# Eureka Math Algebra 1 Module 4 Lesson 2 Answer Key

## Engage NY Eureka Math Algebra 1 Module 4 Lesson 2 Answer Key

### Eureka Math Algebra 1 Module 4 Lesson 2 Example Answer Key

Example 1: Using a Table as an Aid
Use a table to assist in multiplying (x + 7)(x + 3).

x2 + 10x + 21

→ Are there like terms in the table that can be combined?
Yes, the terms in the diagonal can be added to give 10x.
→ After combining like terms, what is the simplified product of the two binomials?
x2 + 7x + 3x + 21 or x2 + 10x + 21

If the leading coefficient for a quadratic expression is not 1, the first step in factoring should be to see if all the terms in the expanded form have a common factor. Then, after factoring out the greatest common factor, it may be possible to factor again.

For example, to factor to 2x3 – 50x completely, begin by finding the GCF.
The GCF of the expression is 2x: 2x(x2 – 25).
Now, factor the difference of squares: 2x(x – 5)(x + 5).

Another example: Follow the steps to factor – 16t2 + 32t + 48 completely.
a. First, factor out the GCF. (Remember: When you factor out a negative number, all the signs on the resulting factor change.)
The GCF is – 16. Hint: Do not leave the negative 1 as the leading coefficient. Factor it out with the 16.
– 16(t2 – 2t – 3)

b. Now look for ways to factor further. (Notice the quadratic expression factors.)
– 16(t – 3)(t + 1)

### Eureka Math Algebra 1 Module 4 Lesson 2 Exercise Answer Key

Exercise 1.
Use a table to aid in finding the product of (2x + 1)(x + 4).

(2x + 1)(x + 4) = 2x2 + x + 8x + 4 = 2x2 + 9x + 4

POLYNOMIAL EXPRESSION: A polynomial expression is either:
(1) A numerical expression or a variable symbol, or
(2) The result of placing two previously generated polynomial expressions into the blanks of the addition operator (__ + __) or the multiplication operator (__× __).
→ What does the definition of a polynomial expression tell us, then, about the constant term of the polynomial x – 7?
That the constant term is actually – 7.
→ While we may write the polynomial as x – 7, the terms of this polynomial are actually x and – 7.

Exercises 2–6
Multiply the following binomials; note that every binomial given in the problems below is a polynomial in one variable, x, with a degree of one. Write the answers in standard form, which in this case takes the form ax2 + bx + c, where a, b, and c are constants.

Exercise 2.
(x + 1)(x – 7)
x2 – 6x – 7

Exercise 3.
(x + 9)(x + 2)
x2 + 11x + 18

Exercise 4.
(x – 5)(x – 3)
x2 – 8x + 15

Exercise 5.
(x + $$\frac{15}{2}$$)(x – 1)
x2 + $$\frac{13}{2}$$ x – $$\frac{15}{2}$$

Exercise 6.
(x – $$\frac{5}{4}$$)(x – $$\frac{3}{4}$$)
x2 – 2x + $$\frac{15}{16}$$

Describe any patterns you noticed as you worked.
All the coefficients for the x2 term are 1. The constant term for the resulting trinomial is the product of constant terms of the two binomials. There are always two terms that are like terms and can be combined; the coefficients of those terms after they are combined is the sum of the constant terms from the two binomials.

Exercises 7–10

Exercise 7.
x2 + 8x + 7
(x + 7)(x + 1)

Exercise 8.
m2 + m – 90
(m + 10)(m – 9)

Exercise 9.
k2 – 13k + 40
(k – 8)(k – 5)

Exercise 10.
v2 + 99v – 100
(v – 1)(v + 100)

### Eureka Math Algebra 1 Module 4 Lesson 2 Problem Set Answer Key

Question 1.
Factor these trinomials as the product of two binomials, and check your answer by multiplying.
a. x2 + 3x + 2
The pair of integers whose product is + 2 and whose sum is + 3 are + 1 and + 2.
So, the factored form is (x + 1)(x + 2).
Check: (x + 1)(x + 2) = x2 + 2x + x + 2 = x2 + 3x + 2

b. x2 – 8x + 15
The pair of integers whose product is + 15 and whose sum is – 8 are – 3 and – 5.
So, the factored form is (x – 3)(x – 5).
Check: (x – 3)(x – 5) = x2 – 5x – 3x + 15 = x2 – 8x + 15

c. x2 + 8x + 15
The pair of integers whose product is + 15 and whose sum is + 8 are + 3 and + 5.
So, the factored form is (x + 3)(x + 5).
Check: (x + 3)(x + 5) = x2 + 5x + 3x + 15 = x2 + 8x + 15

Factor completely.
d. 4m2 – 4n2
The GCF of the terms is 4.
Factor out 4: 4(m2 – n2).
Factor the difference of squares: 4(m – n)(m + n).

e. – 2x3 – 2x2 + 112x
The GCF of the terms is – 2x.
Factor out – 2x: – 2x3 – 2x2 + 112x = – 2x(x2 + x – 56).
Factor the quadratic trinomial: – 2x(x – 7)(x + 8).

f. y8 – 81x4
Factor the difference of squares: y8 – 81x4 = (y4 + 9x2)(y4 – 9x2).
Factor the difference of squares: (y4 + 9x2)(y2 + 3x)(y2 – 3x).

Question 2.
The square parking lot at Gene Simon’s Donut Palace is going to be enlarged so that there will be an additional 30 ft. of parking space in the front of the lot and an additional 30 ft. of parking space on the side of the lot, as shown in the figure below. Write an expression in terms of x that can be used to represent the area of the new parking lot.

We know that the original parking lot is a square. We can let x represent the length of each side, in feet, of the original square. We can represent each side of the new parking lot as x + 30. Using the area formula for a square, Area = s2, we can represent this as
(x + 30)2.
(x + 30)2 = (x + 30)(x + 30)
= x2 + 60x + 900

Explain how your solution is demonstrated in the area model.
The original square in the upper left corner is x “by” x, which results in an area of x2 square feet; each smaller rectangle is 30 “by” x, which results in an area of 30x square feet; there are 2 of them, giving a total of 60x square feet. The smaller square is 30 by 30 square feet, which results in an area of 900 square feet. That gives us the following expression for the area of the new parking lot: x2 + 60x + 900.

### Eureka Math Algebra 1 Module 4 Lesson 2 Exit Ticket Answer Key

Question 1.
Factor completely: 2a2 + 6a + 18
Factor out the GCF: 2(a2 + 3a + 9).

Question 2.
Factor completely: 5x2 – 5
Factor out the GCF: 5(x2 – 1).
Now, factor the difference of perfect squares: 5(x + 1)(x – 1).

Question 3.
Factor completely: 3t3 + 18t2 – 48t