Factoring Polynomials: A Step-by-Step Guide

The question asks us to find the factors of the expression \(x(x + 2) (x + 3) (x + 5) – 72\). To factor this polynomial expression, we can strategically group the terms to simplify the process.

Grouping Terms for Simplification

We can observe that the sum of the constants in the pairs \((x)\) and \((x+5)\) is \(0+5=5\), and the sum of the constants in the pairs \((x+2)\) and \((x+3)\) is \(2+3=5\). This suggests grouping these pairs together:

\( [x(x+5)] [(x+2)(x+3)] - 72 \)

Expanding the Grouped Terms

Now, let's expand each grouped pair:

• Expanding \(x(x+5)\):

  \(x \times x + x \times 5 = x^2 + 5x\)

• Expanding \((x+2)(x+3)\):

  \(x \times x + x \times 3 + 2 \times x + 2 \times 3 = x^2 + 3x + 2x + 6 = x^2 + 5x + 6\)

Substituting these back into the expression, we get:

\((x^2 + 5x) (x^2 + 5x + 6) - 72\)

Substitution to Simplify Further

Notice that the term \(x^2 + 5x\) appears in both parts of the product. Let's substitute \(y = x^2 + 5x\) to make the expression simpler:

\(y (y + 6) - 72\)

Expanding and Factoring the Quadratic

Now, expand this simplified expression:

\(y^2 + 6y - 72\)

This is a quadratic expression in terms of \(y\). We need to factor this quadratic. We look for two numbers that multiply to -72 and add up to 6. These numbers are 12 and -6.

So, we can factor the quadratic as:

\((y + 12)(y - 6)\)

Substituting Back to Find Factors in Terms of x

Now, substitute \(y = x^2 + 5x\) back into the factored expression:

\((x^2 + 5x + 12)(x^2 + 5x - 6)\)

Factoring the Quadratic Factors

We now have two quadratic factors. Let's try to factor each one:

1. Factoring \(x^2 + 5x + 12\):

   We check the discriminant (\(\Delta = b^2 - 4ac\)) for this quadratic (\(ax^2+bx+c\)). Here \(a=1, b=5, c=12\).

   \(\Delta = 5^2 - 4(1)(12) = 25 - 48 = -23\)

   Since the discriminant is negative (\(\Delta < 0\)), this quadratic has no real roots and cannot be factored into linear factors with real coefficients. It is an irreducible quadratic factor over the real numbers.

2. Factoring \(x^2 + 5x - 6\):

   We look for two numbers that multiply to -6 and add up to 5. These numbers are 6 and -1.

   So, \(x^2 + 5x - 6 = (x + 6)(x - 1)\)

Combining the factors, the complete factorization of the original expression is:

\((x^2 + 5x + 12) (x + 6) (x - 1)\)

Rearranging the linear factors for clarity:

\((x - 1) (x + 6) (x^2 + 5x + 12)\)

Comparing with Given Options

Let's compare our derived factors with the options provided:

• Option 1: \(x, (x + 3), (x + 4)\) and \((x – 6)\) - Does not match.
• Option 2: \((x – 1), (x + 6)\) and \((x ²– 2x – 12)\) - The quadratic factor does not match.
• Option 3: \((x – 1), (x + 6)\) and \((x ²+ 5x + 12)\) - This matches our derived factors.
• Option 4: \((x + 1), (x - 6)\) and \((x ²– 5x – 12)\) - The linear factors and the quadratic factor do not match.

Therefore, the correct factors are \((x – 1), (x + 6)\) and \((x^2 + 5x + 12)\).

Revision Table: Key Concepts in Polynomial Factoring

Additional Information: Irreducible Quadratics and Real Factors

An irreducible quadratic over the real numbers is a quadratic expression \(ax^2+bx+c\) where the discriminant \(\Delta = b^2 - 4ac\) is less than zero. Such a quadratic cannot be factored into two linear factors with real coefficients.

In this problem, the factor \(x^2 + 5x + 12\) is irreducible over the real numbers because its discriminant is -23, which is negative. Therefore, the factorization is complete using real coefficients when we have the linear factors \((x-1)\), \((x+6)\), and the irreducible quadratic factor \((x^2+5x+12)\).