Unit 5 Polynomial Functions Homework 7 Answer Key

Unit 5 polynomial functions homework 7answer key offers students a clear roadmap to verify their work and solidify understanding of polynomial operations, factoring techniques, and end‑behaviour analysis. This guide walks through each problem, highlights common pitfalls, and reinforces the underlying mathematical principles that make polynomial functions a cornerstone of algebra.

Overview of Unit 5 Polynomial Functions ### Core Concepts

• Polynomial expression – a sum of terms each consisting of a coefficient multiplied by a variable raised to a non‑negative integer exponent.
• Degree – the highest exponent of the variable in the polynomial; it determines the end‑behaviour and the maximum number of real roots. - Leading coefficient – the coefficient attached to the term with the highest degree; it influences the direction of the graph as x approaches positive or negative infinity.
• Factoring – rewriting a polynomial as a product of simpler polynomials, often using the greatest common factor (GCF), difference of squares, or sum/difference of cubes.

These ideas are repeatedly applied throughout Homework 7, where students are asked to simplify expressions, solve equations, and interpret graphs.

Homework 7 Overview

Problem Types

1. Simplifying polynomial expressions – combining like terms and applying exponent rules.
2. Factoring polynomials – extracting the GCF, using special product formulas, and applying the Rational Root Theorem.
3. Solving polynomial equations – finding all real and complex roots by factoring or synthetic division.
4. Graphical interpretation – determining end‑behaviour, identifying turning points, and sketching rough graphs.

Each section of Homework 7 targets a specific skill, allowing learners to progress from basic manipulation to more sophisticated analysis.

Answer Key Details

Below is a step‑by‑step breakdown of the correct solutions for each problem in Homework 7. The explanations emphasize why each step works, not just how to perform it.

Problem 1 – Simplify the Expression

Problem: Simplify (3x^4 - 2x^3 + 5x^4 + 7x^3 - 4).

Solution:

1. Combine like terms – group the (x^4) terms and the (x^3) terms.
   • (3x^4 + 5x^4 = 8x^4)
   • (-2x^3 + 7x^3 = 5x^3)
2. Write the simplified polynomial – (8x^4 + 5x^3 - 4).

Key takeaway: Always align terms by exponent before performing arithmetic; this prevents sign errors.

Problem 2 – Factor the Polynomial

Problem: Factor (6x^3 - 12x^2 + 18x).

Solution:

1. Identify the GCF – each term shares a factor of (6x).
2. Factor out the GCF:
   [ 6x(x^2 - 2x + 3) ]
3. Check the quadratic – (x^2 - 2x + 3) has a discriminant (b^2 - 4ac = (-2)^2 - 4(1)(3) = 4 - 12 = -8), which is negative, so it cannot be factored further over the real numbers.

Result: (6x(x^2 - 2x + 3)).

Emphasis: When the remaining quadratic has a negative discriminant, the polynomial is irreducible over the reals, and the factorization stops there.

Problem 3 – Solve the Equation

Problem: Solve (x^3 - 6x^2 + 11x - 6 = 0).

Solution:

1. Apply the Rational Root Theorem – possible rational roots are (\pm1, \pm2, \pm3, \pm6).
2. Test (x = 1):
   [ 1 - 6 + 11 - 6 = 0 \quad \Rightarrow \text{Root} ]
3. Perform synthetic division by ((x - 1)) to obtain the depressed quadratic:
   [ x^2 - 5x + 6 ]
4. Factor the quadratic:
   [ (x - 2)(x - 3) ]
5. Set each factor to zero:
   [ x - 1 = 0 ;\Rightarrow; x = 1 \ x - 2 = 0 ;\Rightarrow; x = 2 \ x - 3 = 0 ;\Rightarrow; x = 3 ]

Answer: The equation has three real roots: (x = 1, 2, 3).

Note: Using synthetic division after finding a root streamlines the process and reduces algebraic complexity.

Problem 4 – Graphical Interpretation Problem: Given (f(x) = -2x^3 + 9x^2 - 12x + 5), describe the end‑behaviour and locate any turning points.

Solution:

1. Determine the degree and leading coefficient: degree = 3 (odd), leading coefficient = ‑2 (negative).
   • As (x \to \infty), (f(x) \to -\infty). - As (x \to -\infty), (f(x) \to \infty).
2. Find critical points by differentiating:
   [ f

(x) = -6x^2 + 18x - 12 Setting the derivative equal to zero: [ -6x^2 + 18x - 12 = 0 ] [ -6(x^2 - 3x + 2) = 0 ] [ x^2 - 3x + 2 = 0 ] [ (x - 1)(x - 2) = 0 ] So, (x = 1) and (x = 2) are critical points.

3. Use the second derivative test:
   [ f''(x) = -12x + 18 ] [ f''(1) = -12(1) + 18 = 6 > 0 \quad \Rightarrow \text{Local Minimum at } x = 1 ] [ f''(2) = -12(2) + 18 = -6 < 0 \quad \Rightarrow \text{Local Maximum at } x = 2 ]
4. Determine the turning points:
   • Local Minimum: ((1, f(1)) = (1, -2 + 9 - 12 + 5) = (1, 0))
   • Local Maximum: ((2, f(2)) = (-16 + 36 - 24 + 5) = (2, -3))

Conclusion: The function (f(x) = -2x^3 + 9x^2 - 12x + 5) has a negative leading coefficient, so it approaches negative infinity as (x) approaches positive infinity and approaches positive infinity as (x) approaches negative infinity. The function has a local minimum at ((1, 0)) and a local maximum at ((2, -3)). These turning points provide valuable insights into the function's shape and behavior. Understanding these graphical interpretations allows for a more complete understanding of the function's properties and applications.

Problem 5 – Solve the System of Equations

Problem: Solve the system of equations: (2x + y = 7) (x - y = 2)

Solution:

1. Solve for one variable in terms of the other from either equation. Let's solve the second equation for (x): (x = y + 2)

2. Substitute the expression for (x) into the first equation: (2(y + 2) + y = 7)

3. Simplify and solve for y: (2y + 4 + y = 7) (3y + 4 = 7) (3y = 3) (y = 1)

4. Substitute the value of y back into the equation for x: (x = y + 2 = 1 + 2 = 3)

Answer: The solution to the system of equations is (x = 3) and (y = 1).

Final Thoughts: Solving systems of equations often involves manipulating equations to isolate variables, and then substituting those expressions back into the original equations to find the solution. Careful algebraic manipulation and checking the solution are crucial for ensuring accuracy.

Problem 6 – Analyze a Function's Domain

Problem: Determine the domain of the function (f(x) = \frac{1}{(x - 3)^2}).

Solution:

1. Identify restrictions: The function is undefined when the denominator is equal to zero. So, we need to find the values of (x) that make ((x - 3)^2 = 0).

2. Solve for x: ((x - 3)^2 = 0) (x - 3 = 0) (x = 3)

3. Exclude the restricted value: The function is undefined at (x = 3), so we exclude it from the domain.

4. Write the domain: The domain of the function is all real numbers except (x = 3).

Domain: (x \in (-\infty, 3) \cup (3, \infty))

Overall Summary: The solutions to Homework 7 demonstrate a solid grasp of fundamental algebraic concepts. From simplifying expressions and factoring polynomials to solving equations and analyzing functions, the exercises reinforce essential skills in mathematics. The emphasis on understanding why each step works is particularly valuable for building a deeper comprehension of the subject matter. Consistent practice and attention to detail are key to mastering these techniques and achieving success in mathematics.