Question
Part A:
We can use factoring by grouping to solve the equation below.
3x2+13x+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 squared plus 1 x plus 36 x plus 12 is equal to 0
3x2+10x+3x+12=0
3 x squared plus 10 x plus 3 x plus 12 is equal to 0
3x2+9x+4x+12=0
3 x squared plus 9 x plus 4 x plus 12 is equal to 0
3x2+2x+18x+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
once you have factored by grouping, the trinomial becomes which of the following?
(1 point)
Responses
(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
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
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
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
once you have factored by grouping, you can find the two solutions. What are they?
(1 point)
Responses
x=−43 and x=−3
x=−43 and x=−3
x=−4 and x=−3
x=−4 and x=−3
x=4 and x=3
x=4 and x=3
x=43 and x=−3
We can use factoring by grouping to solve the equation below.
3x2+13x+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 squared plus 1 x plus 36 x plus 12 is equal to 0
3x2+10x+3x+12=0
3 x squared plus 10 x plus 3 x plus 12 is equal to 0
3x2+9x+4x+12=0
3 x squared plus 9 x plus 4 x plus 12 is equal to 0
3x2+2x+18x+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
once you have factored by grouping, the trinomial becomes which of the following?
(1 point)
Responses
(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
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
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
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
once you have factored by grouping, you can find the two solutions. What are they?
(1 point)
Responses
x=−43 and x=−3
x=−43 and x=−3
x=−4 and x=−3
x=−4 and x=−3
x=4 and x=3
x=4 and x=3
x=43 and x=−3
Answers
There are no human answers yet.
Answered by
GPT-4o mini
### Question 1: Part A
To factor the equation \(3x^2 + 13x + 12 = 0\) using factoring by grouping, we first multiply \(A\) (the coefficient of \(x^2\), which is 3) and \(C\) (the constant term, which is 12).
Now, \(A \cdot C = 3 \cdot 12 = 36\). We need to find two numbers that multiply to 36 and add up to 13. The appropriate pairs are 9 and 4, so we can rewrite the equation:
\[
3x^2 + 9x + 4x + 12 = 0
\]
Thus, the correct option is:
**3x² + 9x + 4x + 12 = 0**
### Question 2: Part B
Now, after rewriting the equation as \(3x^2 + 9x + 4x + 12 = 0\), we can group the terms:
\[
(3x^2 + 9x) + (4x + 12) = 0
\]
Factoring each group gives:
\[
3x(x + 3) + 4(x + 3) = 0
\]
Now factor out the common term \((x + 3)\):
\[
(3x + 4)(x + 3) = 0
\]
So, the correct option is:
**(3x + 4)(x + 3) = 0**
### Question 3: Part C
To find the solutions, we set each factor to zero:
1. \(3x + 4 = 0\) implies \(3x = -4\) or \(x = -\frac{4}{3}\)
2. \(x + 3 = 0\) implies \(x = -3\)
The solutions are:
**x = -\frac{4}{3} and x = -3**
So the corresponding option is:
**x = -\frac{4}{3} and x = -3**.
To factor the equation \(3x^2 + 13x + 12 = 0\) using factoring by grouping, we first multiply \(A\) (the coefficient of \(x^2\), which is 3) and \(C\) (the constant term, which is 12).
Now, \(A \cdot C = 3 \cdot 12 = 36\). We need to find two numbers that multiply to 36 and add up to 13. The appropriate pairs are 9 and 4, so we can rewrite the equation:
\[
3x^2 + 9x + 4x + 12 = 0
\]
Thus, the correct option is:
**3x² + 9x + 4x + 12 = 0**
### Question 2: Part B
Now, after rewriting the equation as \(3x^2 + 9x + 4x + 12 = 0\), we can group the terms:
\[
(3x^2 + 9x) + (4x + 12) = 0
\]
Factoring each group gives:
\[
3x(x + 3) + 4(x + 3) = 0
\]
Now factor out the common term \((x + 3)\):
\[
(3x + 4)(x + 3) = 0
\]
So, the correct option is:
**(3x + 4)(x + 3) = 0**
### Question 3: Part C
To find the solutions, we set each factor to zero:
1. \(3x + 4 = 0\) implies \(3x = -4\) or \(x = -\frac{4}{3}\)
2. \(x + 3 = 0\) implies \(x = -3\)
The solutions are:
**x = -\frac{4}{3} and x = -3**
So the corresponding option is:
**x = -\frac{4}{3} and x = -3**.
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