# Eureka Math Algebra 2 Module 1 Lesson 12 Answer Key

## Engage NY Eureka Math Algebra 2 Module 1 Lesson 12 Answer Key

### Eureka Math Algebra 2 Module 1 Lesson 12 Example Answer Key

Example 1.
Find all real solutions to the equation (x2 – 6x + 3)(2x2 – 4x – 7) = 0.
Allow students the opportunity to struggle with factoring these expressions, discuss with their neighbors, and reach the conclusion that neither expression can be factored with integer coefficients.
â†’ We have discovered an obstacle to factoring. The expressions x2 – 6x + 3 and 2x2 – 4x – 7 do not factor as readily as the examples from the previous lesson. Does anybody recall how we might factor them?
Students have completed the square in both Geometry and Algebra I, so give them an opportunity to recall the process.

â†’ When a quadratic expression is not easily factorable, we can either apply a technique called completing the square, or we can use the quadratic formula. Letâ€™s factor the first expression by completing the square.

â†’ We first create some space between the x term and the constant term:
x2 – 6x + —— + 3 = 0.

â†’ The next step is the key step. Take half of the coefficient of the x term, square that number, and add and subtract it in the space we created:
x2 – 6x + (- 3)2 – (- 3)2 + 3 = 0
x2 – 6x + 9 – 9 + 3 = 0.
Discuss the following questions with the class, and give them the opportunity to justify this step.

â†’ Why did we choose 9? Why did we both add and subtract 9? How does this help us solve the equation?
Adding 9 creates a perfect square trinomial in the first three terms.
Adding and subtracting 9 means that we have not changed the value of the expression on the left side of the equation.
Adding and subtracting 9 creates a perfect square trinomial x2 – 6x + 9 = (x – 3)2.

â†’ We cannot just add a number to an expression without changing its value. By adding 9 and subtracting 9, we have essentially added 0 using the additive identity property, which leads to an equivalent expression on the left-hand side of the equation and thus preserves solutions of the equation.

â†’ This process creates a structure that allows us to factor the first three terms of the expression on the left side of the equation and then solve for the variable.

(x – 3)2 – 6 = 0

â†’ Solving for x:
(x – 3)2 = 6
x – 3 = âˆš6 or x – 3 = – âˆš6
x = 3 + âˆš6 or x = 3 – âˆš6.

â†’ Thus, we have found two solutions by setting the first quadratic expression equal to zero, completing the square, and solving the factored equation. Since the leading coefficient of x2 – 6x + 3 is 1, we know from our work in Algebra I that the factored form is
x2 – 6x + 3 = (x -(3 + âˆš6)) (x – (3 – âˆš6).

â†’ Letâ€™s repeat the process with the second equation. What is the first step to completing the square?
2x2 – 4x – 7 = 0
Allow students an opportunity to suggest the first step to completing the square.

â†’ We can only complete the square when the leading coefficient is 1, so our first step is to factor out the 2.
2(x2 – 2x – $$\frac{7}{2}$$) = o

â†’ Now we can complete the square with the expression inside the parentheses.
2(x2 – 2x + ————- + $$\frac{7}{2}$$) = 0
2(x2 – 2x + (- 1)2 – (- 1)2 – $$\frac{7}{2}$$) = 0
2(x2 – 2x + 1 – $$\frac{9}{2}$$) = 0
2((x – 1)2 – $$\frac{9}{2}$$) = 0

â†’ Next, we divide both sides by 2.
(x – 1)2 – $$\frac{9}{2}$$ = 0

â†’ Finally, We solve for x.
(x – 1)2 = $$\frac{9}{2}$$
x = 1 + $$\sqrt{\frac{9}{2}}$$ or x = 1 – $$\sqrt{\frac{9}{2}}$$
x = 1 + $$\frac{3 \sqrt{2}}{2}$$ or x = 1 – $$\frac{3 \sqrt{2}}{2}$$

â†’ Thus, we have found two more solutions to our original fourth-degree equation. We then have 2x2 – 4x – 7 = 2(x2 – 2x – $$\frac{7}{2}$$)
= 2(x – (1 + $$\frac{3 \sqrt{2}}{2}$$)) (x – (1 – $$\frac{3 \sqrt{2}}{2}$$))

â†’ Notice that we needed to multiply the factors by 2 to make the leading coefficients match.
â†’ Finally, we have the factored form of our original polynomial equation:
(x2 – 6x + 3) (2x2 – 4x – 7) = 0 in factored form
2(x – (3 + âˆš6)) (x – (3 – âˆš6)) (x – (1 + $$\frac{3 \sqrt{2}}{2}$$)) (x – (1 – $$\frac{3 \sqrt{2}}{2}$$)) = 0.

