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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

x−7
x minus 7

2x−5

1 answer

To determine which of the given expressions is a factor of \( 2x^2 + 9x - 35 \) by using polynomial long division, we will divide the polynomial by each of the options provided.

Let's use long division to divide \( 2x^2 + 9x - 35 \) by each of the expressions:

Attempt 1: \( x + 5 \)

  1. Divide the leading term: \( \frac{2x^2}{x} = 2x \).
  2. Multiply \( 2x \) by \( x + 5 \): \[ 2x(x + 5) = 2x^2 + 10x \]
  3. Subtract from \( 2x^2 + 9x - 35 \): \[ (2x^2 + 9x - 35) - (2x^2 + 10x) = (9x - 10x) - 35 = -x - 35 \]
  4. Bring down the next term (no term to bring down here).
  5. Divide the leading term: \( \frac{-x}{x} = -1 \).
  6. Multiply \(-1\) by \( x + 5 \): \[ -1(x + 5) = -x - 5 \]
  7. Subtract: \[ (-x - 35) - (-x - 5) = -35 + 5 = -30 \]
  8. The remainder is \(-30\), which means \( x + 5 \) is not a factor.

Attempt 2: \( 2x - 5 \)

  1. Divide the leading term: \( \frac{2x^2}{2x} = x \).
  2. Multiply \( x \) by \( 2x - 5 \): \[ x(2x - 5) = 2x^2 - 5x \]
  3. Subtract from \( 2x^2 + 9x - 35 \): \[ (2x^2 + 9x - 35) - (2x^2 - 5x) = (9x + 5x) - 35 = 14x - 35 \]
  4. Divide the leading term: \( \frac{14x}{2x} = 7 \).
  5. Multiply \( 7 \) by \( 2x - 5 \): \[ 7(2x - 5) = 14x - 35 \]
  6. Subtract: \[ (14x - 35) - (14x - 35) = 0 \]
  7. The remainder is 0, so \( 2x - 5 \) is a factor.

Attempt 3: \( x - 7 \)

  1. Divide the leading term: \( \frac{2x^2}{x} = 2x \).
  2. Multiply \( 2x \) by \( x - 7 \): \[ 2x(x - 7) = 2x^2 - 14x \]
  3. Subtract from \( 2x^2 + 9x - 35 \): \[ (2x^2 + 9x - 35) - (2x^2 - 14x) = (9x + 14x) - 35 = 23x - 35 \]
  4. Divide the leading term: \( \frac{23x}{x} = 23 \).
  5. Multiply \( 23 \) by \( x - 7 \): \[ 23(x - 7) = 23x - 161 \]
  6. Subtract: \[ (23x - 35) - (23x - 161) = 161 - 35 = 126 \]
  7. The remainder is \( 126 \), so \( x - 7 \) is not a factor.

Attempt 4: \( 2x + 7 \)

  1. Divide the leading term: \( \frac{2x^2}{2x} = x \).
  2. Multiply \( x \) by \( 2x + 7 \): \[ x(2x + 7) = 2x^2 + 7x \]
  3. Subtract from \( 2x^2 + 9x - 35 \): \[ (2x^2 + 9x - 35) - (2x^2 + 7x) = (9x - 7x) - 35 = 2x - 35 \]
  4. Divide the leading term: \( \frac{2x}{2x} = 1 \).
  5. Multiply \( 1 \) by \( 2x + 7 \): \[ 1(2x + 7) = 2x + 7 \]
  6. Subtract: \[ (2x - 35) - (2x + 7) = -35 - 7 = -42 \]
  7. The remainder is \( -42 \), so \( 2x + 7 \) is not a factor.

Conclusion

The only factor of \( 2x^2 + 9x - 35 \) from the options provided is:

\( 2x - 5 \)