Use long division to divide 8x3−20x2+18x−11 by 2x2−3 .(1 point)

1 answer

To divide \( 8x^3 - 20x^2 + 18x - 11 \) by \( 2x^2 - 3 \) using long division, follow these steps:

  1. Set up the long division:

    \[ \begin{array}{r|l} 2x^2 - 3 & 8x^3 - 20x^2 + 18x - 11 \ \end{array} \]

  2. Divide the leading term of the dividend \( 8x^3 \) by the leading term of the divisor \( 2x^2 \) to get \( 4x \) (since \( 8x^3 \div 2x^2 = 4x \)).

  3. Multiply \( 4x \) by the entire divisor \( 2x^2 - 3 \):

    \[ 4x(2x^2 - 3) = 8x^3 - 12x \]

  4. Subtract this result from the original polynomial:

    \[ (8x^3 - 20x^2 + 18x - 11) - (8x^3 - 12x) = -20x^2 + 18x + 12x - 11 = -20x^2 + 30x - 11 \]

  5. Now, repeat the process. Divide the leading term of the new polynomial \( -20x^2 \) by the leading term of the divisor \( 2x^2 \):

    \[ -20x^2 \div 2x^2 = -10 \]

  6. Multiply \( -10 \) by the divisor \( 2x^2 - 3 \):

    \[ -10(2x^2 - 3) = -20x^2 + 30 \]

  7. Subtract this from \( -20x^2 + 30x - 11 \):

    \[ (-20x^2 + 30x - 11) - (-20x^2 + 30) = 30x - 11 - 30 = 30x - 41 \]

  8. At this point, we have:

    \[ \begin{array}{r|l} 2x^2 - 3 & 8x^3 - 20x^2 + 18x - 11 \ & \underline{4x} \ & -20x^2 + 30x - 11 \ & \underline{-10} \ & 30x - 41 \ \end{array} \]

  9. Now we can't divide anymore because the degree of the remainder \( 30x - 41 \) is less than the degree of the divisor \( 2x^2 - 3 \).

So, the final result of the division is:

\[ \text{Quotient: } 4x - 10 \] \[ \text{Remainder: } 30x - 41 \]

Therefore, the overall result can be written as:

\[ 8x^3 - 20x^2 + 18x - 11 = (2x^2 - 3)(4x - 10) + (30x - 41) \]

In conclusion, the result of dividing \( 8x^3 - 20x^2 + 18x - 11 \) by \( 2x^2 - 3 \) is:

\[ \text{Quotient: } 4x - 10, \quad \text{Remainder: } 30x - 41 \]