=4\) the left-hand side equals zero. Hence \(x=4\) is indeed a solution. Knowing that \(x=4\) is a root we can state that \((x-4)\) must be a factor of \(P(x)\). Therefore \(P(x)\) can be re-written as a product of a linear and a quadratic term:

\(\qquad P(x)= x^3-17x^2+54x -8 = (x-4) \times \hbox{ (quadratic polynomial)} \).

The quadratic polynomial has already been found in a previous task so we deduce that the given equation can be written

\(\qquad P(x)=x^3-17x^2+54x-8=(x-4)(x^2-13x+2) = 0 \).

In this form we see that \( x-4 = 0 \quad \hbox{ or } \quad x^2 -13x + 2 = 0 \).

The first equation gives \(x=4\) which we already knew.

The second equation must be solved using one of the previously seen methods for solving quadratic equations. For example, using the formula we find

\[\begin{aligned} x &= {-b\pm \sqrt{b^2-4ac}\over 2a}\qquad\text{with}\ a=1,\ b=-13,\ c=2\\ &= {13\pm \sqrt{(-13)^2-4.1.2}\over 2} \\ &= {13 \pm \sqrt{161}\over 2}= {13\pm 12.6886\over 2}. \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \end{aligned}\]

So \(x= 12.8443\) and \(x= 0.1557\) are roots of \(x^2-13x+2\).

Hence the three solutions of \(P(x)=0\) are \(x=4\), \(x=12.8443\) and \(x=0.1557\), to 4 d.p.

Exercise

Verify that the given value \(x=[value]\) is a solution of the equation \([value] = 0\) and hence find all solutions:

\( x = \)

Give your answer as a set of precise numbers.
Feedback: Try taking the factor of \( ([value]) \) out of \([value] = 0\) in order to be left with a quadratic function \( [value] = 0 \) which you can solve using the quadratic formula to obtain all solutions [value].

Solving polynomial equations graphically

Polynomial equations, particularly of high degree, are difficult to solve unless they take a particularly simple form. A useful guide to the approximate values of the solutions can be obtained by sketching the polynomial, and discovering where the curve crosses the \(x\)-axis. The real roots of the polynomial equation \(P(x)=0\) are given by the values of the intercepts of the function \(y=P(x)\) with the \(x\)-axis because on the \(x\)-axis \(y=P(x)\) is zero. Computer software packages and graphics calculators exist which can be used for plotting graphs and hence for solving polynomial equations approximately. Suppose the graph of \(y=P(x)\) is plotted and takes a form similar to that shown below.

[Image: block3fig1]
A polynomial function which cuts the \(x\)-axis at points \(x_1\), \(x_2\), and \(x_3\).

The graph intersects the \(x\)-axis at \(x=x_1\), \(x=x_2\), and \(x=x_3\) and so the equation \(P(x)=0\) has three roots \(x_1\), \(x_2\), and \(x_3\), because \(P(x_1)=0\), \(P(x_2)=0\), and \(P(x_3)=0\).

Example

Plot a graph of the function \(y=4x^4-15x^2+5x+6\) and hence approximately solve the equation \(4x^4-15x^2+5x+6 = 0\).

Solution

The graph has been plotted here with the aid of a computer graph plotting package and is shown in the figure below. By hand, a less accurate result would be produced, of course.

[Image: block3fig2]
Graph of \(y=4x^4-15x^2+5x+6\).

The solutions of the equation are found by looking for where the graph crosses the horizontal axis. Careful examination shows the solutions are at or close to \(x=1\), \(x=1.5\), \(x=-0.5\), \(x=-2\).

An important feature of the graph of a polynomial is that it is continuous. There are never any gaps or jumps in the curve. Polynomial curves never turn back on themselves in the horizontal direction, (unlike a circle). By studying the first graph in this section you will see that if we choose any two values of \(x\), say \(a\) and \(b\), such that \(y(a)\) and \(y(b)\) have opposite signs, then at least one root lies between \(x=a\) and \(x=b\).

