Question

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Factor completely: \(x^{4}-29 x^{2}+100\)

Short answer

(x - 2)(x + 2)(x - 5)(x + 5)

Step-by-step solution

Recognize the Polynomial Form

Notice that the given polynomial, \(x^{4} - 29x^{2} + 100\), is a quadratic in terms of \(x^{2}\). Rewrite it as \((x^{2})^{2} - 29x^{2} + 100\).

Substitute Variable

Let \(y = x^{2}\). Substituting, the equation becomes \(y^{2} - 29y + 100\).

Factor the Quadratic

Factor the quadratic equation \(y^{2} - 29y + 100\). Look for two numbers that multiply to 100 and add up to -29. These numbers are -4 and -25. Thus, the factorization is \((y - 4)(y - 25)\).

Substitute Back

Replace \(y\) with \(x^{2}\). The factored form is \((x^{2} - 4)(x^{2} - 25)\).

Factor Further

Notice that both \(x^{2} - 4\) and \(x^{2} - 25\) are difference of squares. Factor these further: \(x^{2} - 4 = (x - 2)(x + 2)\) and \(x^{2} - 25 = (x - 5)(x + 5)\).

Write Final Answer

Combine all factored parts. The complete factorization is \((x - 2)(x + 2)(x - 5)(x + 5)\).

Key concepts

Quadratic Equations

Quadratic equations are expressions that can be written in the form of \(ax^2 + bx + c = 0 \). This is often called the standard form, where \(a, b, \text{ and } c\) are constants, and \(a eq 0\). To solve a quadratic equation, you typically look for values of \(x \) that satisfy this equation. There are various methods to solve quadratic equations:

• Factoring: Expressing the quadratic into two binomials that multiply together.


• Completing the square: Rearranging the equation to create a perfect square trinomial.


• Quadratic formula: Using the formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \) to find the solutions.

Quadratic equations come up in various contexts, including physics, finance, and engineering, making them essential to understand. In the given exercise, we solved the quadratic \(y^2 - 29y + 100 = 0 \) as part of the polynomial factorization process.

Difference of Squares

The difference of squares is a special factoring pattern. It applies to expressions of the form \(a^2 - b^2\). This can be factored into two binomials:
\[ a^2 - b^2 = (a - b)(a + b) \]
This is because when you expand \((a - b)(a + b)\), you get:
\[ (a - b)(a + b) = a^2 + ab - ab - b^2 = a^2 - b^2 \]
In our exercise, we used this concept to factor both \(x^2 - 4 \) and \(x^2 - 25 \). Specifically, we have:
\[ x^2 - 4 = (x - 2)(x + 2) \]
\[ x^2 - 25 = (x - 5)(x + 5) \]
Recognizing and applying the difference of squares can simplify many algebraic expressions and equations, making larger problems easier to solve. Always look out for expressions that fit this pattern!

Substitution Method

The substitution method is a technique used in algebra to simplify complex expressions and equations. The idea is to replace a part of the equation with a simpler variable. Here's a step-by-step rundown of how it works:

• Identify the substitution: Look for a commonly repeating term and assign it a new variable.


• Replace and simplify: Substituting the new variable into the expression to make it simpler.


• Solve the new equation: This often turns a complex polynomial into a more straightforward quadratic or linear equation.


• Substitute back: Once the simpler equation is solved, replace the variable back with the original term.

In our exercise, we set \(y = x^2\) to transform \(x^4 - 29x^2 + 100 \) into a quadratic equation \(y^2 - 29y + 100 \). This made it possible to factor the polynomial easily. The substitution method helps in breaking down complex problems into manageable steps, leading to efficient and accurate solutions.