(2x^3-3x^2+5x-1)/(x+2)

3 min read Jun 16, 2024
(2x^3-3x^2+5x-1)/(x+2)

Dividing Polynomials: (2x^3 - 3x^2 + 5x - 1) / (x + 2)

This article will guide you through the process of dividing the polynomial 2x^3 - 3x^2 + 5x - 1 by the binomial x + 2. We'll use polynomial long division to achieve this.

Understanding Polynomial Long Division

Polynomial long division is very similar to regular long division with numbers. Here's a breakdown:

  1. Set up the division:

    • Write the dividend (the polynomial being divided) inside the division symbol.
    • Write the divisor (the polynomial dividing) outside the division symbol.
  2. Focus on the leading terms:

    • Divide the leading term of the dividend by the leading term of the divisor.
    • Write the result above the dividend.
  3. Multiply and subtract:

    • Multiply the result from step 2 by the entire divisor.
    • Subtract the result from the dividend.
  4. Bring down the next term:

    • Bring down the next term from the dividend.
  5. Repeat steps 2-4:

    • Repeat the process until there are no more terms in the dividend.

Performing the Division

Let's apply these steps to our problem:

        2x^2 - 7x + 19 
    x + 2 | 2x^3 - 3x^2 + 5x - 1 
           -(2x^3 + 4x^2)
           ------------------
                 -7x^2 + 5x 
                 -(-7x^2 - 14x)
                 ------------------
                         19x - 1
                         -(19x + 38)
                         ---------------
                                -39

Interpreting the Result

The result of the division is:

  • Quotient: 2x^2 - 7x + 19
  • Remainder: -39

This can be written as:

(2x^3 - 3x^2 + 5x - 1) / (x + 2) = 2x^2 - 7x + 19 - 39/(x + 2)

In simpler terms, this means that the polynomial 2x^3 - 3x^2 + 5x - 1 can be expressed as (x + 2) multiplied by 2x^2 - 7x + 19, with a remainder of -39.

Conclusion

Polynomial long division is a powerful tool for dividing polynomials. By following these steps, you can successfully divide any polynomial by another polynomial.

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