Try for yourself: Consider the polynomial \(P(x)=5x^3-47x^2+84x\). By evaluating \(P(2)\) and \(P(3)\) show that at least one root of \(P(x)=0\) lies between \(x=2\) and \(x=3\).

Try for yourself: Without solving the equation or using a graphical calculator, show that \(x^4+4x-1=0\) has a root between \(x=0\) and \(x=1\).

Exercise

Factorise [value] given that [value] is a factor:

\( [value] = ( x + \) \() ( x + \)  \() ( x + \)  \( ) \).
Feedback: We are given the expression \([value]\) for which we know \([value]\) is a factor. Then we have \([value] = ([value])(ax^2 + bx + c)\). Expanding the brackets we obtain \[\begin{aligned} [value] &= ax^3 + bx^2 + cx + [value]ax^2 + [value]bx + [value]c, \\ &= ax^3 + (b + [value]a )x^2 + (c + [value]b )x + [value]c. \end{aligned}\] Comparing coefficients we obtain \(a = 1\), \(b = [value]\), and \(c = [value]\). Therefore we have \([value] = ([value])([value])\). Factorising, we obtain \[[value] = [value].\]

Exercise

Given that \(x=[value]\) is a root of \([value] = 0,\) determine the other roots. Hence record all roots.

Give your answer as a set of precise numbers.
Feedback: By expanding \([value]=([value])({x^2+bx+c})\) and equating coefficients we find \([value]=([value])([value])\). From here we can solve the quadratic and find that the roots of the cubic are [value].

Exercise

Solve the equation [value]:

\( x = \)

Give your answer as a set of precise numbers. Only provide the distinct solutions.
Feedback: Try viewing this as a quadratic function of \(x^2.\) We solve for \(x^2\), from which we can obtain \(x\). We are given the equation [value]. Let \(y = x^2\). Then the equation becomes \[y^2 - [value] + [value] = 0.\] Recognise that this is equivalent to \((y - [value])^2 = 0.\) Therefore \(y = [value]\) which implies \(x = \pm[value].\)

Exercise

Given that two of the roots of [value] have the same modulus but different sign, solve the equation:

\( [value] = ( x + \) \() ( x + \)  \() ( x + \)  \( ) ( x + \)  \( ) \).

List the factors in non-decreasing order in terms of the constant term in the linear factor, repeating roots if appropriate.

Hint: let two of the roots be \(\alpha\) and \(-\alpha\) and use the technique of equating coefficients.
Feedback: We know that both \(a\) and \(-a\) are roots, and that the leading coefficient is \(1\) so we can write the polynomial as \( (x - a) (x + a) (x^2 + bx + c)\). Expanding this gives \[ [value]= [value] .\] By equating coefficients we find from the cubic term \(b=[value]\) and then this information and the linear term gives \(a^2=[value]\). We can then use the constant (or quadratic) term to find \(c\). From here we solve the quadratic \(x^2+bx+c=0\) to determine the unknown roots. In total we have \[[value]=[value].\]

Exercise

Consider the polynomial \(P(x)=5x^3-47x^2+84x\). By evaluating \(P(2)\) and \(P(3)\) show that at least one root of \(P(x)=0\) lies between \(x=2\) and \(x=3\).

Exercise

Factorise [value] given that [value] is a factor:

\( [value] = ( x + \) \() ( x + \)  \() ( x + \)  \( ) ( x + \)  \( ) \).
Feedback: We are given the expression [value] and are told that \([value]\) is a factor of the expression. Then \[[value] = ([value])(ax^3 + bx^2 + cx + d).\] Expand the RHS of the above equation and equate the coefficients to find \[ ax^3 + bx^2 + cx + d = [value].\] From here, try substituting in values of \(x\) to find another root. We could, for example, notice that \([value]\) is actually a factor of this cubic, i.e. it is a factor of multiplicity \(2\) of the original quartic. Then \[[value] = ([value])^2(ax^2 + bx + c).\] Expand the RHS of the above equation and equate the coefficients to find \[ax^2 + bx + c = [value].\] We can then factorise this quadratic as \([value]\). Therefore the original quartic can be expressed as \([value]\).

