Question

Part A:

We can use factoring by grouping to solve the equation below.

3x2+13x+12=0
3
𝑥
2
+
13
𝑥
+
12
=
0

After we multiply A*C, we can pick a factor pair to split up the middle term.

To create groups that will have a common factor, the equation can be rewritten in which of the following ways?

(1 point)
Responses

3x2+1x+36x+12=0
3
𝑥
2
+
1
𝑥
+
36
𝑥
+
12
=
0
3 x squared plus 1 x plus 36 x plus 12 is equal to 0

3x2+10x+3x+12=0
3
𝑥
2
+
10
𝑥
+
3
𝑥
+
12
=
0
3 x squared plus 10 x plus 3 x plus 12 is equal to 0

3x2+9x+4x+12=0
3
𝑥
2
+
9
𝑥
+
4
𝑥
+
12
=
0
3 x squared plus 9 x plus 4 x plus 12 is equal to 0

3x2+2x+18x+12=0
3
𝑥
2
+
2
𝑥
+
18
𝑥
+
12
=
0
3 x squared plus 2 x plus 18 x plus 12 is equal to 0
Question 2
Part B:

Using the same equation, 3x2+13x+12=0
3
𝑥
2
+
13
𝑥
+
12
=
0

once you have factored by grouping, the trinomial becomes which of the following?

(1 point)
Responses

(x+4)(x+3)=0
(
𝑥
+
4
)
(
𝑥
+
3
)
=
0
open paren x plus 4 close paren times open paren x plus 3 close paren is equal to 0

(3x−4)(x−3)=0
(
3
𝑥
−
4
)
(
𝑥
−
3
)
=
0
open paren 3 x minus 4 close paren times open paren x minus 3 close paren is equal to 0

(3x+3)(x+4)=0
(
3
𝑥
+
3
)
(
𝑥
+
4
)
=
0
open paren 3 x plus 3 close paren times open paren x plus 4 close paren is equal to 0

(3x+4)(x+3)=0
(
3
𝑥
+
4
)
(
𝑥
+
3
)
=
0
open paren 3 x plus 4 close paren times open paren x plus 3 close paren is equal to 0
Question 3
Part C:

Using the same equation

3x2+13x+12=0
3
𝑥
2
+
13
𝑥
+
12
=
0

once you have factored by grouping, you can find the two solutions. What are they?

(1 point)
Responses

x=−43 and x=−3
𝑥
=
−
4
3

𝑎
𝑛
𝑑

𝑥
=
−
3
x=−43 and x=−3
𝑥
=
−
4
3

𝑎
𝑛
𝑑

𝑥
=
−
3

x=−4 and x=−3
𝑥
=
−
4

𝑎
𝑛
𝑑

𝑥
=
−
3
x=−4 and x=−3
𝑥
=
−
4

𝑎
𝑛
𝑑

𝑥
=
−
3

x=4 and x=3
𝑥
=
4

𝑎
𝑛
𝑑

𝑥
=
3
x=4 and x=3
𝑥
=
4

𝑎
𝑛
𝑑

𝑥
=
3

x=43 and x=−3

Answer 1

Let's break down the solution step by step.

### Part A:

We want to rewrite the quadratic \(3x^2 + 13x + 12 = 0\) such that we can apply factoring by grouping.

To do this, we first multiply the coefficient of \(x^2\) (which is 3) with the constant term (12):

\[ A \cdot C = 3 \cdot 12 = 36. \]

Next, we need to find two numbers that multiply to 36 and add up to 13.

The pair that works is 9 and 4, because:

\[ 9 + 4 = 13 \quad\text{and}\quad 9 \cdot 4 = 36. \]

Now, we can rewrite the quadratic:

\[ 3x^2 + 9x + 4x + 12 = 0. \]

Now we can group the terms:

1. The correct grouping for the equation is:

\[ 3x^2 + 9x + 4x + 12 = 0. \]

Thus, the correct response for Part A is:

**3x² + 9x + 4x + 12 = 0**

### Part B:

After rewriting, we can now factor by grouping:

1. Group the first two terms and the last two terms:
\[ (3x^2 + 9x) + (4x + 12) = 0. \]

2. Factor out the common factors in each group:

\[ 3x(x + 3) + 4(x + 3) = 0. \]

3. Now factor out \((x + 3)\):

\[ (3x + 4)(x + 3) = 0. \]

The factored form of the equation is:

**(3x + 4)(x + 3) = 0**

### Part C:

To find the solutions, we set each factor equal to zero:

1. From \(3x + 4 = 0\):
\[ 3x = -4 \quad \Rightarrow \quad x = -\frac{4}{3}. \]

2. From \(x + 3 = 0\):
\[ x = -3. \]

Thus, the solutions are:

**x = -4/3 and x = -3.**

The correct answer for Part C is:

**x = -4/3 and x = -3.**