{"id":1378,"date":"2015-11-12T18:35:30","date_gmt":"2015-11-12T18:35:30","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1378"},"modified":"2017-03-31T22:52:00","modified_gmt":"2017-03-31T22:52:00","slug":"evaluate-a-polynomial-using-the-remainder-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/evaluate-a-polynomial-using-the-remainder-theorem\/","title":{"raw":"Evaluate a polynomial using the Remainder Theorem","rendered":"Evaluate a polynomial using the Remainder Theorem"},"content":{"raw":"

In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem<\/strong>. If the polynomial is divided by x<\/em> \u2013\u00a0k<\/em>, the remainder may be found quickly by evaluating the polynomial function at k<\/em>, that is, f<\/em>(k<\/em>)\u00a0Let\u2019s walk through the proof of the theorem.<\/p>\r\n

Recall that the Division Algorithm<\/strong> states that, given a polynomial dividend f<\/em>(x<\/em>)\u00a0and a non-zero polynomial divisor d<\/em>(x<\/em>)\u00a0where the degree of\u00a0d<\/em>(x<\/em>) is less than or equal to the degree of f<\/em>(x<\/em>), there exist unique polynomials q<\/em>(x<\/em>)\u00a0and r<\/em>(x<\/em>)\u00a0such that<\/p>\r\n\r\n

[latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/div>\r\n

If the divisor, d<\/em>(x<\/em>), is x<\/em> \u2013\u00a0k<\/em>, this takes the form<\/p>\r\n\r\n

[latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/div>\r\n

Since the divisor x<\/em> \u2013\u00a0k<\/em>\u00a0is linear, the remainder will be a constant, r<\/em>. And, if we evaluate this for x<\/em> =\u00a0k<\/em>, we have<\/p>\r\n\r\n

[latex]\\begin{cases}f\\left(k\\right)=\\left(k-k\\right)q\\left(k\\right)+r\\hfill \\\\ \\text{ }=0\\cdot q\\left(k\\right)+r\\hfill \\\\ \\text{ }=r\\hfill \\end{cases}[\/latex]<\/div>\r\n

In other words, f<\/em>(k<\/em>)\u00a0is the remainder obtained by dividing f<\/em>(x<\/em>)\u00a0by x<\/em> \u2013\u00a0k<\/em>.<\/p>\r\n\r\n

\r\n

A General Note: The Remainder Theorem<\/h3>\r\n

If a polynomial [latex]f\\left(x\\right)[\/latex] is divided by x<\/em> \u2013\u00a0k<\/em>, then the remainder is the value [latex]f\\left(k\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n

\r\n

How To: Given a polynomial function [latex]f[\/latex], evaluate [latex]f\\left(x\\right)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/h3>\r\n
  1. Use synthetic division to divide the polynomial by [latex]x-k[\/latex].<\/li>\r\n\t
  2. The remainder is the value [latex]f\\left(k\\right)[\/latex].<\/li>\r\n<\/ol><\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 1: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\r\n

    Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7[\/latex]\u00a0at [latex]x=2[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{cases}\\\\ 2\\overline{)\\begin{cases}6\\hfill & -1\\hfill & -15\\hfill & 2\\hfill & -7\\hfill \\\\ \\hfill & 12\\hfill & \\text{ }22\\hfill & 14\\hfill & 32\\hfill \\end{cases}}\\\\ \\begin{cases}\\text{ }6\\hfill & 11\\hfill & \\text{ }7\\hfill & \\text{16}\\hfill & 25\\hfill \\end{cases}\\end{cases}[\/latex]<\/div>\r\n

    The remainder is 25. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Analysis of the Solution<\/h3>\r\n

    We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{cases}f\\left(x\\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) & =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ \\hfill & =25\\hfill \\end{cases}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n
    \r\n

    Try It 1<\/h3>\r\n

    Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2[\/latex]\r\nat [latex]x=-3[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    In the last section, we learned how to divide polynomials. We can now use polynomial division to evaluate polynomials using the Remainder Theorem<\/strong>. If the polynomial is divided by x<\/em> \u2013\u00a0k<\/em>, the remainder may be found quickly by evaluating the polynomial function at k<\/em>, that is, f<\/em>(k<\/em>)\u00a0Let\u2019s walk through the proof of the theorem.<\/p>\n

