Sample Problem 1. Find the sum of the roots, not necessarily distinct, of the quadratic \[f(x)=x^2+3x+1.\]
When looking at this problem, the first solution that comes to mind is just to find the roots and to add them up. By the quadratic formula, we see that the roots of the polynomial are \[\frac{-3\pm \sqrt{3^2-4}}{2}.\] Adding the roots together, we find their sum to be \(-3\). After noticing that this answer is merely \(-1\) times the coefficient of the middle term, we may wonder if this is always true. Indeed, it is (at least it is if the coefficient of the \(x^2\) term is 1), and it is relatively easy to prove.
Sample Problem 2. Find the sum of the roots, not necessarily distinct, of the
quadratic \[f(x)=ax^2+bx+c.\]
Using the same method as we did in the previous problem, we get that the roots are \[\frac{-b \pm\sqrt{b^2-4ac}}{2a}.\] Next, note that \(\frac{-b+\sqrt{b^2-4ac}}{2a} +\frac{-b -\sqrt{b^2-4ac}}{2a}=\frac{-b +\sqrt{b^2-4ac}-b-\sqrt{b^2-4ac}}{2a}=\frac{-2b}{2a}=\frac{-b}{a.}\) We may wonder if this can be extended to the product of the roots.
Sample Problem 3. Find the product of the roots, not necessarily distinct, of
the quadratic \[f(x)=ax^2+bx+c.\]
Just as before, we know that the roots of the quadratic are \[\frac{-b \pm\sqrt{b^2-4ac}}{2a}.\] Next, we multiply and use the difference of squares to find that \[\frac{-b +\sqrt{b^2-4ac}}{2a}\cdot \frac{-b -\sqrt{b^2-4ac}}{2a}=\frac{(-b)^2-(\sqrt{b^2-4ac})^2}{(2a)^2}=\frac{c}{a}.\] Now that both of these expressions turned out nicely, we wonder if there is a nicer explanation that could generalize this to higher degree polynomials.
Sample Problem 4. Relate the sum of the roots and the coefficient of the \(x^2\) term in (x+1)(x+2)(x+3).
To start, we will think about how to expand this in a way that may be slightly different than the way you are used to, in order to not expand the whole thing. When we're expanding a bunch of binomials multiplied by each other, the result will be the sum of each possible combination of choosing one of the two terms in each of the binomials and multiplying the chosen terms together. For example, in this situation, we can choose either the x term or the constant term from each binomial to multiply into a factor. Because we want the \(x^2\) term, we need to choose the x term from two of the binomials and the constant term from the other. We can do this in three ways, depending on which binomial we choose a constant from. Thus, the only relevant terms we get from expanding this are \(x\cdot x \cdot 3\), \(x\cdot 2 \cdot x\), and \(1\cdot x \cdot x\). Thus, we get the coefficient of the \(x^2\) term to be 1+2+3. We note that this coefficient is just -1 times the sum of the roots.
Let's look at expanding a quadratic using the bolded method to make sure you understand it. We will expand (x+1)(x+2). We can choose either a 1 or an x to multiply in from the first binomial and a 2 or an x from the second. Thus, we get the possible combinations of choosing one term from each binomial to be (x)(x), (x)(2), (1)(x), and (1)(2). Adding these together we get the expanded term to be \(x^2+3x+2\). If you don't understand why this method works, think about starting with the binomial (a+b). When we multiply (a+b) by (c+d), we get (a+b)(c+d)=c(a+b)+d(a+b)=ca+cb+da+db. In other words, we can multiply each term of our first binomial by either of the terms in the polynomial we multiplied it by, and then we add up each case.
Let's look at expanding a quadratic using the bolded method to make sure you understand it. We will expand (x+1)(x+2). We can choose either a 1 or an x to multiply in from the first binomial and a 2 or an x from the second. Thus, we get the possible combinations of choosing one term from each binomial to be (x)(x), (x)(2), (1)(x), and (1)(2). Adding these together we get the expanded term to be \(x^2+3x+2\). If you don't understand why this method works, think about starting with the binomial (a+b). When we multiply (a+b) by (c+d), we get (a+b)(c+d)=c(a+b)+d(a+b)=ca+cb+da+db. In other words, we can multiply each term of our first binomial by either of the terms in the polynomial we multiplied it by, and then we add up each case.
Sample Problem 5. Find the sum of the roots, not necessarily distinct, of the
polynomial \[f(x)=a_kx^k+a_{k-1}x^{k-1}+ \ldots + a_1x+a_0.\]
This time, we can't just solve for the roots and add them, so we will need to be clever. When stuck on a problem, it is often helpful try to represent the given information in a different way. Because we are dealing with the roots of a polynomial, we think of using the Fundamental Theorem of Algebra. By the Fundamental Theorem of Algebra (look this up if you want more detail, but the proof of this theorem is beyond the scope of this curriculum), we have a unique factorization of our polynomial \[f(x)=a_k(x-r_1)(x-r_2)(x-r_3)\ldots(x-r_k),\] where each \(r\) is a root of \(f(x)\). Looking at problems 2 and 4, we think we might get \[r_1+r_2+r_3+...+r_k=-\frac{a_{k-1}}{a_k}\] because, in both problems, we found the sum of the roots to be -1 times the second rightmost coefficient divided by the first coefficient. We can note that the \(a_{k-1}\) is the coefficient of the \(x^{k-1}\) term in the first representation of our polynomial. Thus, we want to find the coefficient of the \(x^{k-1}\) term in the second representation of our polynomial. By thinking about what the polynomial would look like if we expanded it and only paying attention to terms in the form of \(cx^{k-1}\) for some constant c, we can see that the coefficient is equal to \(-a_k(r_1+r_2+r_3+\ldots)\). This is because, of the k factors of \(f(x)\) that can contribute an \(x\) or a \(-r_i\), exactly \(k-1\) must contribute an x and the other must contribute a \(-r_i\) for some \(i\) in order to produce one of the terms we are looking at. Therefore, we see that \[-a_k(r_1+r_2+r_3+...+r_k)=a_{k-1},\] and thus, \[r_1+r_2+r_3+...+r_k=-\frac{a_{k-1}}{a_k}.\] Seeing two coefficients represent operations done on the roots, we are curious if we can find more generalizations.
