# Cubic Polynomial Challenge

#### anemone

##### MHB POTW Director
Staff member
Find the constants $$\displaystyle a,\;b, \;c,\; d$$ such that

$$\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.

#### MarkFL

Staff member
Using the triple-angle identity for cosine:

(1) $\displaystyle \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta)$

we may solve the cubic equation:

(2) $\displaystyle 8x^3-6x+\sqrt{3}=0$

To transform equation (2) into a form where the stated identity (1) is useful, we make the substitution $x=a\cos(\theta)$, where $a$ is a constant to be determined. With this substitution, equation (2) can be written:

(3) $\displaystyle 8a^3\cos^3(\theta)-6a\cos(\theta)=-\sqrt{3}$

In equation (3), the coefficient of $\cos^3(\theta)$ is $8a^3$. Since we want this coefficient to be $4$ [as it is in equation (1)], we divide both sides of equation (3) by $2a^3$ to obtain:

(4) $\displaystyle 4\cos^3(\theta)-\frac{3}{a^2}\cos(\theta)=-\frac{\sqrt{3}}{2a^3}$

Next, a comparison of equations (4) and (1) leads us to require that $\displaystyle \frac{3}{a^2}=3$. Thus, $a=\pm1$. For convenience, we choose $a=1$; equation (4) then becomes:

(5) $\displaystyle 4\cos^3(\theta)-3\cos(\theta)=-\frac{\sqrt{3}}{2}$

Comparing equation (5) with the identity in (1) leads us to the equation:

$\displaystyle \cos(3\theta)=-\frac{\sqrt{3}}{2}$

The solutions here are of the form:

$\displaystyle 3\theta=\frac{\pi}{6}(12k+5),\,\frac{\pi}{6}(12k+7)$

$\displaystyle \theta=\frac{\pi}{18}(12k+5),\,\frac{\pi}{18}(12k+7)$

Only 3 of these angles yield distinct values of $\cos(\theta)$, namely:

$\displaystyle \theta=\frac{5\pi}{18},\,\frac{7\pi}{18},\,\frac{17\pi}{18}$

Thus, the solutions of the equation in (2) are:

$\displaystyle x=\cos\left(\frac{5\pi}{18} \right),\,\cos\left(\frac{7\pi}{18} \right),\,\cos\left(\frac{17\pi}{18} \right)$

Hence, $a=4$, and $(b,c,d)$ can be any of the six permutations of the 3 roots listed above.

#### anemone

##### MHB POTW Director
Staff member
Hi MarkFL, thanks for participating and your answer is correct!

Hey, I'm impressed at how fast you were in replying to this problem!

#### MarkFL

Staff member
I struggled at first with Vieta, but then I recalled that this is quite similar to the High School POTW that I submitted for use on 9 June, and so I copied and pasted the solution I had provided, made a few changes, and had it solved.

#### Opalg

##### MHB Oldtimer
Staff member
Find the constants $$\displaystyle a,\;b, \;c,\; d$$ such that

$$\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.
Clearly $a=4$. For the rest ...
$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, so if we put $x=\cos\theta$ then the equation $4x^3-3x+\tfrac{\sqrt{3}}{2}=0$ becomes $\cos(3\theta) = -\tfrac{\sqrt3}2 = \cos150^\circ$, with solutions $\theta = 50^\circ,\;170^\circ,\;290^\circ$ or, if you prefer, $50^\circ,\;70^\circ,\;170^\circ$. Thus we can take $b= \cos50^\circ$, $c = \cos70^\circ$, $d = -\cos10^\circ$.

Edit. Sorry! I took so long writing this that I failed to see that Mark had already replied.

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#### Prove It

##### Well-known member
MHB Math Helper
Find the constants $$\displaystyle a,\;b, \;c,\; d$$ such that

$$\displaystyle 4x^3-3x+\frac{\sqrt{3}}{2}=a(x-b)(x-c)(x-d)$$.
\displaystyle \displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}

Equating the like powers of x we find

\displaystyle \displaystyle \begin{align*} a &= 4 \\ \\ - \left( 4b + 4c + 4d \right) &= 0 \\ b + c + d &= 0 \\ \\ 4\,b\,c + 4\,b\,d + 4\,c\,d &= -3 \\ b\,c + b\,d + c\,d &= -\frac{3}{4} \\ \\ 4\,b\,c\,d &= \frac{\sqrt{3}}{2} \\ b\,c\,d &= \frac{\sqrt{3}}{8} \end{align*}

