Calculating with polynomials

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If \(f(x)\) and \(g(x)\) are two polynomials and \(c\) is a real number, then the following expressions also are polynomials:

  • \(c\cdot {f(x)}\)
  • \(f(x)+g(x)\)
  • \(f(x)-g(x)\)
  • \(f(x)\cdot g(x)\)

Rules of calculation for polynomials

  • Multiplying a polynomial by a constant is equivalent to multiplying each term of the polynomial by that constant
  • Addition of two polynomials in \(x\) is equivalent to adding the coefficients of terms with the same power of \(x\).
  • Subtraction of polynomial \(g(x)\) by a polynomial \(f(x)\) is the same as subtracting the coefficients of terms in \(g(x)\) by the coefficients of the same power of \(x\) in \(f(x)\).
  • Multiplication of two polynomials is obtained by multiplying each term of one polynomial by each term of the other polynomial and adding all the products.

The rules specify how we can add, subtract and multiply polynomials. The quotient \(\frac{f(x)}{g(x)}\) of two polynomials is not always a polynomial, but does result in a rational function.

Degree of polynomials that result from arithmetic operations

Let \(f(x)\) and \(g(x)\) be polynomials of degree respectively \(m\) and \(n\), and let \(c\) be a real number.

  • The degree of \(c\cdot f(x)\) is the degree of \(f(x)\) if \(c\ne0\).
  • The degree of \(f(x)\cdot g(x)\) is the sum of the degrees \(f(x)\) and \(g(x)\).
  • if \(m\gt n\), then the degree of \(f(x)+g(x)\) is equal to the degree of \(f(x)\).
  • If \(m=n\), then the degree of \(f(x)+g(x)\) is less than or equal to the degree of \(f(x)\).

Let’s practice!

What is the product of polynomial \(f(x)=x^3+x^2-4\cdot x-4\) and the constant \(2\)?

In order to compute the product of \(f(x)\) and \(2\) we multiply the coefficient of each power of \(x\) in \(f(x)\) with \(2\):

\(\begin{array}{rcl}2\cdot f(x)&=& 2\cdot \left(x^3+x^2-4\cdot x-4\right)\\ &=&{ -8+{ (2\cdot1)\cdot x^2 }+{ (2\cdot-4)\cdot x }+{ (2\cdot-4) }} \\ &=& 2\cdot x^3+2\cdot x^2-8\cdot x-8\end{array}\)


\(2\cdot x^3+2\cdot x^2-8\cdot x-8\)

The notion of polynomial

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A polynomial is an expression of the form:

\(a_0+a_1x+a_2x^2+\cdots + a_nx^n\tiny,\),

where \(a_2,\ldots,a_n\) are numbers (the coefficients of the polynomial) and \(x\) is the variable.

If \(a_n\ne0\), then \(n\) is called the degree of the polynomial. The number \(a_2\) is then called the leading coefficient of the polynomial.

By way of convention, we say that the polynomial \(0\) is of degree \(-1\).

The above polynomial defines a function \(f\) with rule

\(f(x) =a_0+a_1x+a_2x^2+\cdots + a_nx^n\tiny.\).

Such a function is called a polynomial function.

Let’s practice!

What is the degree of the polynomial \(f(x)=4+7 \cdot x\)? And what is its leading coefficient?


The degree of the polynomial\(f(x)=4+7 \cdot x\) is equal to \(1\). The leading coefficient equals \(7\). Polynomials of degree \(1\) are also known as linear functions.

The corresponding graph is a straight line with slope equal to \(7\) and the intersection with the \(y\)-axis is at \(\left[0,4\right]\).

notion of polynomial.png

Solving equations with factorization

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If we can factor the left hand side of the equation \(ax^2+bx+c=0\), then the solution is easily found:

Consider the quadratic equation with unknown \(x\):


where \(a\), \(b\) and \(c\) are real numbers with \(a\ne0\). If the left hand side can be factored in linear factors:


where \(p\) and \(q\) are real numbers, then the solution of the equation is \(x=p\lor x=q\).

Let’s practice!

Solve the equation for \(x\) below using factorization.

