Factoring $3n^3 - 39n^2 + 360n + 20$

Question

I am wondering how to factor $$f(n) = 3n^3 - 39n^2 + 360n + 20$$ the right way. I think the factors are equal to

$$(n - 39.9762)(n - 12.0791)(n + 0.055248)$$

Answer 1

We want to find a root of $$ 3n^3-39n^2+360n+20=0\tag1 $$ To get rid of the quadratic term, let $n=x+\frac{13}3$; then, we get that $$ x^3+\frac{191}3x+\frac{9826}{27}=0\tag2 $$ To make the linear coefficient $\frac34$, so that we can apply identity $(4)$, let $x=\frac{2\sqrt{191}}3y$; then, we get that $$ y^3+\frac34y=-\frac{4913}{764\sqrt{191}}\tag3 $$ Now we can use the identity $$ \frac14\sinh(3z)=\sinh^3(z)+\frac34\sinh(z)\tag4 $$ by letting $y=\sinh(z)$, and we get that $$ \sinh(3z)=-\frac{4913}{191\sqrt{191}}\tag5 $$ Since $\sinh(-x)=-\sinh(x)$, we get $$ \begin{align} n_1 &=\frac{13}3+\frac{2\sqrt{191}}3\sinh\left(\frac13\sinh^{-1}\left(-\frac{4913}{191\sqrt{191}}\right)\right)\\ &=\frac{13}3-\frac{2\sqrt{191}}3\sinh\left(\frac13\sinh^{-1}\left(\frac{4913}{191\sqrt{191}}\right)\right)\\ &=-0.055223771734\tag6 \end{align} $$

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Once we have a root, we can divide the polynomials $$ \frac{3n^3-39n^2+360n+20}{3(n-n_1)}=n^2+(n_1-13)n+\left(n_1^2-13n_1+120\right)\tag7 $$ Now we can use the quadratic formula to get $$ \begin{align} n_2 &=\frac{13-n_1+i\sqrt{3n_1^2-26n_1+311}}2\\[6pt] &=6.527611885867+i\,8.838056447262\tag8 \end{align} $$ and $$ \begin{align} n_3 &=\frac{13-n_1-i\sqrt{3n_1^2-26n_1+311}}2\\[6pt] &=6.527611885867-i\,8.838056447262\tag9 \end{align} $$

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Therefore, the full factorization of $3n^3-39n^2+360n+20$ over $\mathbb{C}$ would be $$ 3(n-n_1)(n-n_2)(n-n_3)\tag{10} $$ and the factorization over $\mathbb{R}$ would be $$ 3(n-n_1)\left(n^2+(n_1-13)n+\left(n_1^2-13n_1+120\right)\right)\tag{11} $$

Answer 2

There is only one real root which is exactly $$n_1=\frac{13}{3}-\frac{2}{3} \sqrt{191} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{4913}{191 \sqrt{191}}\right)\right)\approx -0.055223771734378147887$$ So, numerically $$f(n) = 3n^3 - 39n^2 + 360n + 20$$ $$f(n)=3(n-n_1)(n^2-13.055223771734378148 n+120.72095869751148663)$$

Answer 3

$\displaystyle (n_1,n_2,n_3)=\\ \Biggl(\dfrac{2}{191} \biggl(-405-2 \sqrt{216010}+\frac{864040 \left(12 \sqrt{216010}+4913\right)}{191 \sqrt[3]{216010 \left(4913 \sqrt{216010}+2592120\right)}+4913 \sqrt{216010}+2592120}\biggr),\\ \dfrac{2}{191} \biggl(-405-2 \sqrt{216010}+\frac{864040 \left(12 \sqrt{216010}+4913\right)}{-191 \sqrt[3]{-216010 \left(4913 \sqrt{216010}+2592120\right)}+4913 \sqrt{216010}+2592120}\biggr),\\ \dfrac{2}{191} \biggl(-405-2 \sqrt{216010}+\frac{864040 \left(12 \sqrt{216010}+4913\right)}{191 (-1)^{2/3} \sqrt[3]{216010 \left(4913 \sqrt{216010}+2592120\right)}+4913 \sqrt{216010}+2592120}\biggr)\Biggr)=\\ ( -0.0552238,\,6.52761 + 8.83806\,i,\, 6.52761 - 8.83806\,i) $

Answer 4

Since the OP tagged the question algebra-precalculus, it is of interest that a serious high school student could show that the OP's function is an injection; see this answer to

$\quad$ What are the conditions on $a, b, c$ so that $x^3+ax^2+bx+c$ is bijective?

Moreover, if you know the intermediate value theorem and (roughly) graph the function

$\tag 1 p(x) = 3x^3 - 39x^2 + 360x + 20$

as $x \to +\infty$ and $x \to -\infty$, you can argue the surjectivity.

