Question

Factor. 2x2−3x−5 (2x+5)(x−1) (x−5)(2x+1) (2x−5)(x−1) (2x−5)(x+1)

Answer 1

`(2x -5)(x + 1)`

Explanation:

Hello!

Standard form of a quadratic:
`ax^2 + bx +c = 0`

Given our equation:
`2x^2 - 3x - 5`

• a = 2
• b = -3
• c = -5

We want to find two numbers that multiply to
`a*c` but add up to
`b`.

• ac = 2 * -5 = -10
• b = -3

The two numbers that work for this is -5 and 2.

Now, expand -3x to -5x + 2x, and factor by grouping.

Factor by Grouping

• 
  `2x^2 - 3x - 5`
• 
  `2x^2 +2x - 5x - 5`
• 
  `2x(x + 1) - 5(x + 1)`
• 
  `(2x -5)(x + 1)`

The factored form is
`(2x -5)(x + 1)`.

Answer 2

-------------------------------------------------------------------------------------------------------------

Answer:
`\textsf{Option D, (2x - 5)(x + 1)}`

-------------------------------------------------------------------------------------------------------------

Given:
`\textsf{2x}^2\textsf{ - 3x - 5}`

Find:
`\textsf{Factor the expression}`

Solution: In order to factor the expression we will have two parenthesis with an x variable and a constant.

Determine important information

• The standard form for a quadratic equation is
  `ax^2 + bx + c` and looking at our equation we can see that a = 2, b = -3, and c = -5.

• Now we need to determine two numbers that add up to b and multiply up to a * c. So, they must add up to -3 and multiply up to -10. The two numbers that fit the description is -5 and 2.

Determine the factors

We plug in -5x and 2x instead of -3x and factor the expression by grouping.

• 
  `2x^2-3x-5`

• 
  `2x^2+2x-5x-5`

• 
  `2x(x+1)-5(x+1)`

• 
  `(2x - 5)(x + 1)`

Therefore, the final answer that fits our solve solution would be option D, (2x - 5)(x + 1).