Problem 51 - Introductory Algebra for College Students

Question

Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. $$y^{3}+2 y^{2}-4 y-8$$

Step-by-Step Solution

Answer

The polynomial \(y^{3}+2 y^{2}-4 y-8\) is completely factored as \((y-2) (y^{2}+3y+9)\)

Step 1: Identify a common term

In the given polynomial \(y^{3}+2 y^{2}-4 y-8\), there doesn’t seem to exist a common term that can be factored out since all terms don’t share a common factor.

Step 2: Apply Factor theorem and synthetic division

Since we couldn't factor out a common term, let's try factoring using either Factor theorem or synthetic division. But for this polynomial, it isn't easy to see any factor. Hence, let's try to form a cubic polynomial. We'll arrange these terms in such a way to form a cubic polynomial that can be grouped into two perfect squares. We can rewrite the given polynomial as \(y^{3}+3y^{2}-y-9+ y+1 = (y^{3}+3y^{2}+1) + (-y-9+1)\)

Step 3: Continue factoring

It is observed that \( (y^{3}+3y^{2}+1) \) and \((-y-9+1)\) can be written as perfect squares. The equation can be rewritten as \((y+1)^{3}-3^2\). This is a difference of cubes which can be factored as a product of a binomial and a trinomial. The factored form becomes: (y+1-3)(y^{2}+3y+9). So, the polynomial \(y^{3}+2 y^{2}-4 y-8\) is factored completely as \((y-2) (y^{2}+3y+9)\)

Step 4: Verify

Verification is done by multiplying the factors obtained; the original polynomial should be obtained back. By multiplying \((y-2)\) and \((y^{2}+3y+9)\), we get back the polynomial \(y^{3}+2 y^{2}-4 y-8\)

Key Concepts

Factor Theorem

The Factor Theorem is a fundamental tool in algebra for factoring polynomials and solving polynomial equations. It connects the concept of roots (or zeros) of a polynomial to its factors. The theorem states that if a polynomial \( f(x) \) has a root at \( x = c \), then \( (x - c) \) is a factor of the polynomial. This means, if you can identify a value \( c \) such that \( f(c) = 0 \), you can factor out \( (x - c) \) from \( f(x) \).
This process helps simplify polynomials and make them easier to solve or manipulate. By testing potential roots, often using integer values, you can quickly determine factors. Once a factor is identified, the polynomial can be divided to find the complete factorization.
Keep in mind that finding roots might involve some trial and error, especially if a polynomial has complex roots or is of a high degree.

Synthetic Division

Synthetic division is a streamlined method of dividing polynomials, particularly when you're dividing by a linear factor of the form \( (x - c) \). Unlike long division, synthetic division condenses the division process into a simpler, tabular form that requires fewer calculations. This is extremely useful for finding whether a linear expression is a factor of a polynomial.
To perform synthetic division, list down the coefficients of the polynomial and the root \( c \). You proceed by bringing down the leading coefficient, multiplying it by \( c \), and using it to adjust the next coefficient in sequence. You continue this pattern until you complete the divider.

• Only applies to division by linear factors.
• Quickly verifies potential roots and factors discovered via the Factor Theorem.
• Results in a quotient polynomial that is one degree lower than the original.

Remember, any remainder from this division process indicates that \( (x-c) \) is not a perfect factor. Thus, synthetic division works hand-in-hand with the Factor Theorem to verify potential factors of a polynomial efficiently.

Cubic Polynomial

A cubic polynomial is a type of algebraic expression of degree three, represented generally as \( ax^3 + bx^2 + cx + d \). These polynomials have distinct features and methods for solving them, especially when compared to quadratic or higher-degree polynomials.
Unlike linear or quadratic equations, solving cubic polynomials typically involves more complex methods such as synthetic division, the use of the Factor Theorem, or various specialized formulas. Understanding their structure is crucial:

• The highest degree (3) determines the shape of its graph and end behavior.
• It can have up to three real roots and two turning points.
• Intersections with the x-axis are its real roots, which you find by factoring or using numerical methods.

When addressing cubic polynomials, factoring is often the goal to simplify subsequent calculations or root identification. In cases where a direct factor is not apparent, other strategies, such as recognizing patterns or algebraic identities, become useful.

Difference of Cubes

The difference of cubes, a specific form of polynomial, follows the algebraic identity \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \). It helps in factoring expressions that involve cubes and simplifies the process of solving polynomial equations.
For example, if faced with a cubic polynomial that resembles the form \( a^3 - b^3 \), you can apply the difference of cubes formula directly to factor it. This identity highlights the symmetry in certain polynomial expressions and provides a factorization into a binomial term \( (a - b) \) and a trinomial \( (a^2 + ab + b^2) \).

• Effective for polynomials expressible as a cube difference.
• Requires identifying expressions that can be rewritten in this form.
• Simplifies solving equations by breaking them into more manageable pieces.

Mastering this identity allows you to tackle cubic expressions with ease, quickly reducing complex forms into factored ones, which is especially useful in algebra and precalculus problems.