Summary
This lesson opens the unit on algebraic functions and polynomials. It presents its nine lessons and the thread that connects them, which is factorization: on it depend the roots, the sign, the inequalities, the domains, the asymptotes and the graphs. It then goes through the historical background of these ideas, from the second-degree problems of the Babylonian tablets and the algebra of al-Khwarizmi to the Italian formulas of the sixteenth century, the notation of Viète and Descartes and the concept of function of Euler and Dirichlet; it formulates precisely the two problems that gave rise to them, solving a polynomial equation and describing a function, and shows with an example how factorization solves both at once. It ends by pointing out where these ideas are used in the rest of the course, in other courses on the site and in science and technology.
Learning objectives
- Describe the content of each lesson of the unit and explain the role that the factorization of polynomials plays in the whole.
- Place the main results of the unit in their historical context, naming their protagonists and the problem each of them was trying to solve.
- Formulate the problem of solving a polynomial equation and that of describing an algebraic function, and recognize in an example how a single factorization answers both.
- Identify where these ideas reappear in the following units, in other courses and in scientific and technical applications.
Entry assessment
Before working through the unit, it is advisable to measure the starting point. This self-assessment is not graded: the first nine questions cover, one by one, the nine lessons of the unit, and the tenth asks you to set up a concrete situation as a function; the same skills are measured again at the end, in the synthesis lesson. The last three check what the unit takes as known from the unit on number systems. Each wrong answer indicates the lesson on which it is advisable to dwell.
Before starting the unit, measure your starting point. It is a short self-assessment: it is not graded and its result is stored only in your browser.
Question 1What is the domain of the function ?
The number has been excluded, as if the root of did not exist. A root of even index requires a nonnegative radicand, not a positive one, when it is in the numerator.The condition on the denominator is missing: at the quotient divides by zero and is not defined.The inequality was reversed when solving: is equivalent to , because moving to the other side leaves .The radicand must be nonnegative, , that is, ; and the denominator cannot vanish, which excludes . At the root equals and the quotient is defined.Question 2The function , restricted to , is bijective onto . What is its inverse?
This is the inverse of the other half of the parabola, the restriction to : it takes values less than or equal to , of which only lies in the domain of the given restriction.The operations have been undone with the sign of the shift changed: to go back from to one must add , not subtract it. For example, , but this formula gives at .This is the multiplicative inverse, not the inverse function: the exponent of is only a notation for the function that undoes .Solving, ; since in the restriction , the number is the principal root of , and not its opposite. One checks that .Review: Function operations, inverse functions and graph transformations
Question 3For what value of is the polynomial divisible by ?
Correct answer:
By the factor theorem, divides the polynomial if and only if the polynomial vanishes at : , whence . Anyone who obtains has evaluated at , which is the root of , not that of .Review: The algebra of real polynomials
Question 4If and are the roots of , what is the value of ?
Correct answer:
By Vieta’s formulas, and ; then . Anyone who obtains has confused with , and anyone who obtains has added instead of subtracting it.Review: Factoring quadratic polynomials and trinomials of degree 2n
Question 5Let and be integers. According to the rational root theorem, which of the following numbers cannot be a root of ?
This number is indeed a candidate: divides and divides . Perhaps the roles of the constant term and the leading coefficient have been swapped.It is a candidate: divides and divides . The theorem does not exclude negative signs, because an integer divides another exactly when its opposite does.It is an integer candidate: divides , and its denominator is , which divides every integer. Whether it is a root will depend on and on .A rational root in lowest terms, , is such that divides the constant term, , and divides the leading coefficient, . In , the numerator does not divide , nor does the denominator divide .Review: Roots of higher-degree polynomials and synthetic division
Question 6Which numbers and satisfy for every real number other than and ?
These are the values of the numerator at the roots, without dividing by the value of the other factor. Covering leaves the factor in view as well, and it is evaluated together with the numerator.The sign of the denominator was lost when evaluating at : the factor equals there, and the quotient of by is positive.The coefficients of the numerator have been shared out between the fractions. The sum has numerator , not .By the cover-up method, is the value at of , which is , and is the value at of , which is . At both sides equal .Question 7What is the solution set of the inequality ?
The number still has to be excluded: there the factor vanishes and the quotient equals , which is not negative. The double root does not change the sign, but it is still a zero.The double root has been treated as if it changed the sign. A factor with an even exponent is positive on both sides of its root, and the sign does not change when crossing .These are the points where the quotient is positive: the direction of the inequality was reversed when reading the sign chart.The quotient has the sign of away from , because the factor is positive there; that quotient is negative between and . At the numerator vanishes, and the strict inequality excludes it.Review: Sign charts and inequalities
Question 8What is the oblique asymptote of the graph of ?
Only the first two terms of the numerator have been divided by , as if the denominator were . One must divide by , and the term of the divisor changes the quotient.This is the quotient of the leading coefficients, which gives a horizontal asymptote only when the degrees are equal. Here the numerator has one degree more, and the asymptote is oblique.This is the vertical asymptote, which corresponds to the zero of the denominator. The oblique asymptote describes the function when tends to or to .Division gives , so that , and the fraction tends to at both ends.Question 9What is the asymptote of the graph of as tends to ?
This is the asymptote as tends to . For negative and large in absolute value, ; moreover, a root is never negative, and the line is negative for .The sign of has been changed without changing that of the constant term. The absolute value changes the sign of the whole expression .The constant term is missing, which comes from completing the square; the difference between the function and tends to , not to .Completing the square, , and the root approaches , which for is . The difference is , which tends to .Question 10A tank contains liters of water with grams of sugar per liter. From a given instant on, liters per minute of a syrup with grams of sugar per liter are added to it, without any liquid leaving, and the mixing is immediate. What values does the sugar concentration, in grams per liter, take from that instant on?
The line is a horizontal asymptote: the fraction never equals , so the mixture never comes to have the concentration of the incoming syrup.The sugar that the tank contained at the start has been forgotten, as if it were pure water: at the initial instant the concentration is , not , and from there it increases.The two concentrations, and , have been averaged, as if equal volumes were mixed. The volume of syrup added grows without bound, and its concentration ends up dominating the mixture.After minutes there are grams of sugar in liters: . The fraction decreases from towards without ever vanishing: the concentration increases from and approaches , that of the incoming syrup, without reaching it.Question 11From earlier unitsWhat is the solution set of the inequality ?
Only has been solved, forgetting that the absolute value also requires .This is the solution of . An absolute value less than describes the points close to , which form an interval.The inequality is equivalent to ; adding and dividing by gives .Question 12From earlier unitsCompute .
Correct answer:
By definition, , and the negative exponent indicates the reciprocal: . Anyone who obtains has read the sign of the exponent as the sign of the result.Question 13From earlier unitsWhat is the solution set of the inequality ?
When dividing by the negative number the direction of the inequality must be reversed, and here it has been kept.The minus sign in front of the second fraction affects its whole numerator: , not .Multiplying by , which is positive, , that is, ; dividing by reverses the direction: .Review: First-degree equations and inequalities, and linear systems
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