Question

Which of the following expressions is a factor of the following polynomial? Use long division to solve.

2x2+9x−35



(1 point)
Responses

x−7

x minus 7

2x−5
2 x minus 5

2x+7
2 x plus 7

x+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.

Let's start by dividing \( 2x^2 + 9x - 35 \) by each candidate factor to see if the result is a polynomial with no remainder.

### Step 1: Long Division by \( x - 7 \)

1. Divide the leading term of the polynomial \( 2x^2 \) by the leading term of \( x - 7 \), which gives \( 2x \).
2. Multiply \( 2x \) by \( x - 7 \) to get \( 2x^2 - 14x \).
3. Subtract this from the original polynomial:

\[
(2x^2 + 9x - 35) - (2x^2 - 14x) = 9x + 14x - 35 = 23x - 35
\]

4. Now divide the leading term \( 23x \) by \( x \) to get \( 23 \).
5. Multiply \( 23 \) by \( x - 7 \) to get \( 23x - 161 \).
6. Subtract \( 23x - 161 \) from \( 23x - 35 \):

\[
(23x - 35) - (23x - 161) = -35 + 161 = 126
\]

So, the result of dividing \( 2x^2 + 9x - 35 \) by \( x - 7 \) is \( 2x + 23 \) with a remainder of \( 126 \). Since there is a remainder, \( x - 7 \) is not a factor.

### Step 2: Long Division by \( 2x - 5 \)

1. Divide the leading term \( 2x^2 \) by \( 2x \) to get \( x \).
2. Multiply \( x \) by \( 2x - 5 \) to get \( 2x^2 - 5x \).
3. Subtract from the original polynomial:

\[
(2x^2 + 9x - 35) - (2x^2 - 5x) = 9x + 5x - 35 = 14x - 35
\]

4. Divide \( 14x \) by \( 2x \) to get \( 7 \).
5. Multiply \( 7 \) by \( 2x - 5 \) to get \( 14x - 35 \).
6. Subtract:

\[
(14x - 35) - (14x - 35) = 0
\]

The polynomial \( 2x^2 + 9x - 35 \) divided by \( 2x - 5 \) gives \( x + 7 \) with no remainder. Therefore, \( 2x - 5 \) is a factor.

### Step 3: Long Division by \( 2x + 7 \)

It is not necessary to test this, but for completion:

1. Divide \( 2x^2 \) by \( 2x \) to get \( x \).
2. Multiply \( x \) by \( 2x + 7 \) to get \( 2x^2 + 7x \).
3. Subtract:

\[
(2x^2 + 9x - 35) - (2x^2 + 7x) = 9x - 7x - 35 = 2x - 35
\]

4. Divide \( 2x \) by \( 2x \) to get \( 1 \).
5. Subtract \( 2x + 7 \):

\[
(2x - 35) - (2x + 7) = -35 - 7 = -42 \neq 0
\]

Thus \( 2x + 7 \) is not a factor.

### Step 4: Long Division by \( x + 5 \)

1. Divide \( 2x^2 \) by \( x \) to get \( 2x \).
2. Multiply \( 2x \) by \( x + 5 \) to get \( 2x^2 + 10x \).
3. Subtract:

\[
(2x^2 + 9x - 35) - (2x^2 + 10x) = 9x - 10x - 35 = -x - 35
\]

4. Divide \( -x \) by \( x \) to get \( -1 \).
5. Subtract:

\[
(-x - 35) - (-x - 5) = -35 + 5 = -30 \neq 0
\]

So, \( x + 5 \) is also not a factor.

### Conclusion

The only polynomial to successfully divide \( 2x^2 + 9x - 35 \) without a remainder is:

\[
\text{Factor: }\ 2x - 5
\]

Thus, the expression **\( 2x - 5 \)** is a factor of the polynomial \( 2x^2 + 9x - 35 \).