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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?

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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 $\pm27$.

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

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

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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 $-3$.

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

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