(7x+19){}(x+1){}(x+5)

Understanding the Expression (7x+19)(x+1)(x+5)

The algebraic expression (7x+19)(x+1)(x+5) is a product of three binomials. Such expressions are fundamental in algebra, often encountered when simplifying, expanding, or factoring polynomial expressions. Grasping how to manipulate this type of expression provides essential skills for solving equations, analyzing functions, and understanding polynomial behavior.

In this article, we will explore the structure of this expression, methods to expand it, techniques for factoring, and real-world applications. Whether you're a student looking to improve your algebra skills or a teacher preparing lesson plans, understanding this expression is a valuable step in mastering algebraic concepts.

Breaking Down the Expression: Components and Structure

The expression consists of three factors:

1. (7x+19) — a linear binomial with a coefficient of 7 for x and a constant term 19.


2. (x+1) — a linear binomial representing a simple shift along the x-axis.


3. (x+5) — another linear binomial, shifted further along the x-axis.

Multiplying these factors together creates a cubic polynomial once fully expanded. This structure makes it a prime example for understanding the process of polynomial expansion.

Step-by-Step Expansion of (7x+19)(x+1)(x+5)

To expand the expression, it's often easiest to proceed step by step:

1. First, multiply two binomials: (7x+19) and (x+1)


2. Then, multiply the resulting binomial by the third binomial: (x+5)

Let's walk through this process.

Step 1: Expand (7x+19)(x+1)

Using distributive property (FOIL method):

• First: 7x x = 7x^2


• Outer: 7x 1 = 7x


• Inner: 19 x = 19x


• Last: 19 1 = 19

Combine like terms:

(7x + 19)(x + 1) = 7x^2 + (7x + 19x) + 19 = 7x^2 + 26x + 19

Step 2: Multiply the result by (x+5)

Now, multiply (7x^2 + 26x + 19) by (x+5):

Distribute each term:

• 7x^2 x = 7x^3


• 7x^2 5 = 35x^2


• 26x x = 26x^2


• 26x 5 = 130x


• 19 x = 19x


• 19 5 = 95

Combine like terms:

7x^3 + (35x^2 + 26x^2) + (130x + 19x) + 95 = 7x^3 + 61x^2 + 149x + 95

Final expanded form:

(7x+19)(x+1)(x+5) = 7x^3 + 61x^2 + 149x + 95

This cubic polynomial represents the original product in standard form.

Factoring the Cubic Polynomial

Factoring polynomials like 7x^3 + 61x^2 + 149x + 95 can be challenging but is vital for solving equations or simplifying expressions.

Methods for Factoring the Polynomial

• Rational Root Theorem: Helps identify possible rational roots.
• Synthetic Division: Used to test potential roots.
• Factor Theorem: If a root r satisfies P(r)=0, then (x - r) is a factor.

Applying the Rational Root Theorem

Possible rational roots are factors of the constant term (95) divided by factors of the leading coefficient (7):

• Factors of 95: ±1, ±5, ±19, ±95


• Factors of 7: ±1, ±7

Possible roots:

±1, ±5, ±19, ±95, ±1/7, ±5/7, ±19/7, ±95/7

Testing these values in the polynomial can identify roots, leading to factors.

Factoring the Polynomial Step-by-Step

Suppose we test x = -1:

P(-1) = 7(-1)^3 + 61(-1)^2 + 149(-1) + 95 = -7 + 61 - 149 + 95 = 0

Since P(-1) = 0, x = -1 is a root, and (x + 1) is a factor.

Now, perform synthetic division to factor out (x + 1):

Set up synthetic division:

| -1 | 7 | 61 | 149 | 95 |
| | -7 | -54 | -95 | 0 |

Bring down the 7:

• Multiply -1 7 = -7; add to 61: 61 - 7 = 54


• Multiply -1 54 = -54; add to 149: 149 - 54 = 95


• Multiply -1 95 = -95; add to 95: 95 - 95 = 0

The quotient coefficients: 7, 54, 95

Thus, the polynomial factors as:

(7x + 1)(x^2 + 54/7 x + 95/7)

Further factoring quadratic:

Multiply through by 7 to clear fractions:

7x^2 + 54x + 95

Calculate discriminant:

D = 54^2 - 4  7  95 = 2916 - 2660 = 256

Square root of D:

√256 = 16

Find roots:

x = [-54 ± 16] / (2  7) = [-54 ± 16] / 14

• For the positive root:

x = (-54 + 16)/14 = (-38)/14 = -19/7

• For the negative root:

x = (-54 - 16)/14 = (-70)/14 = -5

Therefore, quadratic factors as:

(7x + 1)(x + 19/7)(x + 5)

Multiplying back:

(7x + 1)(x + 19/7)(x + 5)

which simplifies to:

(7x + 1)(x + 19/7)(x + 5)

Recognizing that (x + 19/7) is equivalent to (7x + 19)/7, the complete factorization of the original polynomial is:

(7x + 1)(7x + 19)(x + 5) / 7

Multiplying numerator and denominator by 7 gives:

(7x + 1)(7x + 19)(x + 5) / 7

Finally, recalling that the original expression was the product of (7x+19), (x+1), and (x+5), this confirms the polynomial factors perfectly into the original binomials.

Applications of the Polynomial (7x+19)(x+1)(x+5)

Understanding and manipulating such polynomials has many practical applications:

• Solving Real-World Problems: Many physical models involve polynomial equations, such as motion, growth rates, and projectile trajectories.


• Engineering and Design: Polynomial expressions model stress, strain, and other physical phenomena.


• Economics and Finance: Polynomial functions can model cost, revenue, and profit functions over different scenarios.


• Data Fitting and Approximation: Polynomials are often used for curve fitting in statistical analysis.

Conclusion: Mastering Polynomial Expressions

Mastering the expansion and factoring of complex polynomial expressions like (7x+19)(x+1)(x+5) is fundamental in algebra. It builds foundational skills for more advanced topics such as calculus and differential equations. By practicing step-by-step expansion, applying the Rational Root Theorem, and understanding the structure of cubic polynomials, students and professionals can confidently manipulate these expressions to solve a wide range of mathematical problems.

Whether you're simplifying an expression, solving polynomial equations, or applying these concepts in real-world contexts, understanding the structure and techniques related to such expressions enhances your mathematical toolkit. Keep practicing with different binomials and polynomials to strengthen your algebraic proficiency and unlock the full potential of polynomial mathematics.