Which of the following expressions is a factor of 2x^2 + 9x -35 ? Use long division

Question

to solve.
2x + 7 2𝑥 + 7
2x − 5 2𝑥 − 5
x + 5 𝑥 + 5
x − 7 𝑥 − 7

Answer 1

To determine which of the expressions is a factor of \(2x^2 + 9x - 35\), we can use polynomial long division for each of the provided options.

Step 1: Set up the Long Division

Let's start with the first option: \(2x + 7\).

Long Division of \(2x^2 + 9x - 35\) by \(2x + 7\):

1. Divide the leading term: \( \frac{2x^2}{2x} = x \)
2. Multiply: \( x \cdot (2x + 7) = 2x^2 + 7x \)
3. Subtract: \[ (2x^2 + 9x - 35) - (2x^2 + 7x) = 2x - 35 \]
4. Repeat:
   • Divide the leading term: \( \frac{2x}{2x} = 1 \)
   • Multiply: \( 1 \cdot (2x + 7) = 2x + 7 \)
   • Subtract: \[ (2x - 35) - (2x + 7) = -42 \]

The remainder is \(-42\), so \(2x + 7\) is not a factor.

Step 2: Check the next option: \(2x - 5\)

Long Division of \(2x^2 + 9x - 35\) by \(2x - 5\):

1. Divide the leading term: \( \frac{2x^2}{2x} = x \)
2. Multiply: \( x \cdot (2x - 5) = 2x^2 - 5x \)
3. Subtract: \[ (2x^2 + 9x - 35) - (2x^2 - 5x) = 14x - 35 \]
4. Repeat:
   • Divide the leading term: \( \frac{14x}{2x} = 7 \)
   • Multiply: \( 7 \cdot (2x - 5) = 14x - 35 \)
   • Subtract: \[ (14x - 35) - (14x - 35) = 0 \]

The remainder is \(0\), so \(2x - 5\) is a factor.

Step 3: Verify the other options: \(x + 5\) and \(x - 7\)

Long Division of \(2x^2 + 9x - 35\) by \(x + 5\):

1. Divide the leading term: \( \frac{2x^2}{x} = 2x \)
2. Multiply: \( 2x \cdot (x + 5) = 2x^2 + 10x \)
3. Subtract: \[ (2x^2 + 9x - 35) - (2x^2 + 10x) = -x - 35 \]
4. Repeat:
   • Divide the leading term: \( \frac{-x}{x} = -1 \)
   • Multiply: \(-1 \cdot (x + 5) = -x - 5\)
   • Subtract: \[ (-x - 35) - (-x - 5) = -30 \]

The remainder is \(-30\), so \(x + 5\) is not a factor.

Long Division of \(2x^2 + 9x - 35\) by \(x - 7\):

1. Divide the leading term: \( \frac{2x^2}{x} = 2x \)
2. Multiply: \( 2x \cdot (x - 7) = 2x^2 - 14x \)
3. Subtract: \[ (2x^2 + 9x - 35) - (2x^2 - 14x) = 23x - 35 \]
4. Repeat:
   • Divide the leading term: \( \frac{23x}{x} = 23 \)
   • Multiply: \( 23 \cdot (x - 7) = 23x - 161 \)
   • Subtract: \[ (23x - 35) - (23x - 161) = 126 \]

The remainder is \(126\), so \(x - 7\) is not a factor.

Conclusion:

The only expression that is a factor of \(2x^2 + 9x - 35\) is \[ \boxed{2x - 5} \]