Quadratic Squared

Assume that $x^4-2x^3-5x^2+10x-3$ can be factorized as:

$x^4-2x^3-5x^2+10x-3 = (x^2+\alpha x + \beta)(x^2+\gamma x + \delta)$

where $\alpha$ , $\beta$ , $\gamma$ and $\delta$ are integers.

Equating the coefficients on both sides, we have:

$\quad \begin{aligned} \alpha + \beta = -2 &\quad ...(1) \\ \beta + \alpha \gamma + \delta = -5 &\quad ...(2) \\ \alpha \delta + \beta \gamma = 10 &\quad ...(3) \\ \beta \delta = -3 &\quad ...(4) \end{aligned}$

From Eqn 1: $\Rightarrow \alpha \beta = - 2 - \beta^2$

Multiplying Eqn 3 by $\beta^2$ :

$\quad \Rightarrow \beta^3 \gamma = 10\beta^2 - (\alpha \beta) (\beta \delta) = 10\beta^2 - (- 2 - \beta^2 ) (-3) = 7\beta^2-6$

Multiplying Eqn 2 by $\beta^4$ :

$\quad \Rightarrow \beta^5 + (\alpha \beta) ( \beta^3 \gamma) + \beta^3 (\beta \delta ) = -5\beta^4$

$\quad \Rightarrow \beta^5 + (- 2 - \beta^2) (7\beta^2-6) + \beta^3 (-3 ) = -5\beta^4$

$\quad \Rightarrow \beta^5 - 2\beta^4 - 3\beta^3 -8\beta^2 + 12 = 0$

$\quad \Rightarrow \beta = 1\quad \Rightarrow \alpha = -3\quad \Rightarrow \delta = -3\quad \Rightarrow \gamma = 1$ .

Therefore, we have:

$\quad x^4-2x^3-5x^2+10x-3 = (x^2+\alpha x + \beta)(x^2+\gamma x + \delta)$

$\quad = (x^2-3x +1)(x^2+x -3) = ( \frac {3+\sqrt{5}}{2} ) ( \frac {3-\sqrt{5}}{2} ) ( \frac {-1 + \sqrt{13}}{2}) ( \frac {-1 + \sqrt{13}}{2} )$

The smallest positive root is $\dfrac {3 - \sqrt{5}}{2} \quad \Rightarrow a = 3$ , $b=5$ and $c = 2$ , $\quad \Rightarrow a+b+c = \boxed {10}$ .

Let $x=y+\dfrac{1}{2}$ to delete the 3rd degree coefficient:

$\left(y+\dfrac{1}{2}\right)^4-2\left(y+\dfrac{1}{2}\right)^3-5\left(y+\dfrac{1}{2}\right)^2+10\left(y+\dfrac{1}{2}\right)-3=0$

$y^4-\dfrac{13}{2}y^2+4y+\dfrac{9}{16}=0$

Now, try to factor the equation in $y$ into the form $(y^2+py+q)(y^2-py+r)=0$ . Expand and compare the coefficients:

$\quad \begin{aligned} q+r=p^2-\dfrac{13}{2} &\quad ...(1) \\ r-q=\dfrac{4}{p} &\quad ...(2) \\ qr=\dfrac{9}{16} &\quad ...(3) \end{aligned}$

Solve for $q$ and $r$ adding and subtracting $(1)$ and $(2)$ :

$q=\dfrac{p^2-\frac{13}{2}-\frac{4}{p}}{2}$ , $r=\dfrac{p^2-\frac{13}{2}+\frac{4}{p}}{2}$

Substitute them in $(3)$ , expand, simplify and clear denominators:

$p^6-13p^4+40p^2-16=0$

That is a 3rd degree polynomial in $p^2$ , so we can use easily the rational root test, so it factor as:

$(p^2-4)(p^4-9p^2+4)=0$

Since any value of $p$ works, we use $p^2=4 \Longrightarrow p=2$ . Hence,

$q=\dfrac{2^2-\frac{13}{2}-\frac{4}{2}}{2}=-\dfrac{9}{4}$

$r=\dfrac{2^2-\frac{13}{2}+\frac{4}{2}}{2}=-\dfrac{1}{4}$

So the equation in $y$ is: $\left(y^2+2y-\dfrac{9}{4}\right)\left(y^2-2y-\dfrac{1}{4}\right)=0$ .

