# Factorization of a degree three polynomial

So I was doing some Vector Calculus homework and was working with Lagrange Multipliers, but then I came across a polynomial that I either forgot how to factor or never learned. I plugged it into Wolfram and was able to get the real solution, and finished the problem, but I'd like to know if there is a simple way to factor this out and if so what is it?

$3y^3 +y -4 = 0$

I know the real solution to this is $y=1$, but could someone give me a step by step process to receive this answer in the simplest way they know?

Edit: I'm sorry if I didn't make this clear. I Actually didn't know before hand $y=1$ this came after I looked for the solution on wolfram alpha. What I am looking for is actually how to factor that polynomial with no given information of the roots.

Thank You,

Valentino

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Well, since you are in a class (i.e., this is not a polynomial from the wild), if you are expected to find the root of a polynomial of degree 3 or more, it is quite likely that you are expected to start by trying small integer solutions. Zero is clearly out, so the next thing to try is $y=1$. If that hadn't worked, I'd try $y=-1$, and probably $y=2$ and $y=-2$, and then I'd start thinking that I made an error in getting the polynomial. –  Matthew Conroy Feb 16 at 1:07
Do you want to now know how to find the root $\,y=1,\,$ or how to continue once that root is know to find the other roots, or do you want to know how to do both? –  Bill Dubuque Feb 16 at 1:07

I believe the rational root theorem is what you're looking for. It states that the only possible rational roots can be written as a fraction whose numerator is a factor of -4 and whose denominator is a factor of 3. Therefore, you can search for a root among $\pm1,\pm2,\pm4$ (and then if those fail, $\pm4/3$, $\pm2/3$, $\pm1/3$). If you find one root, then as Andrew said, you can divide and be left with a quadratic.

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Yes I believe you are correct. Man I can't believe I forgot about that haha. Thanks though man. I would rate the answer, but I have no reputation –  Valentino Feb 16 at 3:35

If you know that $y = 1$ is a solution, then you can use polynomial division to divide $3y^3+y-4$ by $y-1$. This yields $3y^3+y-4 = (y-1)(3y^2+3y+4)$. From here you can use the quadratic formula to get the remaining complex roots.

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I did not know that y=1 until I checked Wolfram Alpha. So I would not have been able to do this method. –  Valentino Feb 16 at 2:41
@Valentino Read Mark S. link, this is a really common method –  chubakueno Feb 16 at 3:15

Possible roots: $$\pm[1, 2, 4]$$ Plug in each possible root to see if it is an actual root. It is a root if the equation evaluates to 0.
For $y=1$: $$3(1)^3+1-4=0$$ $$3+1-4=0$$ $$0=0$$ $y=1$ is a root of $3y^3+y-4=0$. This means that $(y-1)$ is a factor. Using synthetic division (hopefully you know how to do this; if you don't look it up), we know that the other factor is $(3y^2+3y+4)$. It takes about three seconds of thinking to realize that $(3y^2+3y+4)$ does not have real roots, because its discriminant is less than $0$. Let's use the quadratic formula to find the two complex roots. $$3y^2 + 3y + 4$$ $$y=\dfrac{-3\pm\sqrt{3^2-4(3)(4)}}{2(3)}$$ $$y=\dfrac{-3\pm\sqrt{9-48}}{6}$$ $$y=\dfrac{-3\pm\sqrt{-39}}{6}$$ $$y=\dfrac{-3\pm i\sqrt{39}}{6}$$ The full factoring is: $$(y-1)\left(y+\dfrac{3+i\sqrt{39}}{6}\right)\left(y-\dfrac{i\sqrt{39}-3}{6}\right)$$ I hope I did the factoring right D:

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I like Matthew Conroy's comment above; reminding oneself of the pragmatic context can be either reassuring or maddening (when the answer doesn't come readily enough).

In any case, there's a general solution to a cubic equation, due to Cardano, I believe: Take any monic polynomial of degree three, $x^3 + a_2 x^2 + a_1 x + a_0$, and let $x = z - a_2 / 3$. (Monic just means the leading coefficient is $1$.) Substitute this into your original polynomial; it will simplify to something of the form $z^3 + bz + c$, which is called a "depressed" cubic equation. Then there exists a root of the form

$$z = \sqrt[3]{-\frac{c}{2}+\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}+\sqrt[3]{-\frac{c}{2}-\sqrt{\frac{c^2}{4}+\frac{b^3}{27}}}$$

Needless to say, this will come out messy most of the time; therefore, you should probably start with the rational roots, as an earlier answer recommended. But this will give you an answer.

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When the coefficients add to zero +1 will be a root. If the sum of the coefficients of the even power terms minus the sum of the coefficients of the odd power terms then -1 will be a root.

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