Sample Problem 6. Generalize what we found in sample problem 4 to the
other coefficients of higher degree polynomials.
By looking at our result in problem 3, we can apply similar methods by looking at the \(x^{k-i}\) term (think about how many non-x terms need to multiply together to form one of those terms) for \(i>1\). By using our alternate method of expansion to express each coefficient in terms of the roots of the polynomial, we get the following equations.
\[\bf{r_1+r_2+r_3+...+r_k=-\frac{a_{k-1}}{a_k}}\]
\[\bf{(r_1r_2+r_1r_3+r_1r_4+...+r_1r_k)+(r_2r_3+r_2r_4+r_2r_5+...+r_2r_k)+...+(r_{k-1}r_k)=\frac{a_{k-2}}{a_k}}\]
\[\ldots\]
\[\bf{r_1r_2r_3...r_k=(-1)^k\frac{a_{0}}{a_k}}\]
To be clear, we start with the sum of the roots, then go to the sum of every possible product of 2 roots, then the sum of every possible product of 3 roots, and so on, until we get to the sum of every possible product of k roots. These are formulas are called Vieta's Formulas, and they can be used for a wide variety of problems.
\[\bf{r_1+r_2+r_3+...+r_k=-\frac{a_{k-1}}{a_k}}\]
\[\bf{(r_1r_2+r_1r_3+r_1r_4+...+r_1r_k)+(r_2r_3+r_2r_4+r_2r_5+...+r_2r_k)+...+(r_{k-1}r_k)=\frac{a_{k-2}}{a_k}}\]
\[\ldots\]
\[\bf{r_1r_2r_3...r_k=(-1)^k\frac{a_{0}}{a_k}}\]
To be clear, we start with the sum of the roots, then go to the sum of every possible product of 2 roots, then the sum of every possible product of 3 roots, and so on, until we get to the sum of every possible product of k roots. These are formulas are called Vieta's Formulas, and they can be used for a wide variety of problems.
Exercises
- For positive integer \(n\), the \(nth\) roots (real and complex) of \(1\) are called the \(nth\) roots of unity. For example, the \(4th\) roots of unity are \(i\), \(-i\), \(-1\), and \(1\). Given that there are \(n\) distinct \(nth\) roots of unity (this can be proved relatively easily using the complex number plane; you can find more extensive information on this with a quick google search), find the sum of the \(nth\) roots of unity for \(n>1\).
- Find the product of all possible \(q\) if \(a\),\(b\), and \(c\) are the roots of \[3x^3+5x^2+2x+1\] and \[\frac{q}{a+bc+b+ac+c+ab}+\frac{q^2}{a^2+b^2+c^2+2ab+2bc+2ac+a+b}+\frac{q^3} {a^2bc+b^2ac+c^2ab}+\frac{a^2+b^2+c^2}{a+b+c}=0\] .
Solution 1
Note that the \(nth\) roots of unity are the solutions to \(x^n=1\), or \(x^n-1=0\). By Vieta's Formulas, the sum of the roots of this polynomial is equal to the coefficient of the \(x\) term divided by the coefficient of \(x^n\). However, because \(n>1\), the \(x\) term has a coefficient of \(0\) and \(x^n\) has a coefficient of 1. Thus, the desired sum is \(\frac{0}{1}=0\).\(\blacksquare\)
Solution 2
Note that, by Vieta's Formulas, the product of all of possible values of \(q\) is negative the value of the constant term divided by the coefficient of the \(q^3\) term in the second equation.
First of all, let's try to find the constant term. By Vieta's Formulas on the first equation, we get \(a+b+c=-\frac{5}{3}\). Next, note that \(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=\frac{25}{9}-\frac{4}{3}=\frac{13}{9}\). Thus, the constant term is \(\frac{\frac{13}{9}}{-\frac{5}{3}}=-\frac{13}{15}\).
Next, we will find the \(q^3\) term's coefficient. Notice that \[a^2bc+b^2ac+c^2ab=(a+b+c)(abc)=-\frac{5}{3} \cdot -\frac{1}{3}=\frac{5}{9}.\] Thus, the \(q^3\) term's coefficient is \(\frac{1}{\frac{5}{9}}=\frac{9}{5}\).
Therefore, the final answer is \[-\frac{-\frac{13}{15}}{\frac{9}{5}}=\frac{13}{27}.\blacksquare\]
First of all, let's try to find the constant term. By Vieta's Formulas on the first equation, we get \(a+b+c=-\frac{5}{3}\). Next, note that \(a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=\frac{25}{9}-\frac{4}{3}=\frac{13}{9}\). Thus, the constant term is \(\frac{\frac{13}{9}}{-\frac{5}{3}}=-\frac{13}{15}\).
Next, we will find the \(q^3\) term's coefficient. Notice that \[a^2bc+b^2ac+c^2ab=(a+b+c)(abc)=-\frac{5}{3} \cdot -\frac{1}{3}=\frac{5}{9}.\] Thus, the \(q^3\) term's coefficient is \(\frac{1}{\frac{5}{9}}=\frac{9}{5}\).
Therefore, the final answer is \[-\frac{-\frac{13}{15}}{\frac{9}{5}}=\frac{13}{27}.\blacksquare\]