## 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.

Answer:

Since x^{2}-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.

Answer:

\(\frac{2 x^{4} + 6 x^{3} + 6 x^{2} + 2 x}{3 x^{2} + 3 x}\) = \(\frac{2 x\left(x^{3} + 3 x^{2} + 3 x + 1\right)}{3 x(x + 1)}\) = \(\frac{2 x(x + 1)^{3}}{3 x(x + 1)}\) = \(\frac{2(x + 1)^{2}}{3}\) 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. 9x^{4}-16x^{2}

Answer:

x^{2} (3x + 4)(3x-4)

b. 2x^{3} + 5x^{2}-8x-20

Answer:

(x^{2}-4)(2x + 5) = (x + 2)(x-2)(2x + 5)

c. x^{3} + 3x^{2} + 3x + 1

Answer:

(x + 1)^{3}

d. 8x^{3}-1

Answer:

(2x-1)(4x^{2} + 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. \(\frac{x^{2}-6 x + 5}{x^{2}-3 x-10}\)

Answer:

The denominator factors into x^{2}-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 x^{3} + 2x^{2} + 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 x^{2} + 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. \(\frac{x^{2}-3 x-10}{x^{3} + 6 x^{2} + 12 x + 8}\)

Answer:

The denominator factors into x^{3} + 6x^{2} + 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. \(\frac{x^{3} + 1}{x^{2} + 1}\)

Answer:

While x^{3} + 1 = (x + 1)(x^{2}-x + 1), the polynomial expression in the denominator does not factor. Thus, this expression is already simplified to lowest terms. Since x^{2} + 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) = \(\frac{(x + 2)^{2}(x-3)(x + 1)}{(x + 2)(x + 1)}\)

Answer:

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) = \(\frac{x^{2}-6 x + 9}{x-3}\)

Answer:

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) = \(\frac{3 x^{3}-75 x}{x^{3} + 15 x^{2} + 75 x + 125}\)

Answer:

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

Answer:

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

Answer:

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) = x^{2} + 6x + 8

Answer:

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

Answer:

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) = \(\frac{1}{x + 1}\)

Answer:

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) = x^{2} + 1

Answer:

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) = x^{2}-1

Answer:

Because x^{2} + 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) = \(\frac{x^{3}-1}{x + 1}\)

Answer:

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) = x^{2}

Answer:

Because x^{2} + 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) = \(\frac{x^{4}}{x^{2}}\)

Answer:

The domain of f is all real numbers x with x â‰ 0.

f(x) = \(\frac{x^{4}}{x^{2}}\) = x^{2}, where x â‰ 0

b. f(x) = \(\frac{3 x + 3}{15 x-6}\)

Answer:

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) = \(\frac{x^{2}-x-2}{x^{2} + x}\)

Answer:

Because x^{2} + 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) = \(\frac{8 x^{2} + 2 x-15}{4 x^{2}-4 x-15}\)

Answer:

Because 4x^{2}-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) = \(\frac{2 x^{3}-3 x^{2}-2 x + 3}{x^{3}-x}\)

Answer:

Because x^{3}-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) = \(\frac{3 x^{3} + x^{2} + 3 x + 1}{x^{3} + x}\)

Answer:

Because x^{3} + x = x(x^{2} + 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) = \(\frac{x}{x + 2}\)

Answer:

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

Answer:

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

Question 1.

Identify whether the functions shown are rational:

a. f(x) = \(\frac{x}{x^{2} + 1}\)

Answer:

Yes. Both P(x) = x and Q(x) = x^{2} + 1 are polynomial functions.

b. f(x) = \(\frac{\sqrt{\boldsymbol{x}}}{x^{2} + 1}\)

Answer:

No. The function P(x) = \(\sqrt{x}\) is not a polynomial function.

c. f(x) = \(\frac{x}{x^{0.4} + 1}\)

Answer:

No. The function Q(x) = x^{(0.4)} + 1 is not a polynomial function.

d. f(x) = (\(\frac{x}{x^{2} + 1}\))^{2}

Answer:

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) = \(\frac{\sqrt{2} x}{e x^{2} + \sqrt{\pi}}\)

Answer:

Yes. While the coefficients are not integers, P(x) = \(\sqrt{2}\) x and Q(x) = ex^{2} + \(\sqrt{\pi}\) are both polynomial functions since all the powers of x are whole numbers.

Question 2.

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.

Answer:

She is not correct.

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.