Question

\(\qquad \qquad \qquad \qquad \qquad \huge{\color{Blue}{T}\color{Yellow}{U}\color{Magenta}{T}\color{Pink}{O}\color{MidnightBlue}{R} I\color{Blue}{A}\color{Green}{L}}\)

Solving Cubic Equations of the form:\(Ax^3+Bx^2+Cx+D=0 \quad (A \neq 0)\)
Note: This requires pre-requisite knowledge of synthetic division(or division of polynomials) and solving quadratic equations.

Answer 1

A cubic equation is the equation in x with highest degree as 3(the maximum power of x=3), of the form \(Ax^3+Bx^2+Cx+D=0 \quad (A!=0)\)
The very first thing that should be tried out is to see whether it can be factorized directly in the form \((px^2+q)(rx+s)=0\)
Example:
\(
\quad \boxed{ \begin{align}
\quad & \color{lime}{\Rightarrow} \quad & 2x^3+x^2+6x+3=0 \qquad \\
& \color{lime}{\Rightarrow} & \quad x^2(2x+1)+3(2x+1)=0 \qquad \\
& \color{lime}{\Rightarrow} & \quad (x^2+3)(2x+1)=0 \qquad
\end{align} }
\)
So one root is \(x=\frac{-1}{2}\), solve \(x^2+3=0\) as a quadratic equation.

If you cannot factorize it in such a way, then try to guess one root(Rational Root Theorem might be useful), generally one root is a small integer (not always) say we guess x=a is the root, then verify it by putting x= a in \(Ax^3+Bx^2+Cx+D\), we should get 0;
then a synthetic division using ‘a’ will give a quadratic equation that can be solved using any method \(or\) dividing the polynomial \(Ax^3+Bx^2+Cx+D\) by (x-a) will also give a quadratic polynomial which when equated with 0 will give same quadratic equation.

The rule below will help in determining whether x=1 and/or x=-1 is a root of the cubic equation or not:
\(\textbf{1.Find the sum of coefficients of odd power terms(coefficients of \(x^3\) and x here)}\)
\(\textbf{2.Find the sum of coefficients of even power terms(coefficient of \(x^2\) and constant)}\)
If these two are equal then \(\huge{(x=-1)}\) is the root and \(\huge{(x+1)}\) is the factor.
If the sum of these two=0, then \(\huge{(x=1)}\) is the root and \(\huge{(x-1)}\) is the factor.
Once we get x=+1 and/or x=-1 as roots we can always do synthetic division or divide the given cubic polynomial by \((x \pm 1)\) and find the quadratic equation which gives other two roots.

Example: \(x^3+7x^2+14x+8=0\)

has sum of co-efficients of odd power terms = 1+14=15
sum of co-efficients of even power terms =7+8=15

so, x=-1 is the root or (x+1) is a factor of \(x^3+7x^2+14x+8=0.\)
Then using polynomial division, we get the quadratic equation as \( (x^2+6x+8)=0\)
Which easily gives roots as x=-2,x=-4.
Hence the 3 roots are \(x=-1,-2,-4.\)

Verify that both x=1,x=-1 are the roots of \(x^3+3x^2-x-3=0\)

Note: This rule is valid for an equation with any arbitrary degree. So even \(x^7+x^6+x^5+x^4+x^3+x^2+x+1=0\) or \(x^9+1=0\) have x=-1 as a root or x+1 as a factor.

Now, let’s answer what is a good guess that can be taken to find first root.
The answer can be given by ‘Rational Root Theorem’ (ofcourse assuming that we have one rational root)
If D and A are nonzero, then each rational solution x, when written as a fraction \(x =\frac{ p}{q}\) in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies
 p is an integer factor of the constant term D, and
 q is an integer factor of the leading coefficient A.
Thus, a list of possible rational roots of the equation can be derived using the formula.

Example: \(2x^3-x^2-8x+4=0\)
Here \(\frac{p}{q}=\pm \frac{1,2,4}{1,2} \)
So now your set of one of the possible roots reduces to only \(\pm 1,\pm \frac{1}{2},\pm 4,\pm 2 \)
Incidentally this set gave all the roots \( (2,-2,\frac{1}{2})\) because all roots were rational as this theorem only works for rational roots. But, it will greatly help to solve cubic equations because once we find a root, then getting other roots is just a division and solving quadratic equation away.
________________________________________________________________________________
When options are given, it becomes easier to get the answer. First thing to be done id to substitute one by one, the roots given in options in \(Ax^3+Bx^2+Cx+D\) and check whether it evaluates to 0.
But, this sometimes can become tedious task.
Example: \(x^3-12x^2-45x+616=0\)
Options:
1. -8,-11,-7
2. 8,11,-7
3. 8,-11,7
4. 11,4,14
Here, substituting a root from the option requires lot of calculations.
Instead we can use this: if p,q,r are the roots of \(Ax^3+Bx^2+Cx+D=0\) then
Sum of roots=p+q+r=\(\frac{- B}{A}\)
Product of roots = pqr=\(\frac{- D}{A}\)
pq+pr+qr=\(\frac{C}{A}\)

In the above example only using sum of roots(=12) formula we can get the correct roots as 8,11,-7.

Answer 2

special thanks to @waterineyes for helping me out with some latex part.

Answer 3

A! = 0?

Answer 4

Wouldn't that be an equation of second degree?

Answer 5

ohh,sorry that should be A not equal to 0

Answer 6

\ne <- for latex

Answer 7

Have to go now..

Surely read when I will come..

LATEX is looking nice though...

Answer 8

*(co-efficients of \(x^3\) and x here)
* (co-efficient of \(x^2\) and constant)
the preview for that was showing correctly.....

Answer 9

that should be \(A \neq 0\)

Answer 10

Yeah But when it is posted it requires some extra space..

So your text got outwards..

Answer 11

You can edit the initial question to \ne, by the way.

Answer 12

Yep..

Answer 13

next week tutorial topic is Limits.

Answer 14

good tutorial .

Answer 15

*bookmark

Answer 16

Notes:
Rational root theorem can only be applied to polynomials with integer coefficients, i.e, \(A,B,C,D\in\mathbb{Z}\).
So if you have a polynomial with rational (but not integer) coefficients, you need to multiply both sides (the RHS will cancel, but it's just good practice to remember this) of the cubic equation by the least common multiple of the denominators. In other words, if we have:\[\frac{\pi_1}{\pi_2}x^3+\frac{\rho_1}{\rho_2}x^2+\frac{\sigma_1}{\sigma_2}x+\frac{\tau_1}{\tau_2}=0,~~~~\pi_1,\pi_2,\rho_1,\rho_2,\sigma_1,\sigma_2,\tau_1,\tau_2\in\mathbb{Z}\]Then we multiply the equation by \(\text{lcm} (\pi_2,\rho_2,\sigma_2,\tau_2)\), or, if we're lazy, by \(\pi_2\cdot\rho_2\cdot\sigma_2\cdot\tau_2\), to get a cubic equation with integer coefficients, \(Ax^3+Bx^2+Cx+D=0\), so we can use rational root theorem.

If you have irrational algebraic coefficients, well, you could probably do it, eventually; but if you have transcendental coefficients, you might as well use the general cubic equation or approximate.

Answer 17

otherwise, good tutorial!

Answer 18

Thanks fo making it more accurate @Herp_Derp