Using the quadratic formula we obtain, after undoing the substitution $x=y+\dfrac{1}{2}$ :

$x=\dfrac{-1 \pm \sqrt{13}}{2}$ , $x=\dfrac{3 \pm \sqrt{5}}{2}$

The asked solution is $x=\dfrac{3 - \sqrt{5}}{2}$ , hence $a+b+c=\boxed{10}$ .

I will assume firstly that this polynomial can be factored into the product of two polynomials with degree $2$ and integer coefficients.

$x^{4} - 2x^{3} - 5x^{2} + 10x - 3 = (x^{2} + ax + b)*(x^{2} + cx + d)$

$= x^{4} + (a + c)x^{3} + (b + ac + d)x^{2} + (ad + bc)x + bd$

Thus, we have the following equations:

$\begin{cases} a + c = -2 \\ b + ac + d = -5 \\ ad + bc = 10 \\ bd = -3 \end{cases}$

Now, if all of these coefficients are integer numbers, then by the last equation, $b = \pm 3$ , and $d = \mp 1$ (This can be assumed WLOG, since if it were the other way around then $a$ and $c$ would merely exchange their values.)

Case $I$ : $b = 3$

Then we have $d = -1$ , which changes the equations to:

$\begin{cases} a + c = -2 \\ ac + 2 = -5 \\ -a + 3c = 10 \end{cases}$

Solving for $a$ and $c$ using the first and third equations gives us $c = 2$ and $a = -4$ ; this, however, is not a valid solution, since it doesn't satisfy the second equation as well (We'd have $-6 = -5$ , which is absurd.) Thus, $b \neq 3$ .

Case $II$ : $b = -3$

Then we have $d = 1$ , which give us:

$\begin{cases} a + c = -2 \\ ac - 2 = -5 \\ a - 3c = 10 \end{cases}$

Again, using the first and third equations, we have $c = -3$ and $a = 1$ , which satisfies the second equation as well; thus, we can factor the original equation as: $x^{4} - 2x^{3} - 5x^{2} + 10x - 3 = (x^{2} + x - 3)*(x^{2} - 3x + 1)$

Thus, we have the following solutions for x:

$x_{1} = \frac{-1 + \sqrt{13}}{2}$

$x_{2} = \frac{-1 - \sqrt{13}}{2}$

$x_{3} = \frac{3 + \sqrt{5}}{2}$

$x_{4} = \frac{3 - \sqrt{5}}{2}$

The smallest positive real root is the number $x_{4}$ , thus the number sought $3 + 5 + 2 = 10$ .

$Since\ the\ coefficients \ of\ x^4\ is \ 1\ and\ that \ of\ constant \ is \ -3,\ \\ the\ factors\ would \ be\ \ \ (A)\ \ (X^2+aX+3)(X^2+bX-1).\ \ OR\ \ \ (B)\ \ (X^2+aX-3)(X^2+bX+1).\\ (A)\ Equating\ coefficients\ of\ X^3\ and X\ \ a+b=-2, \ \ 3b-a=10\ \ \implies\ 10= 4b-(a+b)=4b-(-2),\ \ so \ b=2\ and\ a=-4\\ So\ (X^2-4X+3)(X^2+2X-1) Checking \ for \ coefficient \ of\ X^2, \ \ -1+3-8=-6\neq -5. Not \ valid.\\ (B)\ \ Equating\ coefficients\ of\ X^3\ and \ X\ \ a+b=-2, \ \ a - 3b=10\ \ \implies\ \ a+b - 4b=10,\ \ so\ b= - 3\ \ and\ a=1.\\ \therefore\ (X^2+X-3)(X^2-3X+1).\ Checking \ for \ coefficient \ of\ X^2, \ \ 1-3-3=-5 =\ -5.\\ \text{ Since we are interested in smaller root with -tive sign before square root expression } X^2-3X+1=0\ \ gives,\\ \dfrac{3-\sqrt{9-4}} 2=\dfrac{a-\sqrt{b} }c.\\ \therefore\ a+b+c=3+5+2=\Huge\ \ \color{#D61F06}{10}$