4
$\begingroup$

Today, I had this question on a Maths test about Algebra. This was the equation I had to solve:

$$(1-x)(x-5)^3=x-1$$

I worked away the brackets and subtracted $x-1$ from both sides and was left with this:

$$-x^4+16x^3-90x^2+199x-124=0$$

Problem is, I haven't a clue how to solve this? First thing I tried was replacing $x^2$ with another variable like $u$ but that got me no further. Dividing the whole equation by $x^2$ (as is suggested by a lot of sites on this matter) also did not get me any further. I then tried something incredibly ludicrous;

$$(ax+b)(cx^3+dx^2+ex+f)=0$$ $$ \left\{ \begin{aligned} ac &= -1 \\ ad + bc &= 16 \\ ae + bd &= -90 \\ af + be &= 199 \\ bf &= -124 \end{aligned} \right. $$

which got even worse when there were 3 brackets;

$$(ax+b)(gx+h)(ix^2+jx+k)=0$$ $$ \left\{ \begin{aligned} ac &= agi &&= -1 \\ ad + bc &= agj + ahi + bgi &&= 16 \\ ae + bd &= agk + ahj + bgj + bhi &&= -90 \\ af + be &= ahk + bgk + bhj &&= 199 \\ bf &= bhk &&= -124 \end{aligned} \right. $$

only to be left with no result.

When using Wolfram Alpha on this question, it performs a rather strange step I don't understand:

$$-x^4+16x^3-90x^2+199x-124=0$$ $$\downarrow$$ $$-((x-4)(x-1)(x^2-11x+31))=0$$

Could somebody explain how to properly tackle this problem? And if possible, also show me how to get the non-real answers for it. Thanks.

$\endgroup$
1
  • 3
    $\begingroup$ You needn't expand: first, observe that $x=1$ is a solution. Now, looking for other solutions (at most 3 other possible ones), you can divide each side by $x-1$ and obtain the equation $(x-5)^3=-1$ to solve (for values $x\neq 1$). $\endgroup$
    – Clement C.
    May 20, 2015 at 13:13

4 Answers 4

8
$\begingroup$

Euh... I think you overcomplicated things here...

$(1-x)(x-5)^3=x-1$ is equivalent to $(1-x)[(x-5)^3+1]=0$

Either $x=1$ or $(x-5)^3=-1$...

$\endgroup$
3
  • $\begingroup$ I feel so stupid, that I didn't see this in the first place... Thanks for the great answer! $\endgroup$
    – Martijn
    May 20, 2015 at 13:19
  • $\begingroup$ @TitoPiezasIII Done :) $\endgroup$
    – Martijn
    May 20, 2015 at 13:27
  • $\begingroup$ You can also use the identity $a^3+1=(a+1)(...)$ $\endgroup$
    – k1.M
    May 20, 2015 at 14:00
4
$\begingroup$

There are already answers on how to get the real solutions, so I will only show you the non-real solutions.

You have obtained that $(x-5)^3=-1$. Expanding and simplifying we get: $$x^3-15x^2+75x-124=0$$ However, we know that $x=4$ is a solution so we can say that $(x-4)(ax^2+bx+c)=0$. You can equate coefficients or use polynomial division, but as you have already found with Wolfram Alpha, $(ax^2+bx+c)=(x^2-11x+31)$.

To solve, complete the square: $$x^2-11x+31=0$$ $$(x-\frac{11}{2})^2-\frac{121}{4}+31=0$$ $$(x-\frac{11}{2})^2=-\frac{3}{4}$$ $$x-\frac{11}{2}=\sqrt{-\frac{3}{4}}= \pm \frac{\sqrt{3}}{2}i$$ $$x=\frac {11 \pm i \sqrt{3}}{2}$$

$\endgroup$
2
  • 1
    $\begingroup$ Awesome work! I have two kinds of mathematics in the Netherlands and the other kind is about imaginary numbers a.t.m. so this will definitely come in handy, I'm sure! $\endgroup$
    – Martijn
    May 20, 2015 at 13:45
  • $\begingroup$ In general the solutions to $(x-a)^m=-1$ are $x_n=e^{i\pi\frac{2n-1}{m}}+a$, for $n=1,2,\dots m$. $\endgroup$ May 21, 2015 at 4:10
2
$\begingroup$

From the beginning: $$(1-x)(x-5)^3=x-1\\ (1-x)(x-5)^3+1-x=0\\ (1-x)(x-5)^3+(1-x)=0\\ (1-x)[(x-5)^3+1]=0\\$$

This implies $1-x=0$ or $(x-5)^3=-1$. I believe you can solve these.

$\endgroup$
2
  • 4
    $\begingroup$ Isn't there a sign that got flipped between the first and second line? $\endgroup$
    – Clement C.
    May 20, 2015 at 13:18
  • $\begingroup$ Sorry fixed it. @ClementC. $\endgroup$
    – KittyL
    May 20, 2015 at 13:24
2
$\begingroup$

I know that the problem has already been answered but I want to show you a more general method, let's suppose that you don't se how to rewrite the equation:

$-x^4+16x^3-90x^2+199x-124=0$

Or I prefer to write:

$x^4-16x^3+90x^2-199x+124=0$

You can use something called the Ruffini rule: search for integers divisor (both positive and negative) of the constant term and then set $x$ equals to the them and see if one of them is a solution. Starting from one we have:

$1-16+90-199+124=0$

So $x=1$ is a solution, now via Ruffini's rule, which can be seen here, we can rewrite the equation as:

$(x-1)(x^3-15x^2+75x-124)=0$

Now you have $3$ options to end this exercise:

$1.)$ Note that the second factor is a perfect cube;

$2.)$ Use Ruffini's rule again;

$3.)$ Use the general formula for third degree equation (which I'd not advise you to if your interested only in real solution).

This is a more general method so I hope this will help you in the future!

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .