Problem 74 - Algebra A Combined Function

Question

Factor each trinomial completely. See Examples 1 through 7. \(-5+55 x-50 x^{2}\)

Step-by-Step Solution

Answer

Factor the trinomial as \(5(10x + 1)(1-x)\).

Step 1: Write the Trinomial in Standard Form

The given trinomial is \(-5 + 55x - 50x^2\). To write it in standard form, arrange the terms in descending order of their exponents. This gives us \(-50x^2 + 55x - 5\).

Step 2: Factor Out the Greatest Common Factor

First, identify the greatest common factor (GCF) from the coefficients \(-50\), \(55\), and \(-5\). The GCF is \(5\). Factor \(5\) out of the trinomial:\[ 5(-10x^2 + 11x - 1) \]

Step 3: Factor the Quadratic

Now focus on the quadratic expression \(-10x^2 + 11x - 1\). To factor this quadratic, look for two numbers that multiply to \(ac = -10(-1) = 10\) and add to \(b = 11\). These numbers are \(10\) and \(1\).

Step 4: Rewrite and Factor by Grouping

Rewrite the quadratic by splitting the middle term using the numbers found in Step 3:\[-10x^2 + 10x + x - 1\]Now, factor by grouping:\[(-10x^2 + 10x) + (x - 1)\]Factor out the common factors in each group:\[10x(-x + 1) + 1(-x + 1)\]

Step 5: Combine the Terms

Notice that \(-x + 1\) is a common factor in both groups. Factor this out:\[(10x + 1)(-x + 1)\]

Step 6: Final Result

Combine the factored terms with the GCF factored out earlier. The fully factored form of the original trinomial is:\[5(10x + 1)(1 - x)\]

Key Concepts

Greatest Common Factor

The greatest common factor, or GCF, is a crucial concept when simplifying expressions, particularly trinomials. It represents the largest factor that divides the coefficients of each term in the expression without leaving a remainder. Finding the GCF is the first step in factoring any expression because it simplifies the original expression into a more manageable form.

Consider the trinomial given in the exercise:

• Coefficients:
  • -50 for the term \(-50x^2\)
  • 55 for the term \(55x\)
  • -5 for the constant term \(-5\)

To determine the GCF among these coefficients, list all possible factors:

• -50 factors into: 1, 2, 5, 10, 25, 50
• 55 factors into: 1, 5, 11, 55
• -5 factors into: 1, 5

Thus, the greatest number common to all three is 5. By factoring out 5, we simplify the trinomial into a more straightforward expression, setting the stage for further factoring. This is why finding the GCF is vital—it provides an easier pathway to completely factor the polynomial.

Quadratic Expressions

Quadratic expressions appear in the form \(ax^2 + bx + c\). In fact, factoring these expressions is a significant skill in algebra because it allows us to solve equations and simplify expressions. The expression in our problem has the standard form of a quadratic:

• -10 as \(a\), the coefficient of \(x^2\)
• 11 as \(b\), the coefficient of \(x\)
• -1 as \(c\), the constant term

Finding factors that multiply to the product \(ac\) and add up to \(b\) is key to splitting the middle term. In this case, our task was to find two numbers whose product is \(10\) and sum is \(11\). The numbers \(10\) and \(1\) make this possible, allowing us to rewrite the middle term and set up for the next step: factoring by grouping. Breaking down quadratics in this manner makes the process of influencing transformations much simpler and illustrates the importance of quadratic forms in algebra.

Factoring by Grouping

Factoring by grouping is a helpful technique when dealing with expressions that are not straightforward to factor immediately. After rewriting the quadratic with four terms instead of three, we can then focus on grouping terms into pairs to simplify and factor them further. Let's walk through this idea using the expression \(-10x^2 + 10x + x - 1\):

• Group the terms:
  • The first group \(-10x^2 + 10x\)
  • The second group \(x - 1\)
• Factor each group separately:
  • The first group is factored as \(10x(-x + 1)\)
  • The second group is factored as \(1(-x + 1)\)
• The common factor \(-x + 1\) appears in both grouped sections, indicating that it can be factored out again, resulting in \((10x + 1)(-x + 1)\).

By factoring by grouping, we efficiently handle expressions that initially appear difficult to calculate. It reveals a neat way to break down complex expressions using simple algebraic maneuvers. This technique is especially valuable when direct factoring methods don't immediately apply.