If $x^4-8x^3+24x^2-32x-14=0$ has two real roots $x_1, x_2$ and two non-real roots $x_3, x_4$, find the value of $x_1x_2+x_3x_4$

The equation $$x^4-8x^3+24x^2-32x-14=0$$ has two real roots $$x_1, x_2$$ and two non-real roots $$x_3, x_4$$. Find the value of $$x_1x_2+x_3x_4$$.

I took $$x_1, x_2$$ as conjugate pairs and $$x_3$$ and $$x_4$$ conjugate pairs but what I got was not enough to solve the question I then tried by trail and error and got 8 as the answer but I am highly doubtful about it. Please if anyone who can come up with a proper method.

• Hello. I took the liberty of formatting your math equations with latex. Apr 24, 2023 at 11:40
• The two real roots are not conjugate. Apr 24, 2023 at 11:43
• A natural starting point is $$24=x_1 x_2 + x_1 x_3 + x_1 x_4 + x_2 x_3 + x_2 x_4 + x_3 x_4=(x_1x_2+x_3x_4)+(x_1+x_2)(x_3+x_4)$$ but I cannot continue.
– lhf
Apr 24, 2023 at 12:56
• I think the intent of the question is to solve it without finding the roots.
– lhf
Apr 24, 2023 at 13:50
• @lhf I posted an answer that does not require finding the roots. Basically, it's due to the cubic resolvent being reducible over $\mathbb{Q}$, so the Galois group of the irreducible polynomial in question is simply the dihedral group of order $8$. Unexpected thing such as $x_1x_2+x_3x_4\in \mathbb{Q}$, despite not being symmetric, can now happen! Apr 24, 2023 at 15:54

The equation is very similar to $$(x-k)^4$$. You can derive this equation and check the similarity to your equation and $$k$$ comes out to be $$2$$.

You can then use completing square method to solve the equation.

$$x^4-8x^3+24x^2-32x+16-16-14=0$$

$$(x-2)^4-30=0$$

$$x = 2\pm\sqrt[\leftroot{-2}\uproot{2}4]{30}$$ and $$x = 2\pm i\sqrt[\leftroot{-2}\uproot{2}4]{30}$$

You can solve this and the answer comes out to be 8

I will provide a solution that does not require explicitly finding the roots, but let's fix a misconception first.

If $$f$$ is a polynomial with real coefficients, then only the non-real roots of $$f$$ come in conjugate pairs. The theorem is as follows:

Let $$f\in\mathbb{R}[x]$$. For any $$\alpha\in\mathbb{C}$$, we have $$f(\alpha)=0$$ if and only if $$f(\overline\alpha)=0$$.

Apply this to a real root $$\alpha$$, you get nothing. This of course makes sense because you can have $$f=(x-a)(x-b)g$$ where $$a,b\in\mathbb{R}$$ and $$g$$ is an irreducible quadratic real polynomial. Then $$f$$ still has real coefficients, but $$a$$ and $$b$$ are arbitrary.

Back to your problem. Let $$f(x)=x^4-8x^3+24x^2-32x-14$$.

For any polynomial $$g$$ and a polynomial expression $$h(x_1,\dots,x_n)$$ symmetric in the roots $$x_1,\dots,x_n$$ of $$g$$, we know that we can express $$h$$ as a polynomial in the coefficients of $$g$$. This is known as the symmetric function theorem. For example, if $$x_1,x_2$$ are roots of $$x^2-ax+b$$, then we have $$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=a^2-2b$$. For more details, see wiki and an exposition article here.

