# Extracting integral solutions from a quartic equation

The equation $$\begin{equation*} y^{4} + 4y^{3} + 10y^{2} + 12y - 27 = 0 \end{equation*}$$ has two integral roots. Without resorting to the quartic formula, how would one extract the roots from it?

From the rational root theorem: in your case, it tells us that any integer root must divide $$-27$$. Therefore, it has to be one of these numbers: $$\pm1$$, $$\pm3$$, $$\pm9$$, or $$\pm27$$.

• Yes, this is the simplest method to extract an integral root. – A gal named Desire May 20 at 23:10

We can also use an integer polynomial factorisation algorithm , which gives the factorization $$(y^2 + 2y + 9)(y + 3)(y - 1).$$ Of course, the rational root test is much better suited here (and is performed before anyway).

Hint: Try $$y=3$$ or $$y=1$$ both are divisors of $$27$$

Here's one way we might proceed:

There is an old and simple result:

Let $$R$$ be a commutative unital ring and let

$$p(x) = \displaystyle \sum_0^n p_i x^i \in R[x]; \tag 1$$

then $$1$$ is a root of $$p(x)$$ if and only if the sum of the coefficients of $$p(x)$$ is $$0$$; that is,

$$p(1) = 0 \Longleftrightarrow \displaystyle \sum_0^n p_i = 0; \tag 2$$

this is an immediate consequence of

$$p(1) = \displaystyle \sum_0^n p_i 1^i = \sum_0^n p_i, \tag 3$$

as is easy to see.

We can apply this observation to

$$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27, \tag 4$$

and immediately conclude $$1$$ is a zero of $$q(y)$$. Next we synthetically divide $$q(y)$$ by $$y - 1$$, and obtain

$$q(y) = y^4 + 4y^3 + 10y^2 + 12y - 27 = (y - 1)(y^3 + 5y^2 + 15y + 27); \tag 5$$

We Check:

$$(y - 1)(y^3 + 5y^2 + 15y + 27) = y^4 + 5y^3 + 15y^2 + 27y - y^3 - 5y^2 - 15y - 27$$ $$= y^4+ 4y^3 + 10y^2 + 12y - 27 = q(y); \tag 6$$

setting

$$s(y) = y^3 + 5y^2 + 15y + 27, \tag 7$$

we now invoke the rational root theorem to limit the integer candidates for a zero of this cubic polynomial to the (integer) divisors of the constant term, that is, to $$\pm 1$$, $$\pm 3$$, $$\pm 9$$, and $$\pm 27$$; we then make a (hopefully) intelligent guess that

$$s(-3) = 0 \tag 8$$

might hold; and indeed we see that

$$s(-3) = -27 + 45 - 45 + 27 = 0; \tag 9$$

thus $$-3$$ is also a root of $$q(y)$$; now we (again, synthetically) divide $$s(y)$$ by $$y + 3$$:

$$s(y) = y^3 + 5y^2 + 15y + 27 = (y + 3)(y^2 + 2y + 9), \tag{10}$$

and again we may check:

$$(y + 3)(y^2 + 2y + 9) = y^3 + 2y^2 + 9y + 3y^2 + 6y + 27$$ $$= y^3 + 5y^2 + 15y + 27 = s(y); \tag{11}$$

it follows that

$$q(y) = (y - 1)(y + 3)(y^2 + 2y + 9); \tag{12}$$

we have now extracted two integral roots of $$q(y)$$; the remaining zeroes satisfy the quadratic equation

$$y^2 + 2y + 9 = 0; \tag{13}$$

the "formula" yields

$$y = \dfrac{-2 \pm \sqrt{-32}}{2} = \dfrac{-2 \pm 4 \sqrt{-2}}{2} = -1 \pm 2i\sqrt 2; \tag{14}$$

we see the remaining two roots of $$q(y)$$ form a complex conjugate pair, not integers! We have thus verified that $$q(y)$$ has precisely two integral zeroes, $$1$$ and $$-3$$.

• The Rational Root Test is the (convenient) method to extract the root from the given polynomial ... but this is cute. – A gal named Desire May 20 at 23:13
• @AgalnamedDesire: well, shucks, glad you thought it was pretty! – Robert Lewis May 20 at 23:22
• @AgalnamedDesire: I edited in a few words on the RRT; cf. ca. (7)-(8). Cheers! – Robert Lewis May 22 at 18:13