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**How to solve cubic equations using the Factor Theorem?**

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.

f(–1) = –2 + 3 + 11 – 6 ≠ 0

f(2) = 16 + 12 – 22 – 6 = 0

We find that the integer root is 2.

Step 2: Find the other roots either by inspection or by synthetic division.

**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
**What are The Remainder Theorem and the Factor Theorem?**

How to use the Theorems to find the linear factorization of a polynomial?

Factor F(x) = 2x^{3} − 3x^{2} − 5x + 6

**How to use the Factor Theorem to solve a cubic equation?**

Solve the equation 2x^{3} −5x^{2} − 10 = 23x
How to solve a cubic equation?

How to solve Cubic Equations using the Factor theorem and Long Division?

Solve the cubic equation 2x^{3} − 6x^{2} + 7x − 1 = 0
Solving a Cubic Equation

3x^{3} −4x^{2} − 17x = x^{3} + 3x^{2} − 10

You can use the Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.

More Algebra Lessons, More Algebra Worksheets, More Algebra Games

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

**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

Step 1: Use the factor theorem to test the possible values by trial and error.

f(1) = 2 + 3 – 11 – 6 ≠ 0f(–1) = –2 + 3 + 11 – 6 ≠ 0

f(2) = 16 + 12 – 22 – 6 = 0

We find that the integer root is 2.

Step 2: Find the other roots 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

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

1) Factor P(x) = 3x

1) Factor P(x) = 2x

How to use the Theorems to find the linear factorization of a polynomial?

Factor F(x) = 2x

Solve the equation 2x

How to solve Cubic Equations using the Factor theorem and Long Division?

Solve the cubic equation 2x

3x

You can use the Mathway widget below to practice Algebra or other math topics. Try the given examples, or type in your own problem. Then click "Answer" to check your answer.

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