\(x^2-14 x-16=-8 x\tiny.\)

Give your answer in the form of \(x=a\lor x=b\).

This can be solved by means of the following deduction:

\(\begin{array}{rcl} x^2-14 x-16&=&-8 x \\ &&\phantom{xxx}\color{blue}{\text{the original equation}}\\ x^2-6 x-16&=&0 \\ &&\phantom{xxx}\color{blue}{\text{all terms to the left}}\\ (x-8)\cdot (x+2)&=&0 \\ &&\phantom{xxx}\color{blue}{\text{left hand side factored}}\\ x-8=0 &\lor& x+2=0\\ &&\phantom{xxx}\color{blue}{A\cdot B=0 \text{ if and only if }A=0\lor B=0}\\ x=8 &\lor& x=-2\\ &&\phantom{xxx}\color{blue}{\text{constant terms to the right}} \end{array}\)


\(x=8\lor x=-2\)

Quadratic function – Completing the square

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A quadratic function \(f\) is given by \(f(x)=ax^2+bx+c\) with \(a\), \(b\) and \(c\) being real numbers and \(a\neq 0\).
The graph associated with such a function is called a parabola.
When \(a\gt0\), we speak of a parabola opening upwards, and when \(a\lt0\), we speak of a parabola opening downwards.

The significance of these names becomes evident from the graph below of the parabola opening upwards of \(x^2+8\) and the parabola opening downwards of \(-x^2-8\).

Quadratic function image 1.png

To find the zeros of a quadratic function, we need do find the points of intersection of the parabola with the \(x\)-axis. We can do this in three ways:

  • completing the square
  • \(abc\)-formula
  • factorization

In what follows, we will discuss the completing the square method.

An equation like \(\left(x+2\right)^2=3\) can easily be solved by taking the root:

Solve \(x\) in \(\left(x+4\right)^2={{16}\over{25}}\).

Take the root at both sides of the \(=\) sign: \(x+4=\pm {{4}\over{5}}\);

Next, solve the equation: \(x=-4\pm {{4}\over{5}}\);


\(x=-4\pm {{4}\over{5}}\)

However, this will not work in general for a quadratic equation like \(x^2+8x-1=0\), because the unknown \(x\) occurs twice. But there is a method to obtain the first form from the second one:

Completing the square is rewriting a quadratic expression in \(x\) as an expression in which \(x\) only occurs once, in the base of a second power. To be precise, if \(a\), \(b\) and \(c\) are real numbers, then

\(ax^2+bx+c=a\cdot\left( \left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{(2a)^2}\right)\tiny.\)

With this method we cannot only solve quadratic equations, but also determine what the extreme point of a parabola is:

The quadratic polynomial \(ax^2+bx+c\) in which \(a\ne0\), can be written as

\(a\cdot\left(x+\frac{b}{2a}\right)^2 -\frac{b^2-4ac}{4a}\tiny.\)

In particular, \(\left[-\frac{b}{2a},-\frac{{b}^{2}-4a·c}{4a}\right]\) is an extreme:

  • If \(a\gt0\), then the extreme is the lowest point of the parabola opening upwards.
  • If \(a\lt0\), then the extreme is the highest point of the parabola opening downwards.

In other words, the quadratic function \(ax^2+bx+c\) in \(x\) has a minimum or maximum (depending on \(a\gt0\) or \(a\lt0\)) for \(x=-\frac{b}{2a}\), which is \(-\frac{b^2-4ac}{4a}\).

Let’s practice!

Solve \(x\) by completing the square in the equation

\(x^2-6 x-9=0\)

Give your answer in the form \(x=a\lor x=b\).

