Problem 37 - Beginning and Intermediate Algebra

Question

Factor completely by first taking out \(-1\) and then by factoring the trinomial, if possible. Check your answer. $$-h^{2}+2 h+15$$

Step-by-Step Solution

Answer

The completely factored form of the given polynomial is \(-h^2 + 2h + 15 = -(h-5)(h+3)\).

Step 1: Factor out -1

Since the leading term is negative, let's first factor out -1: \(-h^2 + 2h + 15 = -(h^2 - 2h - 15)\)

Step 2: Factor the trinomial

Now we need to factor the trinomial inside the parentheses: \(h^2 - 2h - 15\). To factor a trinomial in the form of \(ax^2 + bx + c\), we need to find two numbers that multiply to give c and add to give b. In this case, we need two numbers that multiply to give -15 and add up to -2. These numbers are -5 and 3. So, \[(h^2 - 2h - 15) = (h-5)(h+3)\]

Step 3: Write the completely factored form

Since we factored out -1 in step 1, we need to include it in our factored form: \[-h^2 + 2h + 15 = -(h-5)(h+3)\]

Step 4: Check the answer

To check our answer, we can expand the factored form and see if it matches the original polynomial. \[-(h-5)(h+3)= -[(h-5)h + (h-5)(3)] = -[h^2 - 5h + 3h - 15] = -[h^2 -2h -15]\] As we see that the expanded form is the same as the original polynomial \(-h^2 + 2h + 15\), our factored form is correct.

Key Concepts

Trinomial Factorization

Trinomial factorization involves breaking down quadratic trinomials into a product of two binomial expressions. For a trinomial of the form \(ax^2 + bx + c\), the goal is to find two numbers that multiply to \(c\) (the constant term) and add up to \(b\) (the linear coefficient).

In the example provided, the expression given was \(-h^2 + 2h + 15\). The first step involves factoring out a \(-1\) because of the negative leading coefficient, turning it into \(-(h^2 - 2h - 15)\).

Simplify the trinomial \(h^2 - 2h - 15\) by finding two numbers that multiply to \(-15\) (constant term) and sum to \(-2\) (linear term). These numbers are \(-5\) and \(3\), allowing us to rewrite it as \((h-5)(h+3)\).

Factoring trinomials is a systematic way to simplify quadratic expressions, making them easier to work with in equations. Practicing this method strengthens algebraic manipulation skills.

Polynomial Expansion

Polynomial expansion refers to the process of expanding a product of binomials back into a polynomial expression. It reverses the factorization process, verifying the factored form's correctness.

To check the factorization from the example \(-(h-5)(h+3)\), the polynomial expansion is performed:

• First, expand \((h-5)(h+3)\) by distributing: \(h(h+3) - 5(h+3) = h^2 + 3h - 5h - 15\).
• Combine like terms to simplify: \(h^2 - 2h - 15\).
• Since we factored out a \(-1\) initially, multiply the expression by \(-1\) again to get \(-h^2 + 2h + 15\).

This matches the original polynomial, confirming the factorization's accuracy.

Understanding polynomial expansion is essential for verifying calculations and understanding the relationship between a polynomial and its factors.

Quadratic Trinomials

Quadratic trinomials are polynomials of degree two, generally written in the form \(ax^2 + bx + c\). These are fundamental components in algebra, representing various problems and solutions.

In the expression \(-h^2 + 2h + 15\), we see a classic quadratic trinomial structure once \(-1\) is factored out. The expression becomes \(h^2 - 2h - 15\).

The core challenge with quadratic trinomials is identifying effective factor pairs, which demands practice with multiplication, factorization, and addition skills to quickly ascertain which numbers satisfy both conditions of multiplying to \(c\) and summing to \(b\).

Quadratic trinomials appear in numerous mathematical contexts, from solving equations to modeling real-world scenarios. Mastery of this topic enhances a student's capacity to handle complex algebraic functions with confidence.