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$$P(x)=x^4-6x^3+4x^2+15x+4$$

Steps I took:

Possible zeros are: $$(x+1)(x-1)(x+2)(x-2)(x+4)(x-4)$$

Used synthetic division to find which zero is an actual zero: (I apologize for the graphical representation, I don't know how else to represent it on here)

$$-1\quad|\quad 1\quad -6\quad 4\quad 15\quad 4\\ ---------------\\ \quad \quad \quad \quad -1\quad 7-11-4\\ ---------------\\ \quad \quad 1\quad -7\quad 11\quad 4\quad 0$$

So the result is: $$(x+1)(x^3-7x^2+11x+4)$$

Now I don't know how to factor this. Either I am not looking at this correctly or I made an amateur mistake somewhere up this point.

Guide me in the right direction please.

Cherry_Developer
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  • Note that $x=4$ is also a solution. – André Nicolas Sep 20 '14 at 02:21
  • Looks good so far. Try some of your remaining choices (you will find $4$ is also a root. Now you have a quadratic remaining, and can use the quadratic formula. – Alex Wertheim Sep 20 '14 at 02:22
  • What's stopping you from checking the possible roots of the cubic? – Andrey Kaipov Sep 20 '14 at 02:23
  • @Andrey I am having trouble factoring it... Like I said. Amatuer – Cherry_Developer Sep 20 '14 at 02:26
  • $(x+1),(x-1),(x+2),(x-2),(x+4),(x-4)$ are not the possible zeros of the polynomial, they are possible divisors or linear factors of the polynomial. Th possible zeros are $+1,-1,+2,-2,+4,-4$. – miracle173 Sep 20 '14 at 02:26
  • @miracle173 I know. I just tried to save time and space by writing them out as linear factors right away. – Cherry_Developer Sep 20 '14 at 02:28
  • sorry, now we have wasted the saved time and space anyway. – miracle173 Sep 20 '14 at 02:30
  • @Cherry_Developer You misunderstood. How did you justify finding the possible roots of the quartic? Rational root theorem? Great. Now notice the constant term of the cubic is also a $4$. So, your possible roots of the cubic are the same as the quartic. If you do your synthetic division magic with the cubic, you will find that $x=4$ is a root of the cubic. Then you have a quadratic and the rest is history. – Andrey Kaipov Sep 20 '14 at 02:52

2 Answers2

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I think I understand your graphical representation. So you have found that $-1$ is a root, which is correct. It is also nice that you have divided out $(x+1)$. Now you have either one or three real roots left to find. Here is how I would recommend proceeding. You have a list of six possible things that could be zeroes. One of them, which I imagine you would inevitably try is $x=4$. This means you can factor out $(x+1)(x-4)$ from your polynomial, and rewrite $$P(x)=(x+1)(x-4)(x^2-3x-1)$$ Now all you have to do is use the quadratic formula to solve $$x^2-3x-1=0$$ for your last two roots and you are done.

graydad
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  • Interesting. For every other problem I have attempted I only needed to find 1 root using synthetic division and then I was able to factor out the quotient and find the rest. I guess this case was different? – Cherry_Developer Sep 20 '14 at 02:43
  • You may have been doing that with cubics in the past, in which case after one round of synthetic division you would have ended up with an easily solvable quartic after? – graydad Sep 20 '14 at 02:47
  • Yes. After one round I would end up with a polynomial that was easily factored – Cherry_Developer Sep 20 '14 at 02:48
  • Ah okay. You can keep doing synthetic division indefinitely so long as you know a zero of your polynomial that has not yet been divided out. Have you encountered complex roots before? – graydad Sep 20 '14 at 02:53
  • Duly noted. I will just keep doing synthetic division next time if a polynomial is unfactorable. I have skimmed the section about them. I will study them tomorrow. – Cherry_Developer Sep 20 '14 at 02:56
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To check that $x=-1$ is a solution, all we need to do is to plug in, we don't need to divide by $x+1$.

Note that $x=4$ also happens to be a solution. Now divide your polynomial by $(x+1)(x-4)$, that is, by $x^2-3x=4$. You will get a quadratic as quotient, and you can use the Quadratic Formula on that.

André Nicolas
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  • I understand what you are saying but why am I unable to proceed in the manner in which I was doing it. I cannot see how I can factor the cubic that I came up with. Why is that? Did I make a mistake somewhere? – Cherry_Developer Sep 20 '14 at 02:40