To find this answer, we complete the square in the following manner:

\(\begin{array}{rcl} (x-3)^2-3^2-9&=&0\\ &&\phantom{xx}\color{blue}{\text{ }x^2-6x\text{ completed to a square}}\\ (x-3)^2&=&3^2+9\\ &&\phantom{xx}\color{blue}{\text{everything outside of the brackets moved to the right hand side}}\\ (x-3)^2&=&18\\ &&\phantom{xx}\color{blue}{\text{simplified the right hand side}}\\ x-3=\sqrt{18} &\lor&x-3=-\sqrt{18}\\ &&\phantom{xx}\color{blue}{\text{the root taken at both sides}}\\ x=3\cdot \sqrt{2}+3&\lor& x=3-3\cdot \sqrt{2}\\ &&\phantom{xx}\color{blue}{-3 \text{ subtracted from both sides}}\\ \end{array}\)


\(x=3\cdot \sqrt{2}+3\lor x=3-3\cdot \sqrt{2}\)

Solving systems of equations by addition

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A general method of solving two linear equations with two unknowns is by eliminating one unknown. Here we discuss a method that is also suitable for larger systems of linear equations, with more unknowns.

The goal is to give the first equation the form \(x = a\) and the second equation the form \(y = b\).

  • The strategy is to edit the equations in such a way that the new system will be equivalent to the old;
  • the new system looks more like a solution than the old one.

Steps that will occur are multiplying all terms in the same equation by the same non-zero number and subtracting one equation over the other.

The addition method for linear equations

A system of two linear equations with unknowns \(x\) and \(y\) can be solved as follows.

  1. Make sure \(x\) occurs in the first equation: if this is not the case, then switch the two equations; this way, \(x\) will occur in the first equation.
  2. Replace the second equation by the difference of this equation with a suitable multiple of the first equation, in such a way that \(x\) no longer occurs in the second equation.
  3. Replace the first equation by the difference of this equation with a suitable multiple of the second equation in such a way that \(y\) no longer occurs in the first equation.
  4. The first equation is now a linear equation with \(x\) as the only unknown, and the second is a linear equation with \(y\) as the only unknown. These equations can be solved with the theory of linear equations with one unknown.

Let’s practice!

Solve the following system of equations with unknowns \(x\) and \(y\).

\(\left\{\begin{array}{l}-5·\left\{x\right\}+12·\left\{y\right\}+3=0\\ 2·\left\{x\right\}-5·y+1=0\end{array}\right.\)

Give the answer in the form \(x=a\land y=b\) for suitable values of \(a\) and \(b\).

There are many ways to get to this solution. We will describe one.

  • To make sure that the unknown \(x\) is present in the first equation, we switch the two equations if this was not the case in the original system: \(-5\cdot x+12\cdot y+3=0\land 2\cdot x-5\cdot y+1=0\).
  • Next we get rid of the term with \(x\) from the second equation by multiplying the first equation with \(\frac{2}{-5}\) and subtracting from the second: \(-5\cdot x+12\cdot y+3=0\land -{{y}\over{5}}=0\).
  • By dividing (left and right hand side) in the second equation by \(-\frac{1}{5}\) we find \(y = 11\). We now have the system \(-5\cdot{x}+12\cdot y+3=0\land y=11\).
  • If we enter the solution of \(y\) (the second equation) in the first equation (or stated different: we subtract \(12\) times the second equation from the first), then we find the system: \(135-5\cdot x=0\land y=11\).
  • The first equation can be solved with the theory of linear equations with one unknown. The result is \(x= 27\land y = 11\).


\(x= 27\land y = 11\)

The equation of a line

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Assume that \(a\), \(b\), and \(c\) are constant real numbers: parameters.

The solution to the equation \(a\cdot x+b\cdot y+c=0\) can be drawn in the plane. They are the points \(\left[x,y\right]\) satisfying \(a\cdot x+b\cdot y+c=0\).  If \(a\ne0\) or \(b\ne0\), then these points form a straight line, or simply just a line.

  • If \(b\ne0\), then the equation can be written as \(y=-\frac{a}{b}\cdot{x}-\frac{c}{b}\). For, these are the solutions if we consider \(x\) as a parameter and \(y\) as unknown. This indicates that for every value of \(x\) there is a point \(\left[x,y\right]\) with \(y\) equal to \(-\frac{a}{b}\cdot{x}-\frac{c}{b}-\frac{a}{b}\cdot{x}-\frac{c}{b}\).
    • If \(a\ne0\), the line is oblique (by oblique we mean neither horizontal nor vertical).
    • If \(a = 0\), then the value of \(y\) is constant, equal to \(-\frac{c}{b}\). In this case the line is horizontal.
  • In the exceptional case \(b = 0\) the equation looks like \(a \cdot{x}+c = 0\).
    • If \(a\ne0\), then the line is vertical.
    • If \(a = 0\) and
      • \(c\ne0\), then there are no solutions
      • \(c = 0\), then each pair of values of \(\left[x,y\right]\) is a solution.