Rearranging the first equation we have \displaystyle \displaystyle \begin{align*} b = -c - d \end{align*} and substituting into the second equation we find

\displaystyle \displaystyle \begin{align*} \left( - c - d \right) c + \left( -c - d \right) d + c\,d &= -\frac{3}{4} \\ -c^2 - c\,d - c\,d - d^2 + c\,d &= -\frac{3}{4} \\ c^2 + c\,d + d^2 &= \frac{3}{4} \\ c^2 + c\,d + \left( \frac{d}{2} \right) ^2 - \left( \frac{d}{2} \right) ^2 + d^2 &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 + \frac{3d^2}{4} &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 &= \frac{ 3 - 3d }{4} \\ c + \frac{d}{2} &= \frac{ \pm \sqrt{ 3 - 3d }}{2} \\ c &= \frac{-d \pm \sqrt{ 3 - 3d } }{2} \end{align*}

and so \displaystyle \displaystyle \begin{align*} b = - \left( \frac{-d \pm \sqrt{ 3 - 3d }}{2} \right) - d = \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \end{align*}

Substituting into the final equation, we find

\displaystyle \displaystyle \begin{align*} \left( \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \right) \left( \frac{-d \pm \sqrt{ 3 - 3d } }{2} \right) d &= \frac{\sqrt{3}}{8} \\ \left( -d \mp \sqrt{ 3 - 3d} \right) \left( -d \pm \sqrt{ 3 - 3d } \right) d &= \frac{\sqrt{3}}{2} \\ \left[ d^2 - \left( 3 - 3d \right) \right] d &= \frac{\sqrt{3}}{2} \\ d^3 + 3d^2 - 3d - \frac{\sqrt{3}}{2} &= 0 \end{align*}

Now making the change of variable \displaystyle \displaystyle \begin{align*} u = d + 1 \end{align*}, we find

\displaystyle \displaystyle \begin{align*} \left( u - 1 \right) ^3 + 3 \left( u - 1 \right) ^2 - 3 \left( u - 1 \right) - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 3u^2 + 3u - 1 + 3u^2 - 6u + 3 - 3u + 3 - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 6u - \frac{\sqrt{3} - 10}{2} &= 0 \end{align*}

and now applying the Cubic Formula, we have

\displaystyle \displaystyle \begin{align*} u &= \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} + \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } + \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} - \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } \\ &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d + 1 &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } - 1 \end{align*}

Back-substituting will enable us to evaluate b and c.

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I have always liked Cardano's method that I learned a long long time ago from a book of my father. Since I still have fond sentimental memories of it, I'm presenting it as well.

We can write the cubic expression in equation form as:
$$x^3 - \frac 3 4 x + \frac {\sqrt 3} 8 = 0 \qquad (1)$$
We substitute
$$x=y+z \qquad\qquad (2)$$
giving us a free choice for either $y$ or $z$:
\begin{array}{lcl}
(y+z)^3 - \frac 3 4 (y+z) + \frac {\sqrt 3} 8
&=&(y^3+z^3) + (3y^2z+3yz^2) - \frac 3 4 (y+z) + \frac {\sqrt 3} 8 = 0 \\
&=&(y^3+z^3) + \left(3yz - \frac 3 4\right)(y+z) + \frac {\sqrt 3} 8
\end{array}
Now we make our choice for z such that $3yz - \frac 3 4 = 0$, which will make the second term vanish.
That is, we substitute:
$$z = \frac 1 {4y} \qquad\qquad (3)$$
The result is:
\begin{array}{lcl}
(y^3+\frac 1 {4^3 y^3}) + \frac {\sqrt 3} 8 &=& 0 \\
y^6 + \frac {\sqrt 3} 8 y^3 + \frac 1 {4^3} &=& 0
\end{array}
Solving the equivalent quadratic equation gives:
$$y^3 = \frac 1 8 \left(-\frac 12 \sqrt 3 \pm \frac 1 2 i\right) = \frac 1 8 \exp\left(\pm \frac 5 6 \pi i\right)$$
The corresponding solutions are:
$$y=\frac 1 2 \exp\left(\pm \frac 5 {18} \pi i\right), \quad \frac 1 2 \exp\left(\pm \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\pm \frac {29} {18} \pi i\right)$$
Substituting in (3) gives us:
$$z=\frac 1 2 \exp\left(\mp \frac 5 {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {29} {18} \pi i\right)$$
In other words, in each case we have $z=\bar y$.
As a result, from (2) we find $x=y+z=y+\bar y=2\Re (y)$, meaning:
$$x=\cos\left( \frac 5 {18} \pi\right), \quad\cos\left( \frac {17} {18} \pi\right), \quad\cos\left( \frac {29} {18} \pi\right)$$

