Question

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

2x+7
2 x plus 7

x+5
x plus 5

2x−5
2 x minus 5

x−7

Answer 1

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

**Step 1: Long Division Using \(x + 5\)**

Let's try dividing \(2x^2 + 9x - 35\) 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 from the original polynomial:
\((2x^2 + 9x - 35) - (2x^2 + 10x) = -x - 35\).
4. Now, divide \(-x\) by \(x\): \(-x \div x = -1\).
5. Multiply \(-1\) by \(x + 5\):
\(-1(x + 5) = -x - 5\).
6. Subtract:
\((-x - 35) - (-x - 5) = -30\).

Since we have a non-zero remainder of \(-30\), we can conclude that \(x + 5\) is **not** a factor.

**Step 2: Long Division Using \(2x + 7\)**

Next, let's try dividing \(2x^2 + 9x - 35\) 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 from the original polynomial:
\((2x^2 + 9x - 35) - (2x^2 + 7x) = 2x - 35\).
4. Now divide \(2x\) by \(2x\): \(2x \div 2x = 1\).
5. Multiply \(1\) by \(2x + 7\):
\(1(2x + 7) = 2x + 7\).
6. Subtract:
\((2x - 35) - (2x + 7) = -42\).

Since we have a non-zero remainder of \(-42\), \(2x + 7\) is **not** a factor.

**Step 3: Long Division Using \(2x - 5\)**

Now, let's divide \(2x^2 + 9x - 35\) 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 from the original polynomial:
\((2x^2 + 9x - 35) - (2x^2 - 5x) = 14x - 35\).
4. Now, divide \(14x\) by \(2x\): \(14x \div 2x = 7\).
5. Multiply \(7\) by \(2x - 5\):
\(7(2x - 5) = 14x - 35\).
6. Subtract:
\((14x - 35) - (14x - 35) = 0\).

Since we have a remainder of 0, \(2x - 5\) is indeed a factor of \(2x^2 + 9x - 35\).

**Step 4: Long Division Using \(x - 7\)**

Finally, let's check if \(x - 7\) is a factor:

1. Divide \(2x^2 + 9x - 35\) by \(x - 7\):
2. Divide the first term: \(2x^2 \div x = 2x\).
3. Multiply \(2x\) by \(x - 7\):
\(2x(x - 7) = 2x^2 - 14x\).
4. Subtract from the original polynomial:
\((2x^2 + 9x - 35) - (2x^2 - 14x) = 23x - 35\).
5. Now, divide \(23x\) by \(x\): \(23x \div x = 23\).
6. Multiply \(23\) by \(x - 7\):
\(23(x - 7) = 23x - 161\).
7. Subtract:
\((23x - 35) - (23x - 161) = 126\).

Since we have a non-zero remainder of 126, \(x - 7\) is also **not** a factor.

**Conclusion**

The only expression that is a factor of \(2x^2 + 9x - 35\) is:

- \( \boxed{2x - 5} \)