So $p(x)$ is a bijection and has exactly one real root, call it $\alpha$.

Using the intermediate value theorem the student can create an algorithm that converges to $\alpha$ (see next section).

Assuming this has been done we arrive at

$\quad \alpha \approx -0.055223771734378147887$

Therefore, there exists coefficients $b,c \in \Bbb R$ such that

$\tag 2 3x^3 - 39x^2 + 360x + 20 = (x - \alpha)\, q(x)$

where $q(x) = 3x^2 + bx + c$ is an irreducible quadratic.

By multiplying $x - \alpha$ and $3x^2 + bx + c$ together and collecting like terms, we can solve for both $b$ and $c$ by using $\text{(2)}$ and equating coefficients.

Since $-\alpha c = 20$,

$\tag 3 c = -\frac{20}{\alpha} \approx \frac{20}{0.055223771734378147887} = 362.1628760925344692429874874$

There are two ways of getting $b$. We choose the relation $-39x^2 = bx^2 - 3\alpha x^2$ (we don't want to use two approximations), giving

$\tag 4 b = 3 \alpha - 39 \approx 3 \times -0.055223771734378147887 - 39 = -39.16567131520313438386438065208494663238525390625$

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Following is a crude Python program that gets an estimate for the root $\alpha$.

Once you observe that $p(-1) = -382$ and $p(0) = 20$, the intermediate value theorem guarantees that the root lies between $-1$ and $0$. So the program does a simple binary search.

Python Program

def y_cubic(x):
    return 3 * x**3 - 39 * x**2 + 360 * x + 20

neg_coord = [-1, y_cubic(-1)]
pos_coord = [0, y_cubic(0)]
it_count = 0


while True:
    P_x = pos_coord[0]
    P_y = pos_coord[1]
    N_x = neg_coord[0]
    N_y = neg_coord[1]
    print(N_x, '< x <', P_x, '        iterations:', it_count)    
    if P_x - N_x < .00000001:
        break
    it_count = it_count + 1
    binSpot = .5 * (N_x + P_x)
    y = y_cubic(binSpot)
    if y == 0:
        print('exact/float answer:', binSpot, it_count)
        break
    if y < 0:
        neg_coord = [binSpot, y]
    else:
        pos_coord = [binSpot, y]

***** OUTPUT *****

-1 < x < 0         iterations: 0
-0.5 < x < 0         iterations: 1
-0.25 < x < 0         iterations: 2
-0.125 < x < 0         iterations: 3
-0.0625 < x < 0         iterations: 4
-0.0625 < x < -0.03125         iterations: 5
-0.0625 < x < -0.046875         iterations: 6
-0.0625 < x < -0.0546875         iterations: 7
-0.05859375 < x < -0.0546875         iterations: 8
-0.056640625 < x < -0.0546875         iterations: 9
-0.0556640625 < x < -0.0546875         iterations: 10
-0.0556640625 < x < -0.05517578125         iterations: 11
-0.055419921875 < x < -0.05517578125         iterations: 12
-0.0552978515625 < x < -0.05517578125         iterations: 13
-0.05523681640625 < x < -0.05517578125         iterations: 14
-0.05523681640625 < x < -0.055206298828125         iterations: 15
-0.05523681640625 < x < -0.0552215576171875         iterations: 16
-0.05522918701171875 < x < -0.0552215576171875         iterations: 17
-0.055225372314453125 < x < -0.0552215576171875         iterations: 18
-0.055225372314453125 < x < -0.05522346496582031         iterations: 19
-0.05522441864013672 < x < -0.05522346496582031         iterations: 20
-0.055223941802978516 < x < -0.05522346496582031         iterations: 21
-0.055223941802978516 < x < -0.055223703384399414         iterations: 22
-0.055223822593688965 < x < -0.055223703384399414         iterations: 23
-0.055223822593688965 < x < -0.05522376298904419         iterations: 24
-0.05522379279136658 < x < -0.05522376298904419         iterations: 25
-0.05522377789020538 < x < -0.05522376298904419         iterations: 26
-0.05522377789020538 < x < -0.055223770439624786         iterations: 27

