To determine which of the given expressions is a factor of the polynomial \(2x^2 + 9x - 35\), we can use polynomial long division. We will divide the polynomial \(2x^2 + 9x - 35\) by each option and check for a zero remainder, which indicates that the expression is a factor.
Option A: \(x - 5\)
- Divide: \(2x^2 \div x = 2x\)
- Multiply: \(2x(x - 5) = 2x^2 - 10x\)
- Subtract: \((2x^2 + 9x - 35) - (2x^2 - 10x) = 19x - 35\)
- Divide again: \(19x \div x = 19\)
- Multiply: \(19(x - 5) = 19x - 95\)
- Subtract again: \((19x - 35) - (19x - 95) = 60\)
Since the remainder is \(60\), \(x - 5\) is not a factor.
Option B: \(2x + 7\)
- Divide: \(2x^2 \div 2x = x\)
- Multiply: \(x(2x + 7) = 2x^2 + 7x\)
- Subtract: \((2x^2 + 9x - 35) - (2x^2 + 7x) = 2x - 35\)
- Divide again: \(2x \div 2x = 1\)
- Multiply: \(1(2x + 7) = 2x + 7\)
- Subtract again: \((2x - 35) - (2x + 7) = -42\)
Since the remainder is \(-42\), \(2x + 7\) is not a factor.
Option C: \(x - 7\)
- Divide: \(2x^2 \div x = 2x\)
- Multiply: \(2x(x - 7) = 2x^2 - 14x\)
- Subtract: \((2x^2 + 9x - 35) - (2x^2 - 14x) = 23x - 35\)
- Divide again: \(23x \div x = 23\)
- Multiply: \(23(x - 7) = 23x - 161\)
- Subtract again: \((23x - 35) - (23x - 161) = 126\)
Since the remainder is \(126\), \(x - 7\) is not a factor.
Option D: \(2x - 5\)
- Divide: \(2x^2 \div 2x = x\)
- Multiply: \(x(2x - 5) = 2x^2 - 5x\)
- Subtract: \((2x^2 + 9x - 35) - (2x^2 - 5x) = 14x - 35\)
- Divide again: \(14x \div 2x = 7\)
- Multiply: \(7(2x - 5) = 14x - 35\)
- Subtract again: \((14x - 35) - (14x - 35) = 0\)
Since the remainder is \(0\), \(2x - 5\) is a factor of \(2x^2 + 9x - 35\).
Therefore, the expression that is a factor of \(2x^2 + 9x - 35\) is:
D. \(2x - 5\).