Now, in your particular problem, the expression $$x_1x_2+x_3x_4$$ is not symmetric (for example, swapping $$x_1$$ and $$x_3$$ results in a different polynomial expression). But also notice that this is not "completely asymmetric". For example, swapping $$x_1$$ and $$x_2$$ does not change the expression. This motivates us to consider how many different expressions there can be under all $$24$$ permutations. Surprisingly, only $$3$$, namely: $$x_1x_2+x_3x_4,\ x_1x_3+x_2x_4,\ x_1x_4+x_2x_3$$ As noted by @lhf, if you sum the $$3$$ expressions, you get $$24$$ because the sum is now symmetric! Can we do better? Sure, consider the following polynomial [1]: $$R_3(x)=(x-(x_1x_2+x_3x_4))(x-(x_1x_3+x_2x_4))(x-(x_1x_4+x_2x_3))$$ Since $$R_3$$ includes all $$3$$ expressions, no matter how you permute $$x_1,\dots,x_4$$, the resulting polynomial will be unchanged. Thus, $$R_3$$ is a polynomial whose coefficients are symmetric polynomials of $$x_1,\dots,x_4$$! This means that we can express the coefficients of $$R_3$$ as polynomials in the coefficients of $$f$$, thanks to the symmetric function theorem.

Let \begin{align*} R_3(x) &= x^3+Ax^2+Bx+C. \end{align*} Then $$A=-(x_1x_2+x_3x_4+x_1x_3+x_2x_4+x_1x_4+x_2x_3)=-24.$$ For $$B$$ and $$C$$, it's long when written out, but the idea is really simple (take advantage of symmetry): \begin{align*} B &= x_{1}^{2} x_{2} x_{3} + x_{1} x_{2}^{2} x_{3} + x_{1} x_{2} x_{3}^{2} + x_{1}^{2} x_{2} x_{4} + x_{1} x_{2}^{2} x_{4} + x_{1}^{2} x_{3} x_{4} \\ &\quad +x_{2}^{2} x_{3} x_{4} + x_{1} x_{3}^{2} x_{4} + x_{2} x_{3}^{2} x_{4} + x_{1} x_{2} x_{4}^{2} + x_{1} x_{3} x_{4}^{2} + x_{2} x_{3} x_{4}^{2} \\ &={\left(x_{1} + x_{2} + x_{3}\right)} x_{1} x_{2} x_{3} +{\left(x_{1} + x_{2} + x_{4}\right)} x_{1} x_{2} x_{4}\\ &\quad +{\left(x_{1} + x_{3} + x_{4}\right)} x_{1} x_{3} x_{4} +{\left(x_{2} + x_{3} + x_{4}\right)} x_{2} x_{3} x_{4} \\ &= (x_1+x_2+x_3+x_4)(x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)-4x_1x_2x_3x_4 \\ &= 8\cdot 32 - 4\cdot (-14) \\ &= 312 \end{align*} and \begin{align*} C &= -x_{1}^{2} x_{2}^{2} x_{3}^{2} - x_{1}^{2} x_{2}^{2} x_{4}^{2} - x_{1}^{2} x_{3}^{2} x_{4}^{2} - x_{2}^{2} x_{3}^{2} x_{4}^{2} \\ &\quad- x_{1} x_{2} x_{3} x_{4}^{3} - x_{1}^{3} x_{2} x_{3} x_{4} - x_{1} x_{2}^{3} x_{3} x_{4} - x_{1} x_{2} x_{3}^{3} x_{4}. \end{align*} Note that \begin{align*} &\quad x_{1}^{2} x_{2}^{2} x_{3}^{2} + x_{1}^{2} x_{2}^{2} x_{4}^{2} + x_{1}^{2} x_{3}^{2} x_{4}^{2} + x_{2}^{2} x_{3}^{2} x_{4}^{2} \\ &= (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)^2\\ &\quad -(2 \, x_{1}^{2} x_{2}^{2} x_{3} x_{4} + 2 \, x_{1}^{2} x_{2} x_{3}^{2} x_{4} + 2 \, x_{1} x_{2}^{2} x_{3}^{2} x_{4} + 2 \, x_{1}^{2} x_{2} x_{3} x_{4}^{2} + 2 \, x_{1} x_{2}^{2} x_{3} x_{4}^{2} + 2 \, x_{1} x_{2} x_{3}^{2} x_{4}^{2}) \\ &= (x_1x_2x_3+x_1x_2x_4+x_1x_3x_4+x_2x_3x_4)^2 \\ &\quad - 2\, x_{1} x_{2} x_{3} x_{4}{\left(x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} + x_{1} x_{4} + x_{2} x_{4} + x_{3} x_{4}\right)} \\ &= 32^2-2\cdot (-14)\cdot 24 \\ &= 1696 \end{align*} and \begin{align*} &\quad x_{1}^{3} x_{2} x_{3} x_{4} + x_{1} x_{2}^{3} x_{3} x_{4} + x_{1} x_{2} x_{3}^{3} x_{4} + x_{1} x_{2} x_{3} x_{4}^{3} \\ &= x_{1} x_{2} x_{3} x_{4}{\left(x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2}\right)} \\ &= x_1x_2x_3x_4 \\ &\quad \cdot ((x_1+x_2+x_3+x_4)^2- 2\, {\left(x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} + x_{1} x_{4} + x_{2} x_{4} + x_{3} x_{4}\right)}) \\ &= -14\cdot (8^2-2\cdot 24) \\ &= -224 \end{align*}

