Problem 23 - Algebra A Combined Function

Question

Factor each trinomial completely. $$ 1+6 x^{2}+x^{4} $$

Step-by-Step Solution

Answer

The expression factors as \((x^2 + 3 - 2\sqrt{2})(x^2 + 3 + 2\sqrt{2})\).

Step 1: Rearrange the terms

The given expression is \( 1 + 6x^2 + x^4 \). To factor it easily, first rearrange the terms in descending order of the exponents: \( x^4 + 6x^2 + 1 \).

Step 2: Recognize it as a quadratic form

Notice that the expression is in the form \( ax^4 + bx^2 + c \), which resembles a quadratic form \( At^2 + Bt + C \), where \( t = x^2 \). Set \( y = x^2 \), then rewrite the expression as \( y^2 + 6y + 1 \).

Step 3: Factor the quadratic expression

Now, factor the quadratic expression \( y^2 + 6y + 1 \) using the quadratic formula: \[ y = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \]where \( A = 1 \), \( B = 6 \), and \( C = 1 \).Calculate the discriminant:\( B^2 - 4AC = 36 - 4 = 32 \), so the roots are \[ y = \frac{-6 \pm \sqrt{32}}{2} \]\[ y = \frac{-6 \pm 4\sqrt{2}}{2} \]\[ y = -3 \pm 2\sqrt{2} \]. So, \( y^2 + 6y + 1 \) factors as \((y + 3 - 2\sqrt{2})(y + 3 + 2\sqrt{2})\).

Step 4: Substitute back to original variable

Recall that \( y = x^2 \). Substitute back to get:\((x^2 + 3 - 2\sqrt{2})(x^2 + 3 + 2\sqrt{2})\). This is the completely factored form of the original expression.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving equations that take a quadratic form. It represents a standard method to find the roots of a quadratic equation, expressed as: - \[ y = \frac{-B \pm \sqrt{B^2-4AC}}{2A} \] In this formula,

• \( A \), \( B \), and \( C \) are coefficients of the quadratic equation \( Ay^2 + By + C = 0 \)
• \( B^2 - 4AC \) is referred to as the discriminant, which determines the nature of the roots.

By using the quadratic formula, we can find the exact solutions for\( y \), which are critical when factoring more complex equations. This formula becomes especially useful when the trinomial cannot be factored easily by observation. It facilitates a straightforward method to derive the roots, even when they are not whole numbers.

Quadratic Form

The quadratic form of an equation is very simple yet extremely useful for identifying potential factorization. In the exercise given, the function initially appears quite complex: - \[ x^4 + 6x^2 + 1 \]. By recognizing that the expression fits into a quadratic form, meaning it resembles: \( ax^{2n} + bx^n + c \), it becomes manageable. Consider the substitution \( y = x^2 \), then the expression simplifies to \( y^2 + 6y + 1 \), making it clear how closely it follows the format of a typical quadratic: \( Ay^2 + By + C \). This substitution transforms higher powers into a more familiar quadratic, making it easier to apply the quadratic formula or other methods to solve or factor them.

Discriminant

The discriminant is a component of the quadratic formula and plays several key roles in understanding quadratic equations. Denoted as \( B^2 - 4AC \), it holds the power to reveal the number and type of roots without actually solving the equation. For the given trinomial, calculating the discriminant helps confirm whether factoring directly is feasible:

• \( B = 6 \), \( A = 1 \), \( C = 1 \)
• \( B^2 - 4AC = 36 - 4 \times 1 \times 1 = 32 \)

A positive discriminant indicates two distinct real roots exist, meaning the quadratic can be factored over the real numbers. If the discriminant were zero, only one real root would exist, and if negative, it would suggest the roots are complex, which requires a different approach. Thus, the discriminant not only aids in choosing the right solution method but also in predicting the solution's nature at a glance.

Factorization Steps

Factoring trinomials involves several methodical steps, often beginning with rearranging terms and making substitutions. For our original expression, \( 1 + 6x^2 + x^4 \):

• **Rearrange the Terms**: Start by writing it as \( x^4 + 6x^2 + 1 \) in descending order of the powers.
• **Recognize a Quadratic Form**: Identify that replacing \( x^2 \) with \( y \) transforms it into a common quadratic \( y^2 + 6y + 1 \).
• **Apply the Quadratic Formula**: Use \( Ay^2 + By + C \) to solve for \( y \) with the quadratic formula, yielding roots.
• **Back-substitution**: Reintroduce the original variable \( x \) back in: \((x^2 + 3 - 2\sqrt{2} )(x^2 + 3 + 2\sqrt{2} )\) are the factors.

Each step incrementally simplifies the challenge of the trinomial into identifiable segments. With patience and precision, what initially appears to be an elaborate expression can be distilled into its contributing factors.