A straight line can be described in different ways.

  1. The solutions \(\left[x,y\right]\) to an equation \(a\cdot x+b\cdot y+c=0\) with unknowns \(x\) and \(y\). Here \(a\), \(b\) and \(c\) are real numbers such that \(a\) and \(b\) are not equal to zero.
  2. The line through two given points in the plane: if \(P = \left[x,y\right]\) and \(Q = \left[s,t\right]\) are points in the plane, then the line through \(P\) and \(Q\) has equation \(a\cdot x+b\cdot y+c=0\) with \(a=q-t\), \(b = s – p\) and \(c = t \cdot {p} – q \cdot{s}\).
  3. The line through a given point, the base point, and a direction, indicated by the number \(-\frac{a}{b}\), where \(a\) and \(b\) are as in the equation given above; this number is called the slope of the line.
  4. The line with function representation \(y = p\cdot x+q\) if \(b\ne0\) and \(x = r\) otherwise; here we have \(p = -\frac{a}{b}\) (the slope), \(q = -\frac{c}{b}\) (the intercept), which is the value of \(y\) for \(x = 0\) and \(r = -\frac{c}{a}\) in terms of the above \(a\), \(b\) and \(c\). This can be seen as a special case of the previous description, with base point \(\left[0,q\right]\). In the case where \(b\ne0\), the variable \(y\) is a function of \(x\), in the other case, \(x\) is a constant function of \(y\).

Let’s practice!

The line segment between the two points \(\left[2,\frac{77}{12}\right]\) and \(\left[6,\frac{63}{4}\right]\) is drawn in the figure below.

equation of a line

Give the function rule for this line in the form \(y = a\cdot{x} + b\).

The line is described by the function rule \(y = a\cdot{x} + b\) where \(a\) is the slope, given as the quotient of the difference of the \(y\)-values with the difference of the \(x\)-values of two points on the line. Hence \(a=\frac{{{63}\over{4}}-{{77}\over{12}}}{6-2}={{7}\over{3}}\). The value of \(b\) follows from \(b = y – a\cdot{x}\), where \(\left[x,y\right]\) is a random point on the line. Hence \(b={{77}\over{12}}-{{7}\over{3}}\cdot{2}={{7}\over{4}}\).




Arithmetic operations for continuity

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Let’s discuss some methods to create a new continuous function from known continuous functions.

Continuity of sums, products, quotients and compositions of continuous functions

Let \(a\) be a real number.

  1. Suppose \(f\) and \(g\) are functions continuous in \(a\). Then, the functions \(f + g\) and \(f \cdot g\) are also continuous in \(a\). If \(g(a)\ne0\), then we have the same for \(\frac{f}{g}\).
  2. Suppose \(f\) is a function in \(a\) and that \(g\) is a function continuous in \(f(a)\). Then the composition \({g}\circ{f}\) is continuous in \(a\).

Continuity of power functions

Let \(d\) be a real number. If \(f\) is a function that is continuous in \(a\) with \(f(a)>0\), then \(f(x)^d\) is also continuous in \(a\).

Let’s practice!

Consider the function rules \(f(x)=x^2\) and \(g(x)=x+1\).

What is the function rule of the composition \(f\circ g\)?

\(\begin{array}{rcl} f \circ g(x) &=& f \left(g(x) \right) \\ &&\phantom{xyzuvw}\color{blue}{\text{composition of functions}} \\ &=& g(x)^2 \\ &&\phantom{xyzuvw}\color{blue}{\text{function rule of }f\text{ entered}} \\ &=&\left(x+1\right)^2 \\ &&\phantom{xyzuvw}\color{blue}{\text{function rule of }g\text{ entered}} \\ \end{array}\)


\(f\circ g(x) = \left(x+1\right)^2\)

Let’s practice once more!