Therefore the solution is
$$a=4, \quad b= \cos\left( \frac 5 {18} \pi\right), \quad c=\cos\left( \frac {17} {18} \pi\right), \quad d=\cos\left( \frac {29} {18} \pi\right). \qquad \blacksquare$$

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#### anemone

##### MHB POTW Director
Staff member
Clearly $a=4$. For the rest ...
$\cos(3\theta) = 4\cos^3\theta - 3\cos\theta$, so if we put $x=\cos\theta$ then the equation $4x^3-3x+\tfrac{\sqrt{3}}{2}=0$ becomes $\cos(3\theta) = -\tfrac{\sqrt3}2 = \cos150^\circ$, with solutions $\theta = 50^\circ,\;170^\circ,\;290^\circ$ or, if you prefer, $50^\circ,\;70^\circ,\;170^\circ$. Thus we can take $b= \cos50^\circ$, $c = \cos70^\circ$, $d = -\cos10^\circ$.

Edit. Sorry! I took so long writing this that I failed to see that Mark had already replied.
Hi Opalg, don't be sorry because I bet you have no idea how much I hope you would reply to all of my challenge problems!

Hi kaliprasad, hmm...so, that blog is yours? Thanks for sharing it with us and I've checked it out and found that you have quite a collection of interesting problems in your blog! My hearty wishes for your blogging success!

\displaystyle \displaystyle \begin{align*} a\left( x-b \right) \left( x-c\right) \left( x-d \right) = a \, x^3 - \left( a\,b + a\,c + a\,d \right) x^2 + \left( a\,b\,c + a\,b\,d + a\,c\,d \right) x - a\,b\,c\,d \end{align*}

Equating the like powers of x we find

\displaystyle \displaystyle \begin{align*} a &= 4 \\ \\ - \left( 4b + 4c + 4d \right) &= 0 \\ b + c + d &= 0 \\ \\ 4\,b\,c + 4\,b\,d + 4\,c\,d &= -3 \\ b\,c + b\,d + c\,d &= -\frac{3}{4} \\ \\ 4\,b\,c\,d &= \frac{\sqrt{3}}{2} \\ b\,c\,d &= \frac{\sqrt{3}}{8} \end{align*}

Rearranging the first equation we have \displaystyle \displaystyle \begin{align*} b = -c - d \end{align*} and substituting into the second equation we find

\displaystyle \displaystyle \begin{align*} \left( - c - d \right) c + \left( -c - d \right) d + c\,d &= -\frac{3}{4} \\ -c^2 - c\,d - c\,d - d^2 + c\,d &= -\frac{3}{4} \\ c^2 + c\,d + d^2 &= \frac{3}{4} \\ c^2 + c\,d + \left( \frac{d}{2} \right) ^2 - \left( \frac{d}{2} \right) ^2 + d^2 &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 + \frac{3d^2}{4} &= \frac{3}{4} \\ \left( c + \frac{d}{2} \right) ^2 &= \frac{ 3 - 3d }{4} \\ c + \frac{d}{2} &= \frac{ \pm \sqrt{ 3 - 3d }}{2} \\ c &= \frac{-d \pm \sqrt{ 3 - 3d } }{2} \end{align*}

and so \displaystyle \displaystyle \begin{align*} b = - \left( \frac{-d \pm \sqrt{ 3 - 3d }}{2} \right) - d = \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \end{align*}

Substituting into the final equation, we find

\displaystyle \displaystyle \begin{align*} \left( \frac{ - d \mp \sqrt{ 3 - 3d }}{2} \right) \left( \frac{-d \pm \sqrt{ 3 - 3d } }{2} \right) d &= \frac{\sqrt{3}}{8} \\ \left( -d \mp \sqrt{ 3 - 3d} \right) \left( -d \pm \sqrt{ 3 - 3d } \right) d &= \frac{\sqrt{3}}{2} \\ \left[ d^2 - \left( 3 - 3d \right) \right] d &= \frac{\sqrt{3}}{2} \\ d^3 + 3d^2 - 3d - \frac{\sqrt{3}}{2} &= 0 \end{align*}

