Part A:

Question

The quadratic equation 15x2+13x+2=0
can be factored into which product?

(1 point)
Responses

(3x+2)(5x+1)=0
open paren 3 x plus 2 close paren times open paren 5 x plus 1 close paren is equal to 0

(3x−2)(5x−1)=0
open paren 3 x minus 2 close paren times open paren 5 x minus 1 close paren is equal to 0

(5x+2)(3x+1)=0
open paren 5 x plus 2 close paren times open paren 3 x plus 1 close paren is equal to 0

(3x−2)(5x+1)=0
open paren 3 x minus 2 close paren times open paren 5 x plus 1 close paren is equal to 0
Question 2
Part B:

The solutions to the quadratic equation 15x2+13x+2=0
are

(1 point)
Responses

x=23 and x=15
x=23 and x=15

x=−23 and x=15
x=−23 and x=15

x=−23 and x=−15
x=−23 and x=−15

x=23 and x=−15

Answer 1

To solve Part A, we need to factor the quadratic equation \( 15x^2 + 13x + 2 = 0 \).

We can factor this by guessing and checking, or by using the method of finding two numbers that multiply to \( ac \) (where \( a = 15 \) and \( c = 2 \), so \( ac = 30 \)) and add to \( b \) (where \( b = 13 \)). The two numbers that satisfy these conditions are \( 3 \) and \( 10 \).

So we can rewrite the quadratic as:

\[ 15x^2 + 3x + 10x + 2 \]

Grouping:

\[ (15x^2 + 3x) + (10x + 2) = 3x(5x + 1) + 2(5x + 1) \]

Factoring gives us:

\[ (3x + 2)(5x + 1) = 0 \]

Thus, the correct factorization is:

\((3x + 2)(5x + 1) = 0\)

Answer for Part A: (3x + 2)(5x + 1) = 0

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For Part B, we need to find the solutions for \( 15x^2 + 13x + 2 = 0 \).

Using the factored form we found:

\[ (3x + 2)(5x + 1) = 0 \]

we can set each factor equal to zero:

1. \( 3x + 2 = 0 \) → \( 3x = -2 \) → \( x = -\frac{2}{3} \)

2. \( 5x + 1 = 0 \) → \( 5x = -1 \) → \( x = -\frac{1}{5} \)

None of the option pairs given (like \( x = 23 \) or \( x = -23 \)) seem correct since we are looking for fractional results. However, if we instead use the quadratic formula for clarity:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Plugging in \( a = 15 \), \( b = 13 \), and \( c = 2 \):

\[ x = \frac{-13 \pm \sqrt{13^2 - 4 \cdot 15 \cdot 2}}{2 \cdot 15} \]

\[ = \frac{-13 \pm \sqrt{169 - 120}}{30} \]

\[ = \frac{-13 \pm \sqrt{49}}{30} \]

\[ = \frac{-13 \pm 7}{30} \]

Calculating the two potential solutions:

1. \( x = \frac{-6}{30} = -\frac{1}{5} \)
2. \( x = \frac{-20}{30} = -\frac{2}{3} \)

Given those results, the solutions are:

• \( x = -\frac{2}{3} \) and \( x = -\frac{1}{5} \)

None of the provided options match directly, suggesting there may be an issue with the choices presented.

However, based on the correct factorization and computation, the solutions are correctly identified as:

• \( x = -\frac{2}{3} \) and \( x = -\frac{1}{5} \)

If I were to select among the options despite their inaccuracy, I would be unable to select something correctly matching without revising the provided answer choices.

In conclusion:

Answer for Part B (based on calculations): x = -\frac{2}{3} and x = -\frac{1}{5}