Exercise

Without solving the equation or using a graphical calculator, show that \(x^4+4x-1=0\) has a root between \(x=0\) and \(x=1\).

Exercise

Verify that the given value \(x=[value]\) is a solution of the equation [value] and hence find all solutions:

\( x = \)  (give your answer as a set of numbers).
Feedback: Try taking the factor of \( ([value]) \) out of [value] in order to be left with a quadratic function \( [value] \) which you can solve using the quadratic formula to obtain all solutions [value].

Exercise

Verify that the given values \(x=[value]\) and \(x=[value]\) are solutions of the equation [value] and hence find all solutions:

\( x = \)

Give your answer as a set of precise numbers.
Feedback: Try taking the factors of \( ([value]) \) and \( ([value]) \) out of [value] in order to be left with a quadratic function \( [value] \) which you can solve using the quadratic formula to obtain all solutions [value].

Exercise

Check for yourself that \(x=[value]\) is a root of [value] and locate the other roots algebraically. Hence record all solutions:

\( x = \)  (give your answer as a set of numbers).
Feedback: We are given the expression \([value]\) for which we know \(x = [value]\) is a root. Therefore \([value]\) is a factor. Then we have \([value] = ([value])(ax^2 + bx + c)\). Expanding the brackets we obtain \[\begin{aligned} [value] &= [value], \\ &= [value]. \end{aligned}\] Comparing coefficients we obtain \(a = 1\), \(b = [value]\), and \(c = [value]\). Therefore we have \([value] = ([value])([value])\). Factorising, we obtain \[[value] = [value].\] Therefore the solutions are \(x - \{ [value], [value], [value] \} \).

Q: Exercise

If \(a\) is root of polynomial \(p(x)\), which of the following is a factor of \(p(x)\)
- \((x-a)\)
- \((x+a)\)
A: \((x-a)\)
Feedback: If \(x=a\) satisfies \(p(x)\) then \(x-a\) is a factor of \(p(x)\).

Exercise

In this exercise we will solve the cubic [value] given that \(x=[value]\) is a root.

The cubic can therefore be expressed as

\( \qquad [value] \)

where \(a\), \(b\), and \(c\) are constants. These can be found by expanding the right-hand side.

Expand the right-hand side:

\( \qquad [value] = (\) \() x^3 + (\) \()x^2 + (\) \()x + \) .

We identify the unknown coefficients by focusing on the greatest and least powers first.

Equate coefficients of \(x^3\) to find \(a\):  .

Equate constant terms to find \(c\):  .

With this we can now solve for \(b\).

Equate coefficients of \(x^2\) to find \(b\):  .

This enables us to write the equation as \([value]\) so \([value]\) or \([value]=0\).

Now solve the quadratic and state all three roots:

Give your answer as a set of precise expressions.
Feedback: We are given the cubic \[[value] = [value].\] If we expand the RHS we obtain \[[value] = [value].\] Collecting like-terms we get \[[value] = [value].\] If we equate the coefficients of the LHS and RHS of the above equation, we obtain \(a = 1\), \(b = [value]\) and \(c=[value]\). This enables us to write the equation as \([value]\) so \([value]\) or \([value]=0\). The quadratic equation can be solved using the quadratic formula to obtain \(x=[value]\) and \(x=[value]\). Thus the three roots of \([value]\) are \(x=[value]\), \(x=[value]\), and \(x=[value].\)

Exercise

Select all valid roots of the equation [value] from the values given below:
Feedback: A number \(a\) is a root of [value] if [value] when \(x=a\) so substitute each given value of \(x\) into [value] to verify whether the value is a root or is not.