Problem 10 - Elementary and Intermediate Algebra

Question

The factorization of \(y^{3}+27\) is \((y+3)\left(y^{2}-3 y+9\right) .\) Is this factored completely, or does \(y^{2}-3 y+9\) factor further?

Step-by-Step Solution

Answer

The factorization is complete; \(y^2 - 3y + 9\) cannot be factored further with real numbers.

Step 1: Identify the Polynomial Type

The given polynomial is \(y^3 + 27\) which is a sum of cubes. It can be written as \(y^3 + 3^3\).

Step 2: Verify the Sum of Cubes Formula

The factored form given is \((y+3)(y^2-3y+9)\). This should match the sum of cubes formula: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\), where \(a = y\) and \(b = 3\).

Step 3: Apply the Sum of Cubes Check

Using \(a = y\) and \(b = 3\), the sum of cubes formula gives us:\((y + 3)(y^2 - 3y + 9)\). This matches the factored form provided.

Step 4: Check for Further Factoring Possibilities

The second factor, \(y^2 - 3y + 9\), is a quadratic expression. To factor it further, check the discriminant \(b^2 - 4ac\): Here \(a = 1\), \(b = -3\), \(c = 9\).Calculate: \((-3)^2 - 4 \times 1 \times 9 = 9 - 36 = -27\).

Step 5: Interpret the Discriminant Results

The discriminant \(b^2 - 4ac = -27\) is negative. A negative discriminant indicates that the quadratic has no real roots and cannot be factored further using real numbers.

Key Concepts

Sum of Cubes

When you're trying to factor a sum of cubes, you're dealing with a specific type of polynomial of the form \(a^3 + b^3\). This can look intimidating, but it follows a special factorization formula that makes things easier: \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).
The polynomial in our exercise is \(y^3 + 27\), which can be expressed as \(y^3 + 3^3\). We identify \(a = y\) and \(b = 3\). This allows us to use the sum of cubes formula directly.
The factorization results in \((y + 3)(y^2 - 3y + 9)\), confirming that it's indeed a sum of cubes. This handy formula breaks our polynomial into simpler parts, making it more manageable.

Quadratic Expression

A quadratic expression is a type of polynomial where the highest exponent of the variable is 2, typically written in the form \(ax^2 + bx + c\).
In our exercise, after factoring the sum of cubes, we're left with the quadratic expression \(y^2 - 3y + 9\).
Quadratics can often be factored further if they have real roots, which means the quadratic has zero, one, or two solutions that can be real numbers. To explore this, we need another tool: the discriminant.

Discriminant

The discriminant is a part of the quadratic formula used to determine the nature of the roots of a quadratic equation. Calculated as \(b^2 - 4ac\) where \(a\), \(b\), and \(c\) are coefficients from \(ax^2 + bx + c\).
In our quadratic \(y^2 - 3y + 9\), we see that \(a = 1\), \(b = -3\), and \(c = 9\). Substituting into the discriminant formula gives us

• \((-3)^2 - 4 \cdot 1 \cdot 9 = 9 - 36 = -27\)

A negative discriminant indicates there are no real roots, meaning this quadratic cannot be factored further using real numbers. Thus, the initial factorization using the sum of cubes is the complete factorization in real numbers.