Now making the change of variable \displaystyle \displaystyle \begin{align*} u = d + 1 \end{align*}, we find

\displaystyle \displaystyle \begin{align*} \left( u - 1 \right) ^3 + 3 \left( u - 1 \right) ^2 - 3 \left( u - 1 \right) - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 3u^2 + 3u - 1 + 3u^2 - 6u + 3 - 3u + 3 - \frac{\sqrt{3}}{2} &= 0 \\ u^3 - 6u - \frac{\sqrt{3} - 10}{2} &= 0 \end{align*}

and now applying the Cubic Formula, we have

\displaystyle \displaystyle \begin{align*} u &= \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} + \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } + \sqrt[3]{ \frac{\frac{\sqrt{3} - 10}{2} - \sqrt{\left( \frac{\sqrt{3} - 10}{2} \right) ^2 - \frac{4 \left( - 6 \right) ^3}{27} }}{2} } \\ &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d + 1 &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } \\ d &= \sqrt[3]{ \frac{\sqrt{3} - 10 + \sqrt{ 231 - 20\sqrt{3} }}{4} } + \sqrt[3]{ \frac{\sqrt{3} - 10 - \sqrt{ 231 - 20\sqrt{3} }}{4} } - 1 \end{align*}

Back-substituting will enable us to evaluate b and c.
Hi Prove It, thanks for providing us this neat and well written solution and I appreciate that you took the time to participate in this problem.

I have always liked Cardano's method that I learned a long long time ago from a book of my father. Since I still have fond sentimental memories of it, I'm presenting it as well.

We can write the cubic expression in equation form as:
$$x^3 - \frac 3 4 x + \frac {\sqrt 3} 8 = 0 \qquad (1)$$
We substitute
$$x=y+z \qquad\qquad (2)$$
giving us a free choice for either $y$ or $z$:
\begin{array}{lcl}
(y+z)^3 - \frac 3 4 (y+z) + \frac {\sqrt 3} 8
&=&(y^3+z^3) + (3y^2z+3yz^2) - \frac 3 4 (y+z) + \frac {\sqrt 3} 8 = 0 \\
&=&(y^3+z^3) + \left(3yz - \frac 3 4\right)(y+z) + \frac {\sqrt 3} 8
\end{array}
Now we make our choice for z such that $3yz - \frac 3 4 = 0$, which will make the second term vanish.
That is, we substitute:
$$z = \frac 1 {4y} \qquad\qquad (3)$$
The result is:
\begin{array}{lcl}
(y^3+\frac 1 {4^3 y^3}) + \frac {\sqrt 3} 8 &=& 0 \\
y^6 + \frac {\sqrt 3} 8 y^3 + \frac 1 {4^3} &=& 0
\end{array}
Solving the equivalent quadratic equation gives:
$$y^3 = \frac 1 8 \left(-\frac 12 \sqrt 3 \pm \frac 1 2 i\right) = \frac 1 8 \exp\left(\pm \frac 5 6 \pi i\right)$$
The corresponding solutions are:
$$y=\frac 1 2 \exp\left(\pm \frac 5 {18} \pi i\right), \quad \frac 1 2 \exp\left(\pm \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\pm \frac {29} {18} \pi i\right)$$
Substituting in (3) gives us:
$$z=\frac 1 2 \exp\left(\mp \frac 5 {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {17} {18} \pi i\right), \quad\frac 1 2 \exp\left(\mp \frac {29} {18} \pi i\right)$$
In other words, in each case we have $z=\bar y$.
As a result, from (2) we find $x=y+z=y+\bar y=2\Re (y)$, meaning:
$$x=\cos\left( \frac 5 {18} \pi\right), \quad\cos\left( \frac {17} {18} \pi\right), \quad\cos\left( \frac {29} {18} \pi\right)$$

Therefore the solution is
$$a=4, \quad b= \cos\left( \frac 5 {18} \pi\right), \quad c=\cos\left( \frac {17} {18} \pi\right), \quad d=\cos\left( \frac {29} {18} \pi\right). \qquad \blacksquare$$
Hi I like Serena, WOW! Thank you kindly for this piece of well thought out strategic plan to tackle this cubic function!

I am so happy that all of you so generously shared the different approaches to solve this problem with me!

I love you guys!