Sum and Product of Roots of Polynomials

Like all cubics, this polynomial is of degree \(3\) so \(n=3\) and the coefficients are: \[a_3 = 2, \ a_2 = -8, \ a_1 = 2, \ a_0 = 12\] so the formula for the sum and product of the roots become: \[\text{Sum of Roots} = \frac{-a_{2}}{a_3} \quad \text{and} \quad \text{Product of Roots} = \begin{pmatrix}-1\end{pmatrix}^n.\frac{a_0}{a_3}\]

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Checking it Works

Before we answer the questions, we notice that (since this is a quadratic) we're dealing with a polynomial of degree \(2\) and its coefficients are: \[a_2 = 3, \ a_1 = 5, \ a_0 = -4\] The formula for the sum and the product of the roots therefore lead to:

Now that we've made a note of these two results, we move on to answering the two questions.

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1. The trick here is to notice that the expansion of \(\begin{pmatrix}p+6\end{pmatrix}\begin{pmatrix}q+6\end{pmatrix}\) leads to an expression that can be written in terms of the sum of the roots as well as the product of the roots \(p\) and \(q\). Indeed: \[\begin{aligned} \begin{pmatrix}p+6\end{pmatrix}\begin{pmatrix}q+6\end{pmatrix} & = pq+6p+6q+36 \\ & = pq + 6\begin{pmatrix}p+q\end{pmatrix} + 36 \end{aligned}\] and so using the fact that \(p+q = \frac{-5}{3}\) and \(pq = \frac{-4}{3}\) we obtain: \[\begin{aligned} \begin{pmatrix}p+6\end{pmatrix}\begin{pmatrix}q+6\end{pmatrix} & = pq + 6\begin{pmatrix}p+q\end{pmatrix} + 36 \\ & = \frac{-4}{3} + 6 \times \begin{pmatrix} \frac{-5}{3}\end{pmatrix} + 36 \\ & = \frac{-4}{3} -\frac{6\times 5}{3} + 36 \\ & = \frac{-4}{3} - \frac{30}{3} + \frac{108}{3} \\ & = \frac{-4-30+108}{3} \\ \begin{pmatrix}p+6\end{pmatrix}\begin{pmatrix}q+6\end{pmatrix} & = \frac{74}{3} \end{aligned}\]

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2. The key result to make a note of for this question is the fact that \(p^2+q^2\) can be written as a combination of the sum and product of \(p\) and \(q\) using: \[p^2+q^2 = \begin{pmatrix} p+q \end{pmatrix}^2-2pq\] Indeed using this we can write: \[\begin{aligned} p^2+q^2 & = \begin{pmatrix} p+q \end{pmatrix}^2-2pq \\ & = \begin{pmatrix} \frac{-5}{3} \end{pmatrix}^2 - 2\times \begin{pmatrix} \frac{-4}{3} \end{pmatrix} \\ & = \frac{(5)^2}{3^2} - 2 \times \begin{pmatrix}\frac{-4}{3} \end{pmatrix} \\ & = \frac{25}{9} - \frac{-8}{3} \\ & = \frac{25}{9} + \frac{8}{3} \\ & = \frac{25}{9} + \frac{24}{9} \\ p^2+q^2 & = \frac{49}{9} \end{aligned}\]

Since we need to find 2 unknowns: we'll need to use 2 distinct equations. Furthermore since we're given all of this polynomial's roots we can use the sum and product of the roots formula. Given \(p(x)=ax^4-15x^3+21x^2+9x+b\) is a polynomial of ddegree \(4\), the leading coefficient is \(a_4=a\) and we can state: \[a_4 = a, \ a_3 = -15, \ a_2 = 21, \ a_1 = 9, \ a_0 = b\]

Finally, now that we have \(a\) and \(b\) we can write the polynomial's full equation: \[p(x)=3x^4-15x^3+21x^2+9x-30\]

