Fill in the blank to show what the quadratic expression would look like when factored. (5 points)

Question

x2−3x−10
= (x−
Response area)(x+
Response area)

3x2−6x+9
= Response area(x2−2x+3)

9x2−25
= (3x−
Response area)(Response area+5
)

Answer 1

Sure! Here are the factored forms for each of the quadratic expressions provided:

1. For \(x^2 - 3x - 10\): \[ x^2 - 3x - 10 = (x - 5)(x + 2) \] So, the answer would be:
   \(x - 5\) and \(x + 2\).

2. For \(3x^2 - 6x + 9\): \[ 3x^2 - 6x + 9 = 3(x^2 - 2x + 3) \] So, the answer would be:
   \(3\) (as a factor).

3. For \(9x^2 - 25\): \[ 9x^2 - 25 = (3x - 5)(3x + 5) \] So, the answer would be:
   \(3x - 5\) and \(3x + 5\).

So, filling in the blanks:

1. \(x - 5\), \(x + 2\)
2. \(3\)
3. \(3x - 5\), \(3x + 5\)

Answer 2

is this correct?

Answer 3

Yes, my previous response outlines the correct factorizations. Here’s a quick verification for each expression:

1. For \(x^2 - 3x - 10\):

   • We want two numbers that multiply to \(-10\) and add to \(-3\), which are \(-5\) and \(2\).
   • Thus, the correct factorization is: \[ x^2 - 3x - 10 = (x - 5)(x + 2) \]

2. For \(3x^2 - 6x + 9\):

   • You can factor out the greatest common factor of \(3\): \[ 3x^2 - 6x + 9 = 3(x^2 - 2x + 3) \]
   • This is already simplified further (the quadratic \(x^2 - 2x + 3\) does not factor into real numbers, as its discriminant is negative).

3. For \(9x^2 - 25\):

   • This is a difference of squares: \[ 9x^2 - 25 = (3x)^2 - (5)^2 = (3x - 5)(3x + 5) \]

So to reiterate:

1. \(x - 5\) and \(x + 2\) for \(x^2 - 3x - 10\).
2. \(3\) for \(3x^2 - 6x + 9\).
3. \(3x - 5\) and \(3x + 5\) for \(9x^2 - 25\).

Your understanding based on that verification is indeed correct! If you have further questions or need clarification, feel free to ask.