Answer 5

Set $$ P(x)=3x^3-39x^2+360 x+20. $$ Then set $x=y+\frac{13}{3}$, then $$ \frac{1}{3}P(x)=P_1(y):=y^3+\frac{191}{3}y+\frac{9826}{27} $$ If $\rho_0=\sqrt[3]{A}+\sqrt[3]{B}$ is the real root of $P_1(y)=0$, then $$ \rho_0^3=A+B+3(AB)^{1/3}(\sqrt[3]{A}+\sqrt[3]{B})=s+3p^{1/3}\rho_0. $$ Hence $s=A+B=-\frac{9826}{27}$, $p=AB=-\left(\frac{191}{9}\right)^3$ and the equation $$ X^2+\frac{9826}{27}X-\left(\frac{191}{9}\right)^3=0, $$ have roots $$ A=\frac{1}{27}(-4913-12\sqrt{216010})\textrm{ , }B=\frac{1}{27}(-4913+12\sqrt{216010}). $$ Hence we find $$ \rho_0=\sqrt[3]{\frac{1}{27}(-4913+12\sqrt{216010})}-\sqrt[3]{\frac{1}{27}(4913+12\sqrt{216010})}. $$ Now $P_1(y)$ have the other two roots such (Vieta) $$ \rho_1+\rho_2=-\rho_0\textrm{ and }\rho_1\rho_2=\frac{9826}{27}\rho_0^{-1} $$ Solving $$ X^2+\rho_0X+\frac{9826}{27}\rho_0^{-1}=0 $$ we get the other two roots of $P_1(y)=0$.

By this way every third degree polynomial equation reduced solving only two degree equations.

NOTE. We have used $\sqrt[3]{-|a|^3}=-|a|$, since the equation $x^3+|a|^3=0$, have solution $x=-|a|$.

Answer 6

This is just a detailed version of the original answer by @robjohn♦

\begin{align} 3n^3-&39n^2+360n+20=0 \\ &\implies n^3-13n^2+120n+\frac{20}{3}=0\\ &\!\implies \left( n-\frac{13}{3}\right)^3-\frac{169}{3}n+\frac{2197}{27}+120n+\frac{20}{3}=0\\ &\!\implies \left( n-\frac{13}{3}\right)^3+\frac{191}{3}\left(n-\frac{13}{3}\right)+\frac{9826}{27}=0\\ &\!\implies \frac{27}{8\times191\sqrt{191}}\left( n-\frac{13}{3}\right)^3+\frac{9}{8\sqrt{191}}\left(n-\frac{13}{3}\right)+\frac{4913}{764\sqrt{191}}\!=\!0\\ &\!\implies \left( \frac{3}{2\sqrt{191}}n-\frac{13}{2\sqrt{191}}\right)^3+\frac{3}{4}\left( \frac{3}{2\sqrt{191}}n-\frac{13}{2\sqrt{191}}\right)+\frac{4913}{764\sqrt{191}}\!=0 \end{align} Now, we can use the identity. $$ \frac14\sinh(3z)=\sinh^3(z)+\frac34\sinh(z)\implies x^3+\frac{3}{4}x=\frac{1}{4}\sinh(3\sinh^{-1}(x)). $$ Thus, we have \begin{align} \frac{1}{4}\sinh\left(3\sinh^{-1}\left( \frac{3}{2\sqrt{191}}n-\frac{13}{2\sqrt{191}}\right)\right) = -\frac{4913}{764\sqrt{191}}\\ \implies n = n_1\triangleq \frac{13}{3}-\frac{2\sqrt{191}}{3}\sinh\left(\frac{1}{3}\sinh^{-1}\left(\frac{4913}{191\sqrt{191}}\right)\right) =-0.055223771734.\tag1 \end{align}

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Once we have a root, we can divide the polynomial as follows: \begin{align} \frac{3n^3-39n^2+360n+20}{3(n-n_1)}&=an^2+bn+c\\ &\hspace{-3cm}\implies (an^2+bn+c)(n-n_1)=n^3-13n^2+120n+\frac{20}{3}\\ &\hspace{-3cm}\implies an^3+(b-an_1)n^2+(c-n_1b)-cn_1=n^3-13n^2+120n+\frac{20}{3} \end{align} We solve for $a,b,c$ as comparing the coefficients: \begin{equation} a=1\hspace{2cm} b=n_1-13 \hspace{2cm} c=120+n_1b=n_1^2-13n_1+120. \end{equation} Now we can use the quadratic formula to solve $An^2+Bn+C$ to get

\begin{align} n&=\frac{-b\pm\sqrt{b-4ac}}{2a}=\frac{13-n_1\pm i\sqrt{3n_1^2-26n_1+311}}{2} \end{align} Thus, the roots are \begin{align} n_2\triangleq\frac{13-n_1+ i\sqrt{3n_1^2-26n_1+311}}{2}=6.527611885867+ i\,8.838056447262\tag2\\ n_3\triangleq\frac{13-n_1- i\sqrt{3n_1^2-26n_1+311}}{2}=6.527611885867-i\,8.838056447262\tag3 \end{align}

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Therefore, the full factorization over $\mathbb{C}$ is $$3n^3-39n^2+360n+20= 3(n-n_1)(n-n_2)(n-n_3), $$ and the factorization over $\mathbb{R}$ is $$ 3(n-n_1)\left(n^2+(n_1-13)n+\left(n_1^2-13n_1+120\right)\right). $$