Problem 88 - Algebra for College Students

Question

Should help you pull together all of the factoring techniques of this chapter. Factor completely each polynomial, and indicate any that are not factorable using integers. $$10 x^{3}+15 x^{2}+20 x$$

Step-by-Step Solution

Answer

The polynomial factors as \(5x(2x^2 + 3x + 4)\), and it cannot be further factored using integers.

Step 1: Identify Common Factors

Look at the polynomial, which is \(10x^3 + 15x^2 + 20x\). First, identify any common factors in the coefficients and the variables of each term. The greatest common factor (GCF) of the coefficients 10, 15, and 20 is 5. Additionally, each term contains at least one \(x\), so \(x\) is also a common factor.

Step 2: Factor Out the GCF

Divide each term in the polynomial by the GCF identified in Step 1, which is \(5x\).\[10x^3 + 15x^2 + 20x = 5x(2x^2 + 3x + 4)\]Now the expression inside the parenthesis, \(2x^2 + 3x + 4\), remains.

Step 3: Check for Further Factorization

Examine the quadratic \(2x^2 + 3x + 4\) inside the parenthesis. Consider attempting further factoring methods, such as factoring by grouping or using the quadratic formula if necessary.Since 2, 3, and 4 do not form a product-sum pair that can factor over integers, \(2x^2 + 3x + 4\) cannot be factored further with integer coefficients.

Step 4: Conclusion

The polynomial \(10x^3 + 15x^2 + 20x\) is completely factored as \(5x(2x^2 + 3x + 4)\). The quadratics inside the parenthesis are irreducible with integer coefficients.

Key Concepts

Greatest Common Factor

The greatest common factor (GCF) is the largest quantity that can exactly divide each term of a polynomial without leaving a remainder. To determine the GCF, one must first identify any common numerical factors across the coefficients of each term. For example, in the polynomial \(10x^3 + 15x^2 + 20x\), examine the coefficients 10, 15, and 20. The highest number that evenly divides all three numbers is 5. Therefore, 5 is a part of the GCF.
Another aspect to consider when identifying the GCF is common variables in each term. In our example, each term contains the variable \(x\). The smallest exponent of \(x\) present in each term is 1, making \(x\) a common factor as well.

• For coefficients 10, 15, 20, the GCF is 5.
• For variables, \(x\) is common with the smallest power being \(x^1\).

Thus, the complete GCF for the polynomial \(10x^3 + 15x^2 + 20x\) is \(5x\). By factoring out the GCF, the polynomial simplifies, revealing more about its structure.

Factoring Methods

Factoring polynomials involves several methods depending on the specific form and complexity. Initially, one always checks for a possible GCF, as we did with \(10x^3 + 15x^2 + 20x\). Once the GCF (\(5x\) here) is factored out, the polynomial may become quite simple.
Different factoring techniques include:

• Factoring by grouping: Useful when terms can be grouped into pairs that share common factors.
• Difference of squares: Applicable for expressions like \(a^2 - b^2 = (a + b)(a - b)\).
• Trinomials: Like \(ax^2 + bx + c\), often factored using trial methods or the quadratic formula.

Back with our example, once \(5x\) is factored out, the expression inside the parenthesis is \(2x^2 + 3x + 4\). We attempted factoring this further, but concluded it as irreducible over the integers since it doesn't conform to simple trinomial patterns. Each step enhances understanding and ensures all methods are exhausted.

Quadratic Equations

Quadratic equations are polynomials of degree two, typically expressed in the standard form \(ax^2 + bx + c = 0\). In our context, once the GCF was factored out, we were left with \(2x^2 + 3x + 4\), which is quadratic.Quadratics can often be further factored, but methods must be applied case-by-case:

• Trinomial factoring: Where the expression can be written as a product of two binomials, often straightforward with easy coefficients.
• Completing the square: Useful for expressions not easily factored otherwise.
• Quadratic Formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), providing roots directly when factoring doesn't yield integer results.

For \(2x^2 + 3x + 4\), attempting factorizing into binomials quickly shows no combination of sums and products yields integer solutions. Hence, this quadratic remains in its current polynomial form when restricted to integer coefficients.