Question

Which of the following expressions is a factor of 2x^2+9x-35 use long division to solve
A.x-5
B.2x+7
C.x-7
D.2x-5

Answer 1

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: \(x - 5\)

1. **Divide**: \(2x^2 \div x = 2x\)
2. **Multiply**: \(2x(x - 5) = 2x^2 - 10x\)
3. **Subtract**: \((2x^2 + 9x - 35) - (2x^2 - 10x) = 19x - 35\)
4. **Divide again**: \(19x \div x = 19\)
5. **Multiply**: \(19(x - 5) = 19x - 95\)
6. **Subtract again**: \((19x - 35) - (19x - 95) = 60\)

Since the remainder is \(60\), \(x - 5\) is **not** a factor.

### Option B: \(2x + 7\)

1. **Divide**: \(2x^2 \div 2x = x\)
2. **Multiply**: \(x(2x + 7) = 2x^2 + 7x\)
3. **Subtract**: \((2x^2 + 9x - 35) - (2x^2 + 7x) = 2x - 35\)
4. **Divide again**: \(2x \div 2x = 1\)
5. **Multiply**: \(1(2x + 7) = 2x + 7\)
6. **Subtract again**: \((2x - 35) - (2x + 7) = -42\)

Since the remainder is \(-42\), \(2x + 7\) is **not** a factor.

### Option C: \(x - 7\)

1. **Divide**: \(2x^2 \div x = 2x\)
2. **Multiply**: \(2x(x - 7) = 2x^2 - 14x\)
3. **Subtract**: \((2x^2 + 9x - 35) - (2x^2 - 14x) = 23x - 35\)
4. **Divide again**: \(23x \div x = 23\)
5. **Multiply**: \(23(x - 7) = 23x - 161\)
6. **Subtract again**: \((23x - 35) - (23x - 161) = 126\)

Since the remainder is \(126\), \(x - 7\) is **not** a factor.

### Option D: \(2x - 5\)

1. **Divide**: \(2x^2 \div 2x = x\)
2. **Multiply**: \(x(2x - 5) = 2x^2 - 5x\)
3. **Subtract**: \((2x^2 + 9x - 35) - (2x^2 - 5x) = 14x - 35\)
4. **Divide again**: \(14x \div 2x = 7\)
5. **Multiply**: \(7(2x - 5) = 14x - 35\)
6. **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. \(2x - 5\)**.