Problem 66 - A Graphical Approach to College Algebra

Question

Completely factor each polynomial by substitution. $$9(a-4)^{2}+30(a-4)+25$$

Step-by-Step Solution

Answer

The expression factors to \((3a - 7)^2\).

Step 1: Recognizing the Structure

The given expression is \(9(a-4)^{2}+30(a-4)+25\). Notice that it follows the form of a quadratic equation \(Ax^2 + Bx + C\), with \(x = (a-4)\). Thus, we can use substitution for simplification.

Step 2: Substituting the Variable

Let \(x = a-4\). Then our expression becomes \(9x^2 + 30x + 25\). It is now a quadratic expression in terms of \(x\).

Step 3: Factoring the Quadratic

To factor \(9x^2 + 30x + 25\), find two numbers that multiply to \(9 \times 25 = 225\) and add to 30. These numbers are 15 and 15.

Step 4: Completing the Factorization

This quadratic can be expressed as a perfect square trinomial:\[9x^2 + 30x + 25 = (3x + 5)^2\]. This means the polynomial factors completely into \((3x + 5)(3x + 5)\) or \((3x + 5)^2\).

Step 5: Substitute Back the Original Expression

Replace \(x\) with \((a-4)\) to revert to the original variable:\[(3(a-4) + 5)^2 = (3a - 12 + 5)^2 = (3a - 7)^2\].

Step 6: Final Result

The completely factored form of \(9(a-4)^{2}+30(a-4)+25\) is \((3a - 7)^2\).

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and appear in the standard form: \[Ax^2 + Bx + C = 0\] where \(A\), \(B\), and \(C\) are constants, and \(x\) represents the variable. They are called 'quadratic' because the highest power of the unknown, \(x\), is squared (\(x^2\)).

Quadratic equations have a parabolic graph, meaning they form a "U" shape when plotted. They can open upwards or downwards, depending on the leading coefficient \(A\): if \(A\) is positive, the parabola opens upwards, and if negative, it opens downwards.

The solutions to a quadratic equation, known as roots, can be found using several methods:

• Factoring
• Completing the square
• The quadratic formula: \[x = \frac{{-B \pm \sqrt{{B^2 - 4AC}}}}{2A\]

Understanding how to rewrite a quadratic expression, identify its components, and manipulate it to reveal its factors is crucial when simplifying and solving these equations.

Substitution in Algebra

Substitution in algebra is a technique used to simplify expressions, allowing you to work with cleaner, more manageable terms. When we substitute, we replace a variable or group of variables with another expression.

In the exercise example, we use substitution to simplify a complex polynomial expression. By recognizing that the polynomial looks like a quadratic in disguise, we set \(x = (a-4)\). This transforms our expression into \[9x^2 + 30x + 25\],which is easier to tackle.

Substitution makes it simpler to use methods of solving quadratic equations. This process involves:

• Identifying a variable or expression within a larger equation
• Replacing it with a single variable to reduce complexity
• Solving or simplifying the new equation
• Remembering to substitute back the original expressions at the end

This approach is particularly effective in making seemingly unwieldy algebra problems more approachable.

Perfect Square Trinomials

A perfect square trinomial is a special type of quadratic expression that can be expressed as the square of a binomial. It takes the form:\[(ax + b)^2 = a^2x^2 + 2abx + b^2\]This implies a strong link between multiplication and squaring. Perfect square trinomials are useful because they simplify the process of factoring polynomials significantly.

In the provided exercise, after substitution, we recognized \[9x^2 + 30x + 25\]as a perfect square trinomial. Here, finding two identical numbers that both sum to \(B = 30\) and multiply to the product of \(9\) and \(25\) (i.e., 225) confirms its status as a perfect square trinomial.

This gives us the simple factorization:\[(3x + 5)^2\], meaning \[3x + 5\times3x + 5\].Recognizing perfect squares helps in simplifying expressions and solving quadratic equations by quickly rewiring the way expressions are perceived and analyzed.