# Eureka Math Precalculus Module 3 Lesson 11 Answer Key

## Engage NY Eureka Math Precalculus Module 3 Lesson 11 Answer Key

### Eureka Math Precalculus Module 3 Lesson 11 Example Answer Key

Example 1.
Simplify the expression  to lowest terms, and identify the value(s) of x that must be excluded to avoid division by zero.
Since x2-5x + 6 = (x-2)(x-3), our original expression can be written as $$\frac{x^{2}-5 x + 6}{x-3}$$ = $$\frac{(x-2)(x-3)}{x-3}$$. To simplify this expression to lowest terms, we need to divide the numerator and denominator by any common factors. The only common factor in this example is x-3. However, we can only divide by x-3 if x-3 â‰  0, which means that we have to exclude 3 as a possible value of x.
Thus, if x â‰  3, we have
$$\frac{x^{2}-5 x + 6}{x-3}$$ = $$\frac{(x-2)(x-3)}{x-3}$$ =
= $$\frac{(x-2)(x-3)}{x-3}$$â‹…$$\frac{\frac{1}{x-3}}{\frac{1}{x-3}}$$
= x-2.
So, as long as x â‰  3, the expressions $$\frac{x^{2}-5 x + 6}{x-3}$$ and x-2 are equivalent.

Example 2.
Let f(x) = $$\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}$$. Simplify the rational expression $$\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}$$ to lowest terms, and use the simplified form to express the rule of f. Be sure to indicate any restrictions on the domain.
$$\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}$$ = $$\frac{2 x\left(x^{3} + 3 x^{2} + 3 x + 1\right)}{3 x(x + 1)}$$ = $$\frac{2 x(x + 1)^{3}}{3 x(x + 1)}$$ = $$\frac{2(x + 1)^{2}}{3}$$ if x â‰  -1 and x â‰  0.
Then, f(x) = $$\frac{2(x + 1)^{2}}{3}$$ for x â‰  0 and x â‰  -1.

### Eureka Math Precalculus Module 3 Lesson 11 Exercise Answer Key

Opening Exercise
Factor each expression completely:
a. 9x4-16x2
x2 (3x + 4)(3x-4)

b. 2x3 + 5x2-8x-20
(x2-4)(2x + 5) = (x + 2)(x-2)(2x + 5)

c. x3 + 3x2 + 3x + 1
(x + 1)3

d. 8x3-1
(2x-1)(4x2 + 2x + 1)

Exercise 1: Simplifying Rational Expressions to Lowest Terms
Exercise 1.
Simplify each rational expression to lowest terms, specifying the values of x that must be excluded to avoid division by zero.
a. $$\frac{x^{2}-6 x + 5}{x^{2}-3 x-10}$$
The denominator factors into x2-3x-10 = (x-5)(x + 2), so to avoid division by zero, we must have
x â‰  5 and x â‰  -2. Thus, $$\frac{x^{2}-6 x + 5}{x^{2}-3 x-10}$$ = $$\frac{(x-5)(x-1)}{(x-5)(x + 2)}$$ = $$\frac{x-1}{x + 2}$$, where x â‰  -2 and x â‰  5.

b. $$\frac{x^{3} + 3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x}$$
The denominator factors into x3 + 2x2 + x = x(x + 1)2, so to avoid division by zero, we must have x â‰  0 and x â‰  -1. Thus, $$\frac{x^{3} + 3 x^{2} + 3 x + 1}{x^{3} + 2 x^{2} + x}$$ = $$\frac{(x + 1)^{3}}{x(x + 1)^{2}}$$ = $$\frac{x + 1}{x}$$, where x â‰  0 and x â‰  1.

c. $$\frac{x^{2}-16}{x^{2} + 2 x-8}$$
The denominator factors into x2 + 2x-8 = (x-2)(x + 4), so to avoid division by zero, we must have x â‰  2 and x â‰  -4. Thus, $$\frac{x^{2}-16}{x^{2} + 2 x-8}$$ = $$\frac{(x-4)(x + 4)}{(x-2)(x + 4)}$$ = $$\frac{x-4}{x-2}$$, where x â‰  2 and x â‰  -4.

d. $$\frac{x^{2}-3 x-10}{x^{3} + 6 x^{2} + 12 x + 8}$$
The denominator factors into x3 + 6x2 + 12x + 8 = (x + 2)3, so to avoid division by zero, we must have x â‰  -2. Thus, $$\frac{x^{2}-3 x-10}{x^{3} + 6 x^{2} + 12 x + 8}$$ = $$\frac{(x-5)(x + 2)}{(x + 2)^{3}}$$ = $$\frac{x-5}{(x + 2)^{2}}$$, where x â‰  -2.

e. $$\frac{x^{3} + 1}{x^{2} + 1}$$
While x3 + 1 = (x + 1)(x2-x + 1), the polynomial expression in the denominator does not factor. Thus, this expression is already simplified to lowest terms. Since x2 + 1â‰¥1 for all values of x, the denominator is never zero. Thus, there are no values of x that need to be excluded.

