Problem 46 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$9 x^{3}-9 x$$

Step-by-Step Solution

Answer

Therefore, the completely factored form of the given polynomial \(9x^{3}-9x\) is given as \(9x(x+1)(x-1)\).

Step 1: Identify Common Factors

First, find a common factor in the given polynomial. Here, the common factor is \(9x\). This is done by observation and understanding the properties of integers. So, the initial factored form would be \(9x(x^{2} - 1)\).

Step 2: Factorize Further

Next step is to factorize \(x^{2} - 1\), which is a difference of squares. A difference of squares can be factored into \((a+b)(a-b)\). In this case, \(a\) is \(x\) and \(b\) is \(1\), leading to the factors of \(x+1\) and \(x-1\). Therefore, further factorization of the polynomial yields \(9x(x+1)(x-1)\).

Step 3: Checking the Factorization

To check the factorization, multiply all the factors. If you get back to the original polynomial, then the factorization is correct. \(9x(x+1)(x-1)\) would be \(9x(x^{2}-1)\), which simplifies to \(9x^{3}-9x\), so the factorization is correct.

Key Concepts

Common Factors

When tackling polynomial factorization, spotting common factors is a great starting point. This means looking for elements that are present in every term of the polynomial. In the example given, the polynomial is \(9x^{3}-9x\). Both terms contain \(9x\) as a common factor.

Once the common factor is identified, you can factor it out of the polynomial. This simplifies the expression and is like peeling off layers to get closer to its basic components. Here, extracting \(9x\) from the polynomial yields \(9x(x^2 - 1)\).

Recognizing common factors makes complicated problems more manageable. They reduce your workload by shrinking the polynomial down to more familiar forms that are easier to handle.

Difference of Squares

The difference of squares is a special algebraic expression, written as \(a^2 - b^2\). It can be instantly factored into \((a - b)(a + b)\). This applies when a polynomial consists of two squared terms with a subtraction in between.

In the polynomial from our exercise, \(x^2 - 1\) is a difference of squares. Here, \(a\) is \(x\) and \(b\) is 1. By applying the formula, you factor it as \((x - 1)(x + 1)\).

This technique is handy for breaking down more complex polynomials and is essential to having a toolkit of algebraic strategies. Learning to identify the difference of squares will turn factorization into a straightforward task.

Multiplication Verification

Once you've factored the polynomial using common factors and special formulas, it's time to verify your work. This is done by multiplying all the factors back together. Checking your factorization this way ensures accuracy and helps catch any mistakes.

For our factored polynomial, \(9x(x + 1)(x - 1)\), we multiply back to see if we arrive at the original polynomial.

• First, calculate \((x + 1)(x - 1)\) to get \(x^2 - 1\).
• Then, multiply \(9x\) by \(x^2 - 1\) to return to \(9x^3 - 9x\).

This check confirms that the factorization was done correctly. Multiplication verification solidifies your understanding and reassures you that you've got the correct solution.