# Factorize polynomial of degree 4 given statements

If $$x^4-x^3-13x^2+26x-8 = (x-a)(x-b)(x-c)(x-d)$$ Such that$$cd=-8\\a>b\\c What are $$a,b,c$$ and $$d$$?

Since the problem gave us the polynomial, I thought we can just expand the $$(x-a)(x-b)(x-c)(x-d)$$ out and match the coefficients, it turned out to be: $$abcd=-8\\a+b+c+d=1\\ab+ac+ad+bc+bd+cd=-13\\abc+acd+abd+bcd=-26$$ Can I solve $$a,b,c,d$$ from there? How can I do so?

• Note that $ab=1$. Along with the given fact that $a \gt b$ you know two roots are not integers. By the rational root theorem they are irrational. – Ross Millikan Jan 28 at 4:32
• Watch your signs. In your last equation you have $26$ where you should have $-26.$ – David K Jan 28 at 4:43

## 2 Answers

You know that $$abcd = -8$$ and $$cd = -8.$$ Therefore $$ab= 1.$$

From the coefficient of $$x$$ in the polynomial, you know that $$abc + acd + abd + bcd = -26$$ (note: not $$+26$$).

But since $$ab=1$$ and $$cd=-8$$, you can see that $$abc + acd + abd + bcd = c - 8a + d - 8b.$$ Therefore

$$-8a - 8b + c + d = -26. \tag1$$

But you also know that

$$a + b + c + d = 1. \tag2$$

Subtract Equation $$(1)$$ from Equation $$(2)$$:

$$9a + 9b = 27.$$

That is, $$a + b = 3.$$ But $$b = \frac1a,$$ so we have

\begin{align} a + \frac1a &= 3, \\ a^2 + 1 &= 3a, \\ a^2 - 3a + 1 &= 0. \\ \end{align}

Apply the quadratic formula to solve $$y^2 - 3y + 1 = 0.$$ Note by symmetry that $$a$$ and $$b$$ both are solutions of this equation. But you are given that $$a > b$$, so you can see how to match $$a$$ and $$b$$ with the two solutions of the quadratic formula.

For $$c$$ and $$d,$$ multiply Equation $$(2)$$ by $$8$$ and add the result to Equation $$(1)$$. You get $$9c + 9d = -18.$$ But also $$d = -\frac8c.$$ Again you can get a quadratic equation out of this and solve it, then use the information that $$c < d$$ to know which root is which.

No guesswork is required, although if you do guess cleverly you can shorten the path a little.

Let $$f(x) = x^4-x^3-13x^2+26x-8$$

Observe that $$f(2) = 0$$ and $$f(-4) = 0$$

Hence $$(x-2)$$ and $$(x+4)$$ are two factors of $$f(x)$$

Also $$f(x)$$ is divisible by $$(x-2)(x+4) = x^2 + 2x - 8$$

Now divide $$f(x)$$ by $$x^2 + 2x - 8$$ to obtain the other quadratic factor $$x^2 - 3x +1$$

If you solve $$x^2 - 3x +1 = 0$$, you'll find $$x = \frac{3}{2} \pm \frac{\sqrt{5}}{2}$$

Finally $$f(x) = \left(x - \left(\frac{3}{2} + \frac{\sqrt{5}}{2}\right) \right) \left(x - \left(\frac{3}{2} - \frac{\sqrt{5}}{2}\right) \right) (x + 4)(x-2)$$

• How did you observe that? – Cyh1368 Jan 28 at 4:22
• Consider the factors of $-8$ – PTDS Jan 28 at 4:25
• @Cyh1368: the rational root theorem tells you that any rational roots are among $\pm1, \pm2, \pm4, \pm8$. In a school problem you are pretty well guaranteed that you can get to a quadratic this way. – Ross Millikan Jan 28 at 4:31