    Recall that the Division Algorithm<\/strong> states that, given a polynomial dividend f<\/em>(x<\/em>)\u00a0and a non-zero polynomial divisor d<\/em>(x<\/em>)\u00a0where the degree of\u00a0d<\/em>(x<\/em>) is less than or equal to the degree of f<\/em>(x<\/em>), there exist unique polynomials q<\/em>(x<\/em>)\u00a0and r<\/em>(x<\/em>)\u00a0such that<\/p>\n

    [latex]f\\left(x\\right)=d\\left(x\\right)q\\left(x\\right)+r\\left(x\\right)[\/latex]<\/div>\n

    If the divisor, d<\/em>(x<\/em>), is x<\/em> \u2013\u00a0k<\/em>, this takes the form<\/p>\n

    [latex]f\\left(x\\right)=\\left(x-k\\right)q\\left(x\\right)+r[\/latex]<\/div>\n

    Since the divisor x<\/em> \u2013\u00a0k<\/em>\u00a0is linear, the remainder will be a constant, r<\/em>. And, if we evaluate this for x<\/em> =\u00a0k<\/em>, we have<\/p>\n

    [latex]\\begin{cases}f\\left(k\\right)=\\left(k-k\\right)q\\left(k\\right)+r\\hfill \\\\ \\text{ }=0\\cdot q\\left(k\\right)+r\\hfill \\\\ \\text{ }=r\\hfill \\end{cases}[\/latex]<\/div>\n

    In other words, f<\/em>(k<\/em>)\u00a0is the remainder obtained by dividing f<\/em>(x<\/em>)\u00a0by x<\/em> \u2013\u00a0k<\/em>.<\/p>\n

    \n

    A General Note: The Remainder Theorem<\/h3>\n

    If a polynomial [latex]f\\left(x\\right)[\/latex] is divided by x<\/em> \u2013\u00a0k<\/em>, then the remainder is the value [latex]f\\left(k\\right)[\/latex].<\/p>\n<\/div>\n

    \n

    How To: Given a polynomial function [latex]f[\/latex], evaluate [latex]f\\left(x\\right)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/h3>\n
      \n
    1. Use synthetic division to divide the polynomial by [latex]x-k[\/latex].<\/li>\n
    2. The remainder is the value [latex]f\\left(k\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 1: Using the Remainder Theorem to Evaluate a Polynomial<\/h3>\n

      Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7[\/latex]\u00a0at [latex]x=2[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      To find the remainder using the Remainder Theorem, use synthetic division to divide the polynomial by [latex]x - 2[\/latex].<\/p>\n

      [latex]\\begin{cases}\\\\ 2\\overline{)\\begin{cases}6\\hfill & -1\\hfill & -15\\hfill & 2\\hfill & -7\\hfill \\\\ \\hfill & 12\\hfill & \\text{ }22\\hfill & 14\\hfill & 32\\hfill \\end{cases}}\\\\ \\begin{cases}\\text{ }6\\hfill & 11\\hfill & \\text{ }7\\hfill & \\text{16}\\hfill & 25\\hfill \\end{cases}\\end{cases}[\/latex]<\/div>\n

      The remainder is 25. Therefore, [latex]f\\left(2\\right)=25[\/latex].<\/p>\n<\/div>\n

      \n

      Analysis of the Solution<\/h3>\n

      We can check our answer by evaluating [latex]f\\left(2\\right)[\/latex].<\/p>\n

      [latex]\\begin{cases}f\\left(x\\right) & =6{x}^{4}-{x}^{3}-15{x}^{2}+2x - 7 \\\\ f\\left(2\\right) & =6{\\left(2\\right)}^{4}-{\\left(2\\right)}^{3}-15{\\left(2\\right)}^{2}+2\\left(2\\right)-7 \\\\ \\hfill & =25\\hfill \\end{cases}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n
      \n

      Try It 1<\/h3>\n

      Use the Remainder Theorem to evaluate [latex]f\\left(x\\right)=2{x}^{5}-3{x}^{4}-9{x}^{3}+8{x}^{2}+2[\/latex]
      \nat [latex]x=-3[\/latex].<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

      \n\t\t\t

      Candela Citations<\/h3>\n\t\t\t\t\t
      \n\t\t\t\t\t\t
      \n\t\t\t\t\t\t\t
      CC licensed content, Shared previously<\/div>