Problem 98 - Algebra and Trigonometry

Question

Factoring \(x^{4}+a x^{2}+b\) A trinomial of the form \(x^{4}+a x^{2}+b\) can sometimes be factored easily. For example \(x^{4}+3 x^{2}-4=\left(x^{2}+4\right)\left(x^{2}-1\right) .\) But \(x^{4}+3 x^{2}+4\) cannot be factored in this way. Instead, we can use the following method. \(x^{4}+3 x^{2}+4=\left(x^{4}+4 x^{2}+4\right)-x^{2} \quad\) Add and subtract \(x^{2}\) \(=\left(x^{2}+2\right)^{2}-x^{2} \quad\) Factor perfect square \(=\left[\left(x^{2}+2\right)-x\right]\left[\left(x^{2}+2\right)+x\right] \quad\) Difference of squares \(=\left(x^{2}-x+2\right)\left(x^{2}+x+2\right)\) Factor the following using whichever method is appropriate. (a) \(x^{4}+x^{2}-2\) (b) \(x^{4}+2 x^{2}+9\) (c) \(x^{4}+4 x^{2}+16\) (d) \(x^{4}+2 x^{2}+1\)

Step-by-Step Solution

Answer

\( (x^2 + 2)(x^2 - 1) = (x^2+2)(x+1)(x-1) \).

Step 1: Substitute u = x^2

Let \( u = x^2 \). The expression becomes \( u^2 + u - 2 \).

Step 2: Factor the quadratic

\( u^2 + u - 2 = (u + 2)(u - 1) \).

Step 3: Back-substitute

\( x^4 + x^2 - 2 = (x^2 + 2)(x^2 - 1) = (x^2 + 2)(x+1)(x-1) \).

Key Concepts

Quadratic Form

A quadratic form of a polynomial is an expression where the highest power of the variable is squared. In the context of polynomials like \(x^{4} + ax^{2} + b\), we view \(x^4\) as \((x^2)^2\), which allows us to analyze it similarly to a more common quadratic expression such as \(ax^2 + bx + c\).
This transformation helps in simplifying complex polynomials by using techniques effective for quadratics.

• The expression \(x^4 + 3x^2 + 4\) can be thought of as a quadratic in \((x^2)\), allowing us to use methods such as substitution or direct factoring strategies.
• By treating \(x^2\) as a single variable, we can apply known quadratic techniques to set the stage for further factoring.

Understanding this approach simplifies factoring and guides us to appropriate methods that may not be immediately obvious without this framework.

Trinomial

A trinomial is a polynomial expression that contains three terms. These terms are generally separated by plus or minus signs. Trinomials like \(x^4 + x^2 - 2\) or \(x^2 + bx + c\) are common, and they require special attention when factoring because they often can be broken down into products of binomials.
When dealing with trinomials, we often look for two numbers that multiply to the constant term and add to the linear coefficient.

• For example, in \(x^4 + x^2 - 2\), think of \(x^4\) as \((x^2)^2\). Then, find two numbers that multiply to \(-2\) and add up to \(1\).
• These numbers help split the middle term and factor the trinomial as a product of two binomials.

Recognizing whether or not a trinomial can be further factored is a crucial skill in simplifying and solving polynomial equations.

Difference of Squares

The difference of squares is a special factorization technique used for binomials that can be expressed as \(a^2 - b^2\). This type of expression can be factored into \((a - b)(a + b)\). This method is particularly useful for expressions where you have the subtraction of two perfect squares.
In the original exercise, when \(x^4 + 3x^2 + 4\) was transformed into \((x^2 + 2)^2 - x^2\), it displayed a typical difference of squares pattern.

• To factor it, we define \(a = (x^2 + 2)\) and \(b = x\), which leads to the result: \( (x^2 + 2 - x)(x^2 + 2 + x) \).

Identifying when an expression can be factored as a difference of squares is crucial for simplifying problems quickly and effectively. It provides a powerful method for dealing with more complicated-looking polynomial expressions.

Perfect Square

A perfect square trinomial is a special type of trinomial that can be expressed as the square of a binomial. This typically appears as \((ax + b)^2 = a^2x^2 + 2abx + b^2\). Recognizing when a trinomial is a perfect square simplifies the factoring process greatly.
In the given problem, transforming the expression \(x^4 + 3x^2 + 4\) into \((x^2 + 2)^2 - x^2\) involves identifying \((x^2 + 2)^2\) as a perfect square.

• Perfect square trinomials can always be factored into a squared binomial, but when paired with a subtractive term (like \(-x^2\) in the example), further techniques, like the difference of squares, may be required to completely factor the expression.
• This method relies heavily on recognizing patterns in the numbers and simplifying complex expressions using elementary algebraic identities.

Spotting perfect squares allows students to quickly reduce polynomials into more manageable pieces, making problem-solving easier.