Difference between revisions of "Cubic polynomial"
m (Added a category) |
|||
| (One intermediate revision by one other user not shown) | |||
| Line 14: | Line 14: | ||
<math>w^3 + rw^{-3} = q</math> | <math>w^3 + rw^{-3} = q</math> | ||
| − | We can now turn this into a quadratic in terms of <math>w^3</math>, solve for <math>w</math>, and then solve for <math>y</math> and finally <math>x</math>. | + | We can now turn this into a quadratic in terms of <math>w^3</math>, solve for <math>w</math>, and then solve for <math>y</math> and finally <math>x</math>. ([[Cubic Equation|Here]] is another way to do it.) |
Because cubic polynomials have an odd degree, their end behaviors go in opposite directions, and therefore all cubic polynomials must have at least one real root. | Because cubic polynomials have an odd degree, their end behaviors go in opposite directions, and therefore all cubic polynomials must have at least one real root. | ||
| Line 21: | Line 21: | ||
Note: The above method will work for any cubic, but the cubic polynomials in these problems have special tricks to solve them. | Note: The above method will work for any cubic, but the cubic polynomials in these problems have special tricks to solve them. | ||
| − | [[2013 AIME I Problems/Problem 5]] The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. | + | *[[2013 AIME I Problems/Problem 5]] The real root of the equation <math>8x^3 - 3x^2 - 3x - 1 = 0</math> can be written in the form <math>\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}</math>, where <math>a</math>, <math>b</math>, and <math>c</math> are positive integers. Find <math>a+b+c</math>. |
| − | [[2014 AIME I Problems/Problem 9]] Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>. | + | *[[2014 AIME I Problems/Problem 9]] Let <math>x_1<x_2<x_3</math> be the three real roots of the equation <math>\sqrt{2014}x^3-4029x^2+2=0</math>. Find <math>x_2(x_1+x_3)</math>. |
| − | [[2010 AIME II Problems/Problem 7]] <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=z^3+az^2+bz+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> | + | *[[2010 AIME II Problems/Problem 7]] <!-- don't remove the following tag, for PoTW on the Wiki front page--><onlyinclude>Let <math>P(z)=z^3+az^2+bz+c</math>, where a, b, and c are real. There exists a complex number <math>w</math> such that the three roots of <math>P(z)</math> are <math>w+3i</math>, <math>w+9i</math>, and <math>2w-4</math>, where <math>i^2=-1</math>. Find <math>|a+b+c|</math>.<!-- don't remove the following tag, for PoTW on the Wiki front page--></onlyinclude> |
| + | |||
| + | [[category:Algebra]] | ||
Latest revision as of 11:16, 27 September 2024
A cubic polynomial is a polynomial of the form 
.
A cubic polynomial is a polynomial of degree 3. A univariate cubic polynomial has the form . An equation involving a cubic polynomial is called a cubic equation. A closed-form solution known as the cubic formula exists for the solutions of an arbitrary cubic equation.
Solving a cubic
If the leading coefficient of the cubic is not 1, then divide both sides by the leading coefficient so it is 1. You get an equation of the form
Now, we will make a change in variables to get rid of the 
term. If we do the substitution 
, this does the trick. Our new equation is of the form
We do another substitution 
, and our new equation is of the form
We can now turn this into a quadratic in terms of 
, solve for 
, and then solve for 
and finally 
. (Here is another way to do it.)
Because cubic polynomials have an odd degree, their end behaviors go in opposite directions, and therefore all cubic polynomials must have at least one real root.
Problems
Note: The above method will work for any cubic, but the cubic polynomials in these problems have special tricks to solve them.
- 2013 AIME I Problems/Problem 5 The real root of the equation
can be written in the form ![$\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$](//latex.artofproblemsolving.com/2/3/5/235e83f864e3b413c0b9b5eac719eea71dbcd5d1.png)
, where 
, 
, and 
are positive integers. Find 
.
can be written in the form ![$\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$](http://latex.artofproblemsolving.com/2/3/5/235e83f864e3b413c0b9b5eac719eea71dbcd5d1.png)
, where 
, 
, and 
are positive integers. Find 
.- 2014 AIME I Problems/Problem 9 Let
be the three real roots of the equation 
. Find 
.
be the three real roots of the equation 
. Find 
.- 2010 AIME II Problems/Problem 7 Let
, where a, b, and c are real. There exists a complex number 
such that the three roots of 
are 
, 
, and 
, where 
. Find 
.
, where a, b, and c are real. There exists a complex number 
are 
, 
, and 
, where 
. Find 
.