Substituting this back into $$C$$, we have \begin{align*} C = -1696 - (-224) = -1472. \end{align*}

The dust settles, and we are left with: $$R_3(x) = x^3-24x^2+312x-1472.$$ We want to see if we can factorize $$R_3$$. We can see if there is a rational root of $$R_3$$. By rational root theorem, any rational root must be a divisor of $$1472=2^6\cdot 23$$, and hence an integer. If $$\alpha$$ is a such a root, then we have \begin{align*} \alpha^3&\equiv 0\ (\text{mod}\ 2) \\ \alpha^3-2&\equiv 0\ (\text{mod}\ 3) \\ \alpha^3-\alpha^2+2\alpha-2&\equiv 0\ (\text{mod}\ 5) \end{align*} After a few testings, we have: \begin{align*} \alpha&\equiv 0\ (\text{mod}\ 2) \\ \alpha&\equiv 2\ (\text{mod}\ 3) \\ \alpha&\equiv 3\ (\text{mod}\ 5) \end{align*} Using Chinese remainder theorem, it's not hard to find that the smallest positive solution is $$\alpha=8$$, which happens to be a root of $$R_3$$!

Then we have $$R_3(x)={\left(x - 8\right)}{\left(x^{2} - 16 \, x + 184\right)}.$$ The discriminant of the quadratic factor is $$16^2-4\cdot 184=-480<0$$, so $$x=8$$ is the only real root of $$R_3$$.

Finally, recall that the roots of $$R_3$$ are: $$x_1x_2+x_3x_4,\ x_1x_3+x_2x_4,\ x_1x_4+x_2x_3$$ Since we know that $$x_1x_2+x_3x_4$$ is real, it has to be the unique real root of $$R_3$$, so we have $$x_1x_2+x_3x_4=8$$.

[1]: This is known as the cubic resolvent of $$f$$. If you know a little Galois theory, you will immediately see that the coefficients of $$R_3$$ can be expressed as polynomials of coefficients of $$f$$.

• (+1) I worked it essentially the same way and got $(x-(ab+cd))(x-(ac+bd))(x-(ad+bc))=x^3-\beta x^2+\left(\alpha\gamma-4\delta\right)x-\left(\gamma^2+\alpha^2\delta-4\beta\delta\right)$ where $\{a,b,c,d\}$ are roots of $x^4+\alpha x^3+\beta x^2+\gamma x+\delta$
– robjohn
Apr 24, 2023 at 20:58
• As in your answer, this gives $x^3-24x^2+312x-1472=(x-8)\left((x-8)^2+120\right)$
– robjohn
Apr 24, 2023 at 21:19