We're dealing with a cubic polynomial, so the degree is \(3\), and the coefficients are: \[a_3 = 1, \ a_2 = -k, \ a_1 = 0, \ a_0 = -3k\] It's worth pointing out that the information, given in the question, is the product of two of the three roots: \[\alpha \beta = -\frac{k^2}{4}\] This gives us the product \(\alpha \beta \) in terms of \(k\), which suggests we could find \(k\) in two steps:

• Step 1: write the third root \(\alpha + \beta \) in terms of \(k\) and solve it.
• Step 2: use the fact that the product of the roots, \(\alpha \beta \begin{pmatrix}\alpha + \beta \end{pmatrix}\) is equal to \(\frac{(-1)^3a_0}{a_3}\) to write an equation for \(k\)

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For step 1 we consider the sum of the roots.

Sum of the Roots

\[\begin{aligned} \text{Sum of Roots} & = \alpha + \beta + \alpha + \beta \\ \text{Sum of Roots} & = 2\begin{pmatrix}\alpha + \beta \end{pmatrix} \quad \text{(E1)} \end{aligned}\] and we also know that: \[\begin{aligned} \text{Sum of Roots} & = \frac{-a_2}{a_3} \\ & = \frac{-(-k)}{1} \\ & = \frac{k}{1} \\ \text{Sum of Roots} & = k \quad \text{(E2)} \end{aligned}\] Equating (E1) and (E2), we obtain: \[2\begin{pmatrix}\alpha + \beta \end{pmatrix} = k\] And we can divide both sides by \(2\) to find an expression for the root \(\alpha + \beta\) in terms of \(k\): \[\begin{pmatrix}\alpha + \beta \end{pmatrix} = \frac{k}{2} \quad \text{(E3)}\]

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Product of the Roots

\[ \text{Product of the Roots} = \alpha \beta \begin{pmatrix} \alpha + \beta \end{pmatrix} \] Now using the fact that we are told (in the question) that \(\alpha \beta = -\frac{k^2}{4}\) and the result found in (E3), we can write the product of the roots in terms of \(k\) as: \[\begin{aligned} \text{Product of the Roots} & = \underbrace{\frac{-k^2}{4}}_{\alpha \beta} \times \underbrace{\frac{k}{2}}_{\alpha + \beta}\\ & = \frac{-k^3}{8} \quad \text{(E4)} \end{aligned}\] And the product of the roots can also be written in terms of the, given, polynomial's coefficients: \[\begin{aligned} \text{Product of Roots}& = \frac{(-1)^3a_0}{a_3} \\ & = \frac{(-1)^3 \times (3k)}{1} \\ & = \frac{-1\times (3k)}{1} \\ & = \frac{-3k}{1} \\ \text{Product of Roots} & = -3k \quad \text{(E5)} \end{aligned}\] Equating (E4) and (E5) we obtain an equation for \(k\): \[\frac{-k^3}{8} = -3k \] Now we solve: \[\begin{aligned} & \frac{-k^3}{8} = -3k \quad \text{(multiply both sides by \(-8\))} \\ & k^3 = 24k \quad \text{(subtract \(24k\) from boths sides)} \\ & k^3 - 24 k = 0 \quad \text{(write in factored form)} \\ & k\begin{pmatrix}k^2 - 24 \end{pmatrix} = 0 \end{aligned}\] For this product to equal zero, we must either have:

• \(k=0\), which won't do since we're told \(k>0\) in the question, or
• \(k^2 - 24 = 0\), which leads to: \[\begin{aligned} & k^2 - 24 = 0 \\ & k^2 = 24 \\ & k = \pm \sqrt{24} \\ & k = \pm 2\sqrt{6} \end{aligned}\] This leads to two values for \(k\), \(-2\sqrt{6}\) and \(2\sqrt{6}\) but since we know that \(k>0\) only the second option will do.

Finally we can state the answer: \[k=2\sqrt{6}\]