â†’ Thus, the solutions to the equation (x2 – 6x + 3) (2x2 – 4x – 7) = 0 are the four values 3 + ?6, 3 – ?6, 1 + $$\frac{3 \sqrt{2}}{2}$$, and 1 – $$\frac{3 \sqrt{2}}{2}$$.

â†’ Similarly, we could have applied the quadratic formula to find the solutions to each quadratic equation in the previous example. Recall the quadratic formula.

â†’ The two solutions to the quadratic equation ax2 + bx + c = 0 are $$\frac{-b+\sqrt{b^{2}-4 a c}}{2 a}$$ and $$\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}$$

Example 2.
Find all solutions to x3 + 3x2 – 9x – 27 = 0 by factoring the equation.
â†’ Letâ€™s start with our original equation x3 + 3x2 – 9x – 27 = 0. Is there a greatest common factor (GCF) for all four terms on the left-hand side we can factor out?
No, the GCF is 1.

â†’ Letâ€™s group the terms of the left-hand side as follows:
x3 + 3x2 – 9x – 27 = (x3 + 3x2 )- (9x + 27).

â†’ Can we factor out a GCF from each set of parentheses independently?
Yes, x2 can be factored out of the first piece and 9 out of the second.

Factor the GCF out of each part. Have students do as much of this work as possible.
x3 + 3x2 – 9x – 27 = x2 (x + 3) – 9(x + 3)

â†’ Do you notice anything interesting about the right side of the above equation?
I noticed that x + 3 is a common factor.

â†’ Since both terms have a factor of (x + 3), we have found a quantity that can be factored out.
x3 + 3x2 – 9x – 27 = (x + 3) (x2 – 9)

â†’ And as we saw above, we can take this one step further.
x3 + 3x2 – 9x – 27 = (x + 3) (x + 3) (x – 3)

â†’ Because of the zero property, the original problem is now easy to solve because x3 + 3x2 – 9x – 27 = 0 exactly when (x + 3)2(x – 3) = 0. What are the solutions to the original equation?
The solutions to x3 + 3x2 – 9x -27 = 0 are x = – 3 and x = 3.

â†’ The process you just completed is often called factoring by grouping, and it works only on certain 3rd degree polynomial expressions, such as x3 + 3x2 – 9x – 27.

### Eureka Math Algebra 2 Module 1 Lesson 12 Exercise Answer Key

Exercise 1.
Factor and find all real solutions to the equation (x2 – 2x – 4) (3x2 + 8x – 3) = 0.
Ask half of the students to apply the quadratic formula to solve x2 – 2x – 4 = 0 and the other half to apply the quadratic formula to solve 3x2 + 8x – 3 = 0.

The quadratic formula gives solutions 1 + âˆš5 and 1 – âˆš5 for the first equation and – 3 and $$\frac{1}{3}$$ for the second equation.

Since 1 + âˆš5 and 1 – âˆš5 are the two solutions to x2 – 2x – 4 = 0 found by the quadratic formula, we know from work in Algebra I that (x – (1 + âˆš5)) (x – (1 – âˆš5)) = x2 – 2x – 4. However, we need to be more careful when using the solutions to factor the second quadratic expression. The leading coefficient of (x + 3) (x – $$\frac{1}{3}$$) = x2 + $$\frac{8}{3}$$x – 1 is 1, and the leading coefficient of 3x2 + 8x – 3 is 3, so we need to multiply our factors by 3:
3x2 + 8x – 3 = 3 (x + 3) (x – $$\frac{1}{3}$$).

Thus, the factored form of the original equation is
(x2 – 2x – 4) (3x2 + 8x – 3) = 3(x – (1 + âˆš5) (x – (1 – âˆš5)) (x + 3) (x – $$\frac{1}{3}$$) = 0,
and the four solutions to (x2 – 2x – 4) (3x2 + 8x – 3) = 0 are 1 + âˆš5, 1 – âˆš5, – 3, and $$\frac{1}{3}$$.

