{"id":949,"date":"2025-07-17T18:04:49","date_gmt":"2025-07-17T18:04:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=949"},"modified":"2026-03-10T16:34:21","modified_gmt":"2026-03-10T16:34:21","slug":"dividing-polynomials-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/dividing-polynomials-learn-it-2\/","title":{"raw":"Dividing Polynomials: Learn It 2","rendered":"Dividing Polynomials: Learn It 2"},"content":{"raw":"

Synthetic Division<\/h2>\r\nAs we've seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division<\/strong> is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is [latex]1[\/latex].\r\n\r\n
\r\n
\r\n

synthetic division<\/h3>\r\nSynthetic division<\/strong> is a shortcut that can be used when the divisor is a binomial in the form [latex]x\u2212k[\/latex] where [latex]k[\/latex] is a real number. In synthetic division, only the coefficients are used in the division process.\r\n\r\n<\/div>\r\n<\/section>To illustrate the process, let's look at an example.\r\n\r\n
Divide [latex]2x^3 - 3x^2 + 4x + 5[\/latex] by [latex]x + 2[\/latex] using the long division algorithm.The final form of the process looked like this:
\"Synthetic<\/center>There is a lot of repetition in the table. If we don't write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.
\"Synthetic<\/center>Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by [latex]2[\/latex], as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the \u201cdivisor\u201d to [latex]\u20132[\/latex], multiply and add. The process starts by bringing down the leading coefficient.
\"Synthetic<\/center>We then multiply it by the \u201cdivisor\u201d and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is [latex]2x^2 - 7x + 18[\/latex] and the remainder is [latex]-31[\/latex].\r\n\r\n<\/section>
How to: Given two polynomials, use synthetic division to divide.<\/strong>\r\n
    \r\n \t
  1. Write [latex]k[\/latex]\u00a0for the divisor.<\/li>\r\n \t
  2. Write the coefficients of the dividend.<\/li>\r\n \t
  3. Bring the lead coefficient down.<\/li>\r\n \t
  4. Multiply the lead coefficient by [latex]k[\/latex]. \u00a0Write the product in the next column.<\/li>\r\n \t
  5. Add the terms of the second column.<\/li>\r\n \t
  6. Multiply the result by [latex]k[\/latex].\u00a0Write the product in the next column.<\/li>\r\n \t
  7. Repeat steps 5 and 6 for the remaining columns.<\/li>\r\n \t
  8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree [latex]0[\/latex], the next number from the right has degree [latex]1[\/latex], the next number from the right has degree [latex]2[\/latex], and so on.<\/li>\r\n<\/ol>\r\n<\/section>Try an example on your own.\r\n\r\n
    Use synthetic division to divide [latex]4x^3 + 10x^2 - 6x - 20[\/latex] by [latex]x + 2[\/latex]. [reveal-answer q=\"981343\"]Show Solution[\/reveal-answer] [hidden-answer a=\"981343\"]The binomial divisor is [latex]x + 2[\/latex] so [latex]k = -2[\/latex]. Add each column, multiply the result by [latex]-2[\/latex], and repeat until the last column is reached.
    \"Synthetic<\/center>The result is [latex]4x^2 + 2x - 10[\/latex]. The remainder is [latex]0[\/latex]. Thus, [latex]x + 2[\/latex] is a factor of [latex]4x^3 + 10x^2 - 6x - 20[\/latex].Analysis<\/strong>[latex]\\\\[\/latex]The graph of the polynomial function [latex]f(x) = 4x^3 + 10x^2 - 6x - 20[\/latex] in the figure below shows a zero at [latex]x = k = -2[\/latex]. This confirms that [latex]x + 2[\/latex] is a factor of [latex]4x^3 + 10x^2 - 6x - 20[\/latex].
    \"Synthetic<\/center>[\/hidden-answer]<\/section>
    [ohm_question hide_question_numbers=1]318858[\/ohm_question]<\/section>
    [ohm_question hide_question_numbers=1]318860[\/ohm_question]<\/section>
    \r\n

    Use synthetic division to divide [latex]-9{x}^{4}+10{x}^{3}+7{x}^{2}-6[\/latex]\u00a0by [latex]x - 1[\/latex].<\/p>\r\n[reveal-answer q=\"565402\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"565402\"]\r\n

    Notice there is no x<\/em>-term. We will use a zero as the coefficient for that term.\r\n\"Synthetic<\/span><\/p>\r\n

    The result is [latex]-9{x}^{3}+{x}^{2}+8x+8+\\frac{2}{x - 1}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]318861[\/ohm_question]<\/section>","rendered":"

    Synthetic Division<\/h2>\n

    As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division<\/strong> is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is [latex]1[\/latex].<\/p>\n

    \n
    \n

    synthetic division<\/h3>\n

    Synthetic division<\/strong> is a shortcut that can be used when the divisor is a binomial in the form [latex]x\u2212k[\/latex] where [latex]k[\/latex] is a real number. In synthetic division, only the coefficients are used in the division process.<\/p>\n<\/div>\n<\/section>\n

    To illustrate the process, let’s look at an example.<\/p>\n

    Divide [latex]2x^3 - 3x^2 + 4x + 5[\/latex] by [latex]x + 2[\/latex] using the long division algorithm.The final form of the process looked like this:<\/p>\n
    \"Synthetic<\/div>\n

    There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.<\/p>\n

    \"Synthetic<\/div>\n

    Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by [latex]2[\/latex], as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the \u201cdivisor\u201d to [latex]\u20132[\/latex], multiply and add. The process starts by bringing down the leading coefficient.<\/p>\n

    \"Synthetic<\/div>\n

    We then multiply it by the \u201cdivisor\u201d and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is [latex]2x^2 - 7x + 18[\/latex] and the remainder is [latex]-31[\/latex].<\/p>\n<\/section>\n

    How to: Given two polynomials, use synthetic division to divide.<\/strong><\/p>\n
      \n
    1. Write [latex]k[\/latex]\u00a0for the divisor.<\/li>\n
    2. Write the coefficients of the dividend.<\/li>\n
    3. Bring the lead coefficient down.<\/li>\n
    4. Multiply the lead coefficient by [latex]k[\/latex]. \u00a0Write the product in the next column.<\/li>\n
    5. Add the terms of the second column.<\/li>\n
    6. Multiply the result by [latex]k[\/latex].\u00a0Write the product in the next column.<\/li>\n
    7. Repeat steps 5 and 6 for the remaining columns.<\/li>\n
    8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree [latex]0[\/latex], the next number from the right has degree [latex]1[\/latex], the next number from the right has degree [latex]2[\/latex], and so on.<\/li>\n<\/ol>\n<\/section>\n

      Try an example on your own.<\/p>\n

      Use synthetic division to divide [latex]4x^3 + 10x^2 - 6x - 20[\/latex] by [latex]x + 2[\/latex]. <\/p>\n