Which of the following expressions is a factor of 2x2+9x−35 ? Use long division to solve.(1 point)

Question

Responses

2x+7
2 x plus 7

x+5
x plus 5

x−7
x minus 7

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 or synthetic division. However, a more straightforward approach for this problem involves checking the potential factors by directly performing the division.

Step 1: Check \( x + 5 \)

Perform long division of \( 2x^2 + 9x - 35 \) by \( x + 5 \).

1. Divide the first term: \[ 2x^2 \div x = 2x \]

2. Multiply \( 2x \) by \( x + 5 \): \[ 2x(x + 5) = 2x^2 + 10x \]

3. Subtract this from the original polynomial: \[ (2x^2 + 9x - 35) - (2x^2 + 10x) = -x - 35 \]

4. Now, divide the first term of the remainder: \[ -x \div x = -1 \]

5. Multiply \(-1\) by \(x + 5\): \[ -1(x + 5) = -x - 5 \]

6. Subtract: \[ (-x - 35) - (-x - 5) = -30 \]

The result is a remainder of \(-30\), which means \( x + 5 \) is not a factor.

Step 2: Check \( x - 7 \)

Perform long division of \( 2x^2 + 9x - 35 \) by \( x - 7 \).

1. Divide the first term: \[ 2x^2 \div x = 2x \]

2. Multiply \( 2x \) by \( x - 7 \): \[ 2x(x - 7) = 2x^2 - 14x \]

3. Subtract this from the original polynomial: \[ (2x^2 + 9x - 35) - (2x^2 - 14x) = 23x - 35 \]

4. Now, divide the first term of the remainder: \[ 23x \div x = 23 \]

5. Multiply \( 23 \) by \( x - 7 \): \[ 23(x - 7) = 23x - 161 \]

6. Subtract: \[ (23x - 35) - (23x - 161) = 126 \]

Since the remainder is also non-zero, \( x - 7 \) is not a factor.

Step 3: Check \( 2x + 7 \)

Perform long division of \( 2x^2 + 9x - 35 \) by \( 2x + 7 \).

1. Divide the first term: \[ 2x^2 \div 2x = x \]

2. Multiply \( x \) by \( 2x + 7 \): \[ x(2x + 7) = 2x^2 + 7x \]

3. Subtract this from the original polynomial: \[ (2x^2 + 9x - 35) - (2x^2 + 7x) = 2x - 35 \]

4. Now divide the first term of the remainder: \[ 2x \div 2x = 1 \]

5. Multiply \( 1 \) by \( 2x + 7 \): \[ 1(2x + 7) = 2x + 7 \]

6. Subtract: \[ (2x - 35) - (2x + 7) = -42 \]

Again, the remainder is non-zero, so \( 2x + 7 \) is not a factor.

Step 4: Check \( 2x - 5 \)

Lastly, perform long division of \( 2x^2 + 9x - 35 \) by \( 2x - 5 \).

1. Divide the first term: \[ 2x^2 \div 2x = x \]

2. Multiply \( x \) by \( 2x - 5 \): \[ x(2x - 5) = 2x^2 - 5x \]

3. Subtract this from the original polynomial: \[ (2x^2 + 9x - 35) - (2x^2 - 5x) = 14x - 35 \]

4. Now divide the first term of the remainder: \[ 14x \div 2x = 7 \]

5. Multiply \( 7 \) by \( 2x - 5 \): \[ 7(2x - 5) = 14x - 35 \]

6. Subtract: \[ (14x - 35) - (14x - 35) = 0 \]

Since the remainder is \( 0 \), that means \( 2x - 5 \) is a factor of \( 2x^2 + 9x - 35 \).

Final Answer

The correct answer is: \( 2x - 5 \).