Exercise 2.
Determine the domain of each rational function, and express the rule for each function in an equivalent form in lowest terms.
a. f(x) = $$\frac{(x + 2)^{2}(x-3)(x + 1)}{(x + 2)(x + 1)}$$
The domain is all real numbers x so that x â‰  -1 and x â‰  -2.
f(x) = $$\frac{(x + 2)^{2}(x-3)(x + 1)}{(x + 2)(x + 1)}$$ = (x + 2)(x-3) for x â‰  -1 and x â‰  -2

b. f(x) = $$\frac{x^{2}-6 x + 9}{x-3}$$
The domain is all real numbers x so that x â‰  3.
f(x) = $$\frac{(x-3)^{2}}{x-3}$$ = x-3 for x â‰  3

c. f(x) = $$\frac{3 x^{3}-75 x}{x^{3} + 15 x^{2} + 75 x + 125}$$
The domain is all real numbers x so that x â‰  -5.
f(x) = $$\frac{3 x\left(x^{2}-25\right)}{(x + 5)^{3}}$$ = $$\frac{3 x(x + 5)(x-5)}{(x + 5)^{3}}$$ = $$\frac{3 x(x-5)}{(x + 5)^{2}}$$ for x â‰  -5

### Eureka Math Precalculus Module 3 Lesson 11 Problem Set Answer Key

Question 1.
For each pair of functions f and g, find the domain of f and the domain of g. Indicate whether f and g are the same function.
a. f(x) = $$\frac{x^{2}}{x}$$, g(x) = x
The domain of f is all real numbers x with x â‰  0. The domain of g is all real numbers x.
No, functions f and g are not the same function.

b. f(x) = $$\frac{x}{x}$$, g(x) = 1
The domain of f is all real numbers x with x â‰  0. The domain of g is all real numbers x.
No, functions f and g are not the same function.

c. f(x) = $$\frac{2 x^{2} + 6 x + 8}{2}$$, g(x) = x2 + 6x + 8
The domain of f is all real numbers x. The domain of g is all real numbers x.
Yes, functions f and g are the same function.

d. f(x) = $$\frac{x^{2} + 3 x + 2}{x + 2}$$, g(x) = x + 1
The domain of f is all real numbers x with x â‰  -2. The domain of g is all real numbers x.
No, functions f and g are not the same function.

e. f(x) = $$\frac{x + 2}{x^{2} + 3 x + 2}$$, g(x) = $$\frac{1}{x + 1}$$
The domain of f is all real numbers x with x â‰  -2 and x â‰  -1. The domain of g is all real numbers x with x â‰  -1.
No, functions f and g are not the same function.

f. f(x) = $$\frac{x^{4}-1}{x^{2}-1}$$, g(x) = x2 + 1
The domain of f is all real numbers x with x â‰  1 and x â‰  -1. The domain of g is all real numbers x.
No, functions f and g are not the same function.

g. f(x) = $$\frac{x^{4}-1}{x^{2} + 1}$$, g(x) = x2-1
Because x2 + 1 is never zero, the domain of f is all real numbers x. The domain of g is all real numbers x.
Yes, functions f and g are the same function.

h. f(x) = $$\frac{x^{4}-x}{x^{2} + x}$$, g(x) = $$\frac{x^{3}-1}{x + 1}$$
The domain of f is all real numbers x with x â‰  0 and x â‰  -1. The domain of g is all real numbers x with x â‰  -1.
No, functions f and g are not the same function.

i. f(x) = $$\frac{x^{4} + x^{3} + x^{2}}{x^{2} + x + 1}$$, g(x) = x2
Because x2 + x + 1 doesnâ€™t factor, the denominator of f is never zero, and the domain of f is all real numbers x. The domain of g is also all real numbers x.
Yes, functions f and g are the same function.

