Solving Cubic Equations

In this lesson, we will consider how to solve cubic equations of the form
px3 + qx2 + rx + s = 0 where p, q, r and s are constants by using the Factor Theorem and Synthetic Division.

Related Topics:
More Algebra Lessons

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Example:

Find the roots of f(x) = 2x3 + 3x2 – 11x – 6 = 0, given that it has at least one integer root.

Solution:

Since the constant in the given equation is a 6, we know that the integer root must be a factor of 6. The possible values are

We can use the factor theorem to test the possible values by trial and error.

f(1) = 2 + 3 – 11 – 6 ≠ 0

f(–1) = –2 + 3 + 11 – 6 ≠ 0

f(2) = 16 + 12 – 22 – 6 = 0

We find that the integer root is 2. Now we need to find the other roots. We can do that either by inspection or by synthetic division.

2x3 + 3x2 – 11x – 6
= (x – 2)(ax2 + bx + c)
= (x – 2)(2x2 + bx + 3)
= (x – 2)(2x2 + 7x + 3)
= (x – 2)(2x + 1)(x +3)

So, the roots are

Example:

Solve the cubic equation x3 – 7x2 + 4x + 12 = 0

Solution:

Let f(x) = x3 – 7x2 + 4x + 12

The possible values are

We find that f(–1) = –1 – 7 – 4 + 12 = 0

So, (x + 1) is a factor of f(x)

x3 – 7x2 + 4x + 12
= (x + 1)(x2 – 8x + 12)
= (x + 1)(x – 2)(x – 6)

So, the roots are –1, 2, 6

Videos

This video demonstrates how to use the Factor Theorem to factor polynomials.
1) Factor P(x) = 3x3 − x2 − 19x + 8
1) Factor P(x) = 2x3 − 9x2 + x + 12

 

The Remainder Theorem and the Factor Theorem :
What the theorems are and how they can be used to find the linear factorization of a polynomial.
Factor F(x) = 2x3 − 3x2 − 5x + 6

Using the Factor Theorem to solve a cubic equation
Solve the equation 2x3 −5x2 − 10 = 23x

Solving cubics with comparing coefficients and polynomial long division cubic equations.
Solve x3 −4x2 + x + 6 = 0







Solving Cubic Equations using the Factor theorem and Long Division
Solve the cubic equation 2x3 − 6x2 + 7x − 1 = 0



Solving a Cubic Equation
3x3 −4x2 − 17x = x3 + 3x2 − 10







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