To summarize, if we have a fourth-degree polynomial already factored into two quadratic expressions, we can try to factor the entire polynomial by completing the square on one or both quadratic expressions,or by using the quadratic formula to find the roots of the quadratic polynomials and then constructing the factored form of each quadratic polynomial.

â†’ Discussion
We have overcome the obstacle of difficult-to-factor quadratic expressions. Letâ€™s look next at the obstacles encountered when attempting to solve a third-degree polynomial equation such as the following:
x3 + 3x2 – 9x – 27 = 0.

â†’ How might we begin to solve this equation?

Allow students an opportunity to brainstorm as a class, in pairs, or in table groups. Students may note that coefficients are powers of 3 but may not be sure how that helps. Let them know they are seeing something important that they ma be able to use. Stronger students might even try to group the components.

â†’ While we have made some interesting observations, we have not quite found a way to factor this expression. What if we know that x + 3 is one factor?

If students do not come up with polynomial division, point them in that direction through a numerical example:
Suppose we want the factors of 210, and we know that one factor is 3. How do we find the other factors?

Have students perform the polynomial division

â†’ Since x3 + 3x3 – 9x – 27 = (x + 3) (x2 – 9), we know that
x3 + 3x2 – 9x – 27 = (x + 3) (x – 3) (x + 3) = (x + 3)2 (x – 3).
â†’ By the zero product property, the solutions to x3 + 3x2 – 9x – 27 = 0 are – 3 and 3.
â†’ But, how do we start if we donâ€™t know any of the factors in advance?

Exercise 2.
Find all real solutions to x3 – 5x2 – 4x + 20 = 0.
x3 – 5x2 – 4x + 20 = 0Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Alternate approach:
x3 (x – 5) – 4 (x – 5) = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  x3 – 5x2 – 4x + 20 = 0
(x – 5) (x2 – 4) = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  x(x3 – 4) – 5(x2 – 4) = 0
(x – 5) (x – 2) (x + 2) = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (x – 5)(x2 – 4) = 0
Thus, the solutions are 5, 2, and – 2.Â  Â  Â  Â  Â  Â  Â  Â  (x – 5) (x – 2) (x + 2) = 0

Exercise 3.
Find all real solutions to x3 – 8x2 – 2x + 16 = 0.
x3 – 8x2 – 2x + 16 = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  x3 – 8x – 2x + 16 = 0
x2 (x – 8) – 2 (x – 8) = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  x(x2 – 2) – 8(x2 – 2) = 0
(x – 8) (x2 – 2) = 0 Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (x – 8) (x2 – 2) = 0
Thus, the solutions are 8, âˆš2 and -âˆš2.

### Eureka Math Algebra 2 Module 1 Lesson 12 Problem Set Answer Key

Question 1.
Solve each of the following equations by completing the square.
a. x2 – 6x + 2 = 0
3 + âˆš7, 3 – âˆš7

b. x2 – 4x = – 1
2 + âˆš3, 2 – âˆš3

c. x2 + x – $$\frac{3}{4}$$ = 0
$$\frac{1}{2}$$, $$-\frac{3}{2}$$

d. 3x2 – 9x = – 6
2, 1

e. (2x2 – 5x + 2) (3x2 – 4x + 1) = 0
2, $$\frac{1}{2}$$, 1, $$\frac{1}{3}$$

f. x4 – 4x2 + 2 = 0

Question 2.
Solve each of the following equations using the quadratic formula.
a. x2 – 5x – 3 = 0
$$\frac{5}{2}+\frac{\sqrt{37}}{2}, \frac{5}{2}-\frac{\sqrt{37}}{2}$$

b. (6x2 – 7x + 2) (x2 – 5x + 5) = 0
$$\frac{1}{2}, \frac{2}{3}, \frac{1}{2}(5+\sqrt{5}), \frac{1}{2}(5-\sqrt{5})$$

c. (3x2 – 13x + 14) (x2 – 4x + 1) = 0
2, $$\frac{7}{3}$$, 2 + âˆš3, 2 – âˆš3

Question 3.
Not all of the expressions in the equations below can be factored using the techniques discussed so far in this course. First, determine if the expression can be factored with real coefficients. If so, factor the expression, and find all real solutions to the equation.
a. x2 – 5x – 24 = 0
Can be factored: (x – 8)(x + 3) = 0.
Solutions: 8, – 3

b. 3x2 + 5x – 2 = 0
Can be factored: (3x – 1)(x + 2) = 0.
Solutions: $$\frac{1}{3}$$, – 2

c. x2 + 2x + 4 = 0
Cannot be factored with real number coefficients.