Question 2.
Determine the domain of each rational function, and express the rule for each function in an equivalent form in lowest terms.
a. f(x) = $$\frac{x^{4}}{x^{2}}$$
The domain of f is all real numbers x with x â‰  0.
f(x) = $$\frac{x^{4}}{x^{2}}$$ = x2, where x â‰  0

b. f(x) = $$\frac{3 x + 3}{15 x-6}$$
Because 15x-6 = 3(5x-2), the domain of f is all real numbers x with x â‰  $$\frac{2}{5}$$.
f(x) = $$\frac{3(x + 1)}{3(5 x-2)}$$ = $$\frac{x + 1}{5 x-2}$$, where x â‰  $$\frac{2}{5}$$

c. f(x) = $$\frac{x^{2}-x-2}{x^{2} + x}$$
Because x2 + x = x(x + 1), the domain of f is all real numbers x with x â‰  0 and x â‰  -1.
f(x) = $$\frac{x^{2}-x-2}{x^{2} + x}$$ = $$\frac{(x-2)(x + 1)}{x(x + 1)}$$ = $$\frac{x-2}{x}$$, where x â‰  0 and x â‰  -1

d. f(x) = $$\frac{8 x^{2} + 2 x-15}{4 x^{2}-4 x-15}$$
Because 4x2-4x-15 = (2x + 3)(2x-5), the domain of f is all real numbers x with x â‰  –$$\frac{3}{2}$$ and x â‰  $$\frac{5}{2}$$.
f(x) = $$\frac{8 x^{2} + 2 x-15}{4 x^{2}-4 x-15}$$ = $$\frac{(2 x + 3)(4 x-5)}{(2 x + 3)(2 x-5)}$$ = $$\frac{(4 x-5)}{(2 x-5)}$$, where x â‰  $$\frac{5}{2}$$ and x â‰  –$$\frac{3}{2}$$

e. f(x) = $$\frac{2 x^{3}-3 x^{2}-2 x + 3}{x^{3}-x}$$
Because x3-x = x(x-1)(x + 1), the domain of f is all real numbers x with x â‰  0, x â‰  1 and x â‰  -1.
f(x) = $$\frac{2 x^{3}-3 x^{2}-2 x + 3}{x^{3}-x}$$ = $$\frac{(2 x-3)\left(x^{2}-1\right)}{x\left(x^{2}-1\right)}$$ = $$\frac{2 x-3}{x}$$, where x â‰  0 and x â‰  Â±1

f. f(x) = $$\frac{3 x^{3} + x^{2} + 3 x + 1}{x^{3} + x}$$
Because x3 + x = x(x2 + 1), the domain of f is all real numbers x with x â‰  0.
f(x) = $$\frac{3 x^{3} + x^{2} + 3 x + 1}{x^{3} + x}$$ = $$\frac{(3 x + 1)\left(x^{2} + 1\right)}{x\left(x^{2} + 1\right)}$$ = $$\frac{3 x + 1}{x}$$, where x â‰  0

Question 3.
For each pair of functions below, calculate f(x) + g(x), f(x)-g(x), f(x)â‹…g(x), and $$\frac{f(x)}{g(x)}$$. Indicate restrictions on the domain of the resulting functions.
a. f(x) = $$\frac{2}{x}$$, g(x) = $$\frac{x}{x + 2}$$

b. f(x) = $$\frac{3}{x + 1}$$, g(x) = $$\frac{x}{x^{3} + 1}$$

### Eureka Math Precalculus Module 3 Lesson 11 Exit Ticket Answer Key

Question 1.
Identify whether the functions shown are rational:
a. f(x) = $$\frac{x}{x^{2} + 1}$$
Yes. Both P(x) = x and Q(x) = x2 + 1 are polynomial functions.

b. f(x) = $$\frac{\sqrt{\boldsymbol{x}}}{x^{2} + 1}$$
No. The function P(x) = $$\sqrt{x}$$ is not a polynomial function.

c. f(x) = $$\frac{x}{x^{0.4} + 1}$$
No. The function Q(x) = x(0.4) + 1 is not a polynomial function.

d. f(x) = ($$\frac{x}{x^{2} + 1}$$)2
Yes. When multiplied out, f(x) = $$\frac{x^{2}}{x^{4} + 2 x^{2} + 1}$$, so f is the quotient of two polynomial functions.

e. f(x) = $$\frac{\sqrt{2} x}{e x^{2} + \sqrt{\pi}}$$
Yes. While the coefficients are not integers, P(x) = $$\sqrt{2}$$ x and Q(x) = ex2 + $$\sqrt{\pi}$$ are both polynomial functions since all the powers of x are whole numbers.
Anmol says f(x) = $$\frac{x + 1}{x^{2}-1}$$ and g(x) = $$\frac{1}{x-1}$$represent the same function. Is she correct? Justify your answer.
The function f(x) = $$\frac{x + 1}{x^{2}-1}$$ is not defined for x = 1 and x = -1. However, the function g(x) = $$\frac{1}{x-1}$$ is not defined for x = 1. These two functions do not have the same domain, so they are not the same function.