Problem 47 - Algebra A Combined Function

Question

Factor each trinomial by grouping. Exercises 9 through 12 are broken into parts to help you get started. $$ 35+12 x+x^{2} $$

Step-by-Step Solution

Answer

The factored form is \((x+5)(x+7)\).

Step 1: Rewrite the trinomial in standard form

First, arrange the terms of the trinomial in descending order of the variable's power. The expression is given as \( 35 + 12x + x^2 \). Rearrange it to \( x^2 + 12x + 35 \).

Step 2: Identify the coefficients

In the standard form \( ax^2 + bx + c \), identify \( a = 1 \), \( b = 12 \), and \( c = 35 \).

Step 3: Find two numbers that multiply to ac and add to b

We need two numbers that multiply to \( a \times c = 1 \times 35 = 35 \) and add to \( b = 12 \). These two numbers are 5 and 7 because \( 5 \times 7 = 35 \) and \( 5 + 7 = 12 \).

Step 4: Rewrite the middle term using the two numbers

Express \( 12x \) as \( 5x + 7x \). The expression becomes \( x^2 + 5x + 7x + 35 \).

Step 5: Group the terms

Group the terms into two binomials: \( (x^2 + 5x) + (7x + 35) \).

Step 6: Factor out the greatest common factor from each group

In the first group \( x^2 + 5x \), factor out \( x \): \( x(x + 5) \). In the second group \( 7x + 35 \), factor out \( 7 \): \( 7(x + 5) \). The expression becomes \( x(x + 5) + 7(x + 5) \).

Step 7: Factor out the common binomial

Both groups contain the common binomial factor \( x + 5 \). Factor \( x + 5 \) from both terms to get \( (x + 5)(x + 7) \).

Key Concepts

Algebraic Expressions

An algebraic expression combines numbers, variables (often represented by letters), and operations (such as addition, subtraction, multiplication, and division). It serves as the foundation for more complex mathematical concepts. When dealing with algebraic expressions like the trinomial in our exercise, it's important to understand that each component has a specific role.

In the expression \(35 + 12x + x^{2}\), you have a constant term, a linear term, and a quadratic term.

• The constant term "35" stands alone and has no variable attached.
• The linear term "12x" includes a variable raised to the first power.
• The quadratic term "x^2" features the variable squared, indicating it has the highest degree among the terms.

Recognizing these elements is essential when rewriting the expression and performing operations, such as factoring. Understanding how these parts interact helps us manipulate and simplify expressions effectively.

Polynomial Factoring

Polynomial factoring is the process of breaking down a polynomial into simpler components, called factors, that when multiplied together give the original polynomial. It's like figuring out what numbers multiply to produce a specific number.

For the trinomial \(x^2 + 12x + 35\), factoring allows us to express it as a product of two binomials. This process is immensely useful for solving equations and finding solutions that a polynomial equation holds.

The factors of a polynomial can reveal important information, such as the roots or zeroes of the polynomial, by setting the factors equal to zero. The exercise shows that after factoring, we obtain \((x + 5)(x + 7)\). This step not only factors the polynomial but helps solve the equation \(x^2 + 12x + 35 = 0\) by revealing roots \(x = -5\) and \(x = -7\). Remember that factoring transforms complex expressions into manageable pieces.

Binomial Grouping

Binomial grouping is a strategic method used in polynomial factoring. It's particularly useful when working with trinomials that might be challenging to factor directly. This method involves grouping terms to create common factors.

In our trinomial \(x^2 + 12x + 35\), after rewriting \(12x\) as \(5x + 7x\), we group the expression into \((x^2 + 5x) + (7x + 35)\). This grouping allows us to factor each segment separately, unearthing common factors in each part.

• For \(x^2 + 5x\), the greatest common factor is \(x\).
• For \(7x + 35\), the common factor is \(7\).

Through binomial grouping, we simplify the expression into a form where factoring becomes straightforward, resulting in \((x + 5)(x + 7)\). This clarity in factoring adds simplicity and reveals potential solutions for polynomial equations.