Factoring Polynomials Using the Factor Theorem and Polynomial Long Division

Polynomial factorization is a fundamental skill in algebra that is widely used in various fields, including engineering, physics, and computer science. This article will guide you through the process of determining if a binomial is a factor of a polynomial and then fully factor the polynomial. We will use the example tx x3 - 5x2 - 9x 45 to demonstrate the method, and we will also explain the underlying principles behind these techniques. This guide is designed to be comprehensive and useful for anyone looking to enhance their algebraic problem-solving skills.

Factor Theorem and Its Application

To verify if x - 5 is a factor of the polynomial tx x3 - 5x2 - 9x 45, we use the Factor Theorem. The factor theorem states that a polynomial f(x) has a factor x - c if and only if f(c) 0. In our case, c 5. Let's calculate t(5):

t(5) 53 - 5(52) - 9(5) 45

Multiplying and simplifying:

t(5) 125 - 125 - 45 45 0

Since t(5) 0, we confirm that x - 5 is indeed a factor of tx.

Polynomial Long Division

Once we have established that x - 5 is a factor, we can perform polynomial long division to factor the polynomial completely. The steps are as follows:

Step 1: Set Up for Division

Divide the leading term of the dividend by the leading term of the divisor:

deg(dividend)/deg(divisor) x3/x x2

Multiply x2 by the divisor x - 5:

x2(x - 5) x3 - 5x2

Subtract this from the original polynomial:

x3 - 5x2 - 9x 45 - (x3 - 5x2) -9x 45

Step 2: Repeat the Division Process

Divide the leading term of the new polynomial by the leading term of the divisor:

-9x/x -9

Multiply -9 by the divisor x - 5:

-9(x - 5) -9x 45

Subtract this from the new polynomial:

-9x 45 - (-9x 45) 0

The division yields the quotient x2 - 9.

Therefore, we can write the original polynomial as:

tx (x - 5)(x2 - 9)

Factoring the Quadratic Polynomial

The quadratic polynomial x2 - 9 can be factored using the difference of squares formula:

x2 - 9 (x - 3)(x 3)

Thus, the complete factorization of the original polynomial is:

tx (x - 5)(x - 3)(x 3)

Using the Fundamental Theorem of Algebra

According to the Fundamental Theorem of Algebra, a polynomial of degree n has exactly n roots, counting multiplicities. In our polynomial tx x3 - 5x2 - 9x 45, the polynomial can be expressed in the form:

tx (x - 5)(x - 3)(x 3)

We already know one root 5, and by factoring the quadratic, we find the other two roots 3 and -3. We can verify this:

tx (x - 5)x - 3x 3

Multiplying out should give us back our original polynomial:

(x - 5)(x - 3)(x 3)

Expanding this, we get:

(x - 5)(x2 - 9) x3 - 9x - 5x2 45 x3 - 5x2 - 9x 45

So, we have confirmed our factorization is correct.

Conclusion

In conclusion, we have shown that x - 5 is a factor of tx and the complete factorization of the polynomial is tx (x - 5)(x - 3)(x 3). Understanding these techniques is crucial for solving more complex polynomial problems.