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In these lessons, we will consider how to solve cubic equations of the form

*px*^{3} + *qx*^{2} + *rx* + *s* = 0 where *p, q, r * and *s* are constants by using the Factor Theorem and Synthetic Division.

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More Algebra Lessons

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**Example:**

Find the roots of f(*x*) = 2x^{3} + 3*x*^{2} – 11*x* – 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.

2x^{3} + 3*x*^{2} – 11*x* – 6

= (*x* – 2)(*ax*^{2} + *bx + c*)

= (*x* – 2)(2*x*^{2} + *bx + *3)

= (*x* – 2)(2*x*^{2} + 7*x + *3)

= (*x* – 2)(2*x* + 1)(*x* +3)

So, the roots are

** Example: **

Solve the cubic equation *x*^{3} – 7*x*^{2} + 4*x* + 12 = 0

** Solution: **

Let f(*x*) = *x*^{3} – 7*x*^{2} + 4*x* + 12

The possible values are

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

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

* **x*^{3} – 7*x*^{2} + 4*x* + 12

= (*x* + 1)(*x*^{2} – 8*x* + 12)

= (*x* + 1)(*x* – 2)(*x* – 6)

So, the roots are –1, 2, 6

This video demonstrates how to use the Factor Theorem to factor polynomials.

1) Factor P(x) =
3x^{3} − x^{2} − 19x + 8

1) Factor P(x) =
2x^{3} − 9x^{2} + 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) =
2x^{3} − 3x^{2} − 5x + 6

Using the Factor Theorem to solve a cubic equation

Solve the equation 2x^{3} −5x^{2} − 10 = 23x

Solving Cubic Equations using the Factor theorem and Long Division

Solve the cubic equation 2x

Solving a Cubic Equation

3x

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