d. x3 + 3x2 – 2x + 6 = 0
Cannot be factored with real number coefficients.

e. x3 + 3x2 + 2x + 6 = 0
Can be factored: (x + 3) (x2 + 2) = 0.
SolutIon: – 3

f. 2x3 + x2 – 6x – 3 = 0
Can be factored: (2x + 1)(x – âˆš3)(x + âˆš3) = 0
Solutions: –$$\frac{1}{2}$$, âˆš3, – âˆš3

i. 4x3 + 2x2 – 36x – 18 = 0
Can be factored: 2(2x + 1)(x – 3)(x + 3) = 0.
Solutions: –$$\frac{1}{2}$$, 3, – 3

j. x2 –$$\frac{1}{2}$$x – $$\frac{15}{2}$$ = 0
Can be factored: (x +)(x – 3) = 0.
Solutions: – $$\frac{5}{2}$$, 3

Question 4.
Solve the following equations by bringing all terms to one side of the equation and factoring out the greatest common factor.
a. (x – 2) (x – 1) = (x – 2) (x + 1)
(x – 2) (x + 1) – (x – 2) (x – 1) = 0
(x – 2) (x + 1 – (x – 1)) = 0
(x – 2) (2) = 0
x = 2
So, the only solution to (x – 2)(x – 1) = (x – 2)(x + 1) is 2.

b. (2x + 3) (x – 4) = (2x + 3) (x + 5)
(2x + 3) (x – 4) – (2x + 3) (x + 5) = 0
(2x + 3) (x – 4 – (x + 5)) = 0
(2x + 3) (- 9) = 0
x = –$$\frac{3}{2}$$
So, the only solution to (2x + 3)(x – 4) = (2x + 3)(x + 5) is –$$\frac{3}{2}$$.

c. (x – 1) (2x + 3) = (x – 1) (x + 2)
(x – 1) (2x + 3) – (x – 1) (x + 2) = 0
(x – 1) (2x + 3 – (x + 2)) = 0
(x – 1) (x + 1) = 0
x = 1 or x = -1
The solutions to (x – 1) (2x + 3) = (x – 1) (x + 2) are 1 and – 1.

d. (x2 + 1) (3x – 7) = (x2 + 1) (3x + 2)
(x2 + 1) (3x – 7) – (x2 + 1) (3x + 2) = 0
(x2 + 1) (3x – 7 – (3x + 2)) = 0
(x2 + 1) (- 9) = 0
x2 + 1 = 0
There are no real number solutions to (x2 + 1) (3x -7) = (x2 + 1) (3x + 2).

e. (x + 3) (2x2 + 7) = (x + 3) (x2 + 8)
(x + 3) (2x2 + 7) – (x + 3) (x2 + 8) = 0
(x + 3) (2x2 + 7 – (x2 + 8)) = 0
(x + 3) (x2 – 1) = 0
(x + 3) (x – 1) (x + 1) = 0
The three solutions to (x + 3) (2x2 + 7) = (x + 3) (x2 + 8) are – 3, – 1, and 1.

Question 5.
Consider the expression x4 + 1. Since x2 + 1 does not factor with real number coefficients, we might expect that x4 + 1 also does not factor with real number coefficients. In this exercise, we investigate the possibility of factoring
x4 + 1.
a. Simplify the expression (x2 + 1)2 – 2x2.
(x2 + 1)2 – 2x2 = x4 + 1

b. Factor (x2 + 1)2 – 2x2 as a difference of squares.
(x2 + 1)2 – 2x2 = ((x2 + 1) – âˆš2x) ((x2 + 1) + âˆš2x)

c. Is it possible to factor x4 + 1 with real number coefficients? Explain.
Yes. x4 + 1 = ((x2 + 1) – âˆš2x) ((x2 + 1) + âˆš2x)
In an equivalent but more conventional form, we have
x4 + 1 (x2 – âˆš2x + 1) (x2 + âˆš2x + 1).

### Eureka Math Algebra 2 Module 1 Lesson 12 Exit Ticket Answer Key

Question 1.
Solve the following equation, and explain your solution method.
x3 + 7x2 – x – 7 = 0