Difference between revisions of "Cubic polynomial"

Revision as of 00:19, 18 July 2018

A cubic polynomial is a polynomial of the form $ax^3+bx^2+cx+d=0$ . 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

$x^3 + ax^2 + bx + c = 0$

Now, we will make a change in variables to get rid of the $x^2$ term. If we do the substitution $x = y - \frac{a}{3}$ , this does the trick. Our new equation is of the form

$y^3 + py = q$

We do another substitution $y = w - \frac{p}{3w}$ , and our new equation is of the form

$w^3 + rw^{-3} = q$

We can now turn this into a quadratic in terms of $w^3$ , solve for $w$ , and then solve for $y$ and finally $x$ .

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 $8x^3 - 3x^2 - 3x - 1 = 0$ can be written in the form $\frac{\sqrt[3]a + \sqrt[3]b + 1}{c}$ , where $a$ , $b$ , and $c$ are positive integers. Find $a+b+c$ .

2014 AIME I Problems/Problem 9 Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ .

2010 AIME II Problems/Problem 7 Let $P(z)=z^3+az^2+bz+c$ , where a, b, and c are real. There exists a complex number $w$ such that the three roots of $P(z)$ are $w+3i$ , $w+9i$ , and $2w-4$ , where $i^2=-1$ . Find $|a+b+c|$ .

@@ Line 17: / Line 17: @@
 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 <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>.
+[[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>

Page

Toolbox

Search

Difference between revisions of "Cubic polynomial"

Revision as of 00:19, 18 July 2018

Solving a cubic

Problems