Let \(f\) be the function with rule \(f(x) = {{1}\over{x^2+4}}\) and \(g\) the function with rule \(g(x)=x^2\).

Determine the function rule for \({f – g}\).

\(\left({f – g}\right)(x)=f(x)-g(x)=\left({{1}\over{x^2+4}}\right)-\left(x^2\right)={{-x^4-4\cdot x^2+1}\over{x^2+4}}\)


\(\left({f – g}\right)(x)\,=\, {{-x^4-4\cdot x^2+1}\over{x^2+4}}\)


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Let \(f\) be a real function defined on an open interval around point \(a\).

If \(\displaystyle \lim_{x\to a}f(x) =f(a)\), then \(f\) is called continuous at \(a\). If not, then \(f\) is called discontinuous at \(a\).

If \(f\) is continuous at every point of an open interval \((c.d)\), then \(f\) is called continuous on \((c.d)\).

In other words, a function is continuous if the graph can be drawn without lifting the pen from the paper.

Let’s practice!

Consider the function \(f\) given by

\(\displaystyle f(x)=\begin{cases} 8 x+9&\text{if }x\lt1\\ 9 x^2+8&\text{if }x\geq1\end{cases}\)

Is this function continuous at \(1\)?

The function \(f\) is left continuous at \(x = 1\):

\(\begin{array}{rcl} \displaystyle\lim_{x\uparrow1}f(x)&=&\lim_{x\uparrow1}( 8x+9)=8+9=17=f(1)\end{array}\)

The function \(f\) is right continuous at \(x = 1\):

\(\begin{array}{rcl}\displaystyle\lim_{x\downarrow1}f(x)&=&\lim_{x\downarrow1} (9 x^2+8)=9+8=17=f(1)\end{array}\)

Since \(f\) is both right continuous and left continuous at \(1\), it is continuous at \(1\).

The graph of \(f\) is drawn below. It has no jump at \(1\).

Continuity image 1.png


Yes, this function is continuous at \(1\)

The notion of limit

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Let \(a\) and \(b\) be real numbers and let \(f\) be a real function that is defined on an open interval containing \(a\).

We say that \(f\) has limit \(b\) at \(a\) if \(f(x)\) comes closer to \(b\) as \(x\) comes closer to \(a\).

In this case, we write \(\textstyle\lim_{x\to a} f(x) = b\) or \(\displaystyle\lim_{x\to a} f(x) = b\).

Let’s practice!

The rational function \(f(x) = \frac{2\cdot x-16}{x-8}\) is defined everywhere except at \(8\).

What is the limit of \(f\) at \(8\)?

For every value of \(x\), \(f(x)=2\) is close to (but distinct from) \(8\).


\(\lim_{x\to 8}f(x)= 2\)

The range of a function

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Consider the real function \(f\).

The set of all values \(f(x)\) for \(x\) in the domain of \(f\) is called the range of \(f\).

If \(p\) is a real number in the domain of \(f\) with \(f(p)=0\), then \(p\) is called a zero of \(f\).

The range of a function depends on the domain of the function: the larger we choose the domain, the larger in general the range will be.

Let’s practice!

Consider the real function \(f(x) = \sqrt{x-9}+4\).

The largest possible domain of \(f\) has the form \(x\ge a\) for a certain number \(a\); with other words: it is of the form \(\left[a,\infty \right]\).

The range of \(f\) on this domain is of the form \(\left[b,\infty \right]\) for a certain number \(b\). Hence, it consists of all \(y\) with \(y\ge b\).

Determine \(a\) and \(b\).

The largest possible domain of \(f\) is the set of values of \(x\) at which \(f(x)\) is defined. Because the argument of the root has to be greater than or equal to zero, the function \(f(x) = \sqrt{x-9}+4\) is only defined for \(x-9\ge0\), or \(x\ge 9\). Therefore, the largest possible domain of \(f\) is \(x\ge 9\).

The range is the set of values that \(f\) can take on this domain. The smallest value \(f\) can take \(f(9) = 4\). Hence, the range of \(f\) is \(y\ge 4\).


\(a = 9\) and \(b = 4\)