It might be remarked that if one missed the "binomial-fourth-power" coefficients in $$\ f(x) \ = \ x^4 - 8x^3 + 24x^2 - 32x - 14 \ \$$ (which leads to Vinanth S Bharadwaj's very direct argument), "depressing" the polynomial takes us to a similar place: $$f(x+2) \ \ = \ \ F(y) \ \ = \ \ y^4 \ - \ 30 \ \ ,$$ for which the zeroes are arranged on the real and imaginary axes at a radius $$\ \rho \ \ . \$$ While it is easy to obtain $$\ \rho \ = \ 30^{1/4} \ \ , \$$ we don't actually care what its value is. Returning to $$\ f(x) \ \ , \$$ we find that the products of the real and imaginary zeroes are $$\ x_1·x_2 \ = \ (2 + \rho)·(2 - \rho) \ = \ 4 - \rho^2 \ \$$ and $$\ x_3·x_4 \ = \ (2 + i\rho)·(2 - i\rho)$$ $$= \ 4 + \rho^2 \ \ . \$$ The sum we seek is then $$\ x_1x_2 \ + \ x_3x_4 \ = \ 8 \ \ .$$

Going to somewhat more trouble, we could use the complex-conjugacy of $$\ x_3 \ , \ x_4 = \overline{x_3} \ = \ \gamma \pm i·\delta \ \$$ to write two quadratic factors for $$\ f(x) \ = \ (x^2 + ax + b)·(x^2 - 2·\gamma·x + c) \ \ , \$$ with $$\ c \ = \ z_{3,4}·\overline{z_{3,4} } \ \ .$$ Without presenting the ensuing calculations for a system of four equations, we obtain $$f(x) \ = \ (x^2 - 4x + 4 - \sqrt{30})·(x^2 - 4x + 4 + \sqrt{30}) \ \ .$$ Once again, we don't need to determine the zeroes, but can simply observe that $$\ x_1·x_2 \$$ and $$\ x_3·x_4 \$$ are equal to the constant terms of these factors (we won't even need to say which pair* corresponds to which constant). The sum of said constant terms is $$\ 8 \ \ .$$

$$\ ^{*}$$ though we see that the second factor is irreducible over $$\ \mathbb{R} \ .$$

For the sake of completeness, the following is a computer-aided way to brute force it.

Let $$\,x_3 = z\,$$ be one of the (non-real) complex roots, and let $$\,t=|z|^2 \ge 0\,$$. Because the polynomial has real coefficients, the other complex root is $$\,x_4=\bar z =\dfrac{t}{z}\,$$. Therefore the equalities hold true:

\begin{cases} \begin{align} \;\; z^4 - 8 z^3 + 24 z^2 - 32 z - 14 &= 0 \\ \;\; t^4 - 8 t^3 z + 24 t^2 z^2 - 32 t z^3 - 14 z^4 &= 0 \end{align} \end{cases} \tag{*}

Eliminating $$\,z\,$$ between the two equations using polynomial resultants gives (courtesy WA):

$$(t^2 - 8 t - 14)^2 (t^4 - 16 t^3 + 36 t^2 - 1696 t + 196) (t^4 - 16 t^3 + 156 t^2 + 224 t + 196)^2 = 0$$

Since equations $$(*)$$ have two roots in common, $$z$$ and $$\bar z$$, $$\,t\,$$ must be a double root of the resultant. The last factor has no real roots, so $$\,t\,$$ must be a root of the quadratic factor i.e. $$\,t=4 \pm \sqrt{30}\,$$, and $$\,t \ge 0 \implies t = 4 + \sqrt{30}\,$$.

Then from Vieta's relations $$\,x_1x_2 = -\dfrac{14}{x_3x_4}=-\dfrac{14}{t}=-\dfrac{14}{4+\sqrt{30}}=4-\sqrt{30}\,$$, and therefore $$\,x_1x_2+x_3x_4$$ $$\require{cancel}= (4-\cancel{\sqrt{30}})+(4+\cancel{\sqrt{30}})=8\,$$.