{"id":9456,"date":"2023-10-20T19:48:47","date_gmt":"2023-10-20T19:48:47","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=9456"},"modified":"2023-12-18T13:22:05","modified_gmt":"2023-12-18T13:22:05","slug":"polynomial-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/polynomial-functions-learn-it-3\/","title":{"raw":"Polynomial Functions: Learn It 3","rendered":"Polynomial Functions: Learn It 3"},"content":{"raw":"

Evaluate Polynomials<\/h2>\r\n

Evaluating a Polynomial Using the Remainder Theorem<\/h3>\r\n

Previously, 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 [latex]x - k[\/latex], the remainder may be found quickly by evaluating the polynomial function at [latex]k[\/latex], that is, [latex]f(k)[\/latex].<\/p>\r\n

\r\n
\r\n

the remainder theorem<\/h3>\r\n

If a polynomial [latex]f(x)[\/latex] is divided by [latex]x\u2212k[\/latex], then the remainder is the value [latex]f(k)[\/latex].<\/p>\r\n<\/div>\r\n<\/section>\r\n

Let's walk through the proof of the theorem.<\/p>\r\n

\r\n

Recall that the Division Algorithm states that, given a polynomial dividend [latex]f(x)[\/latex] and a non-zero polynomial divisor [latex]d(x)[\/latex], there exist unique polynomials [latex]g(x)[\/latex] and [latex]r(x)[\/latex] such that<\/p>\r\n

[latex]f(x) = d(x)g(x) + r(x)[\/latex]<\/p>\r\n

and either [latex]r(x) = 0[\/latex] or the degree of [latex]r(x)[\/latex] is less than the degree of [latex]d(x)[\/latex]. In practice divisors, [latex]d(x)[\/latex] will have degrees less than or equal to the degree of [latex]f(x)[\/latex]. If the divisor, [latex]d(x)[\/latex], is [latex]x - k[\/latex], this takes the form<\/p>\r\n

[latex]f(x) = (x - k)g(x) + r[\/latex]<\/p>\r\n

Since the divisor [latex]x - k[\/latex] is linear, the remainder will be a constant, [latex]r[\/latex]. And, if we evaluate this for [latex]x = k[\/latex], we have<\/p>\r\n

[latex]\\begin{array}{rl} f(k) & = (k - k)g(k) + r \\\\ & = 0 \\cdot g(k) + r \\\\ & = r \\end{array}[\/latex]<\/p>\r\n

In other words, [latex]f(k)[\/latex] is the remainder obtained by dividing [latex]f(x)[\/latex] by [latex]x - k[\/latex].<\/p>\r\n<\/section>\r\n

How to: Given a polynomial function [latex]f[\/latex], evaluate [latex]f(x)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/strong>\r\n
    \r\n\t
  1. Use synthetic division to divide the polynomial by [latex]x\u2212k[\/latex].<\/li>\r\n\t
  2. The remainder is the value [latex]f(k)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>\r\n
    \r\n

    Use the remainder theorem to evaluate [latex]f(x) = 6x^4 - x^3 - 15x^2 + 2x - 7[\/latex] at [latex]x = 2[\/latex].<\/p>\r\n\r\n\r\n[reveal-answer q=\"981343\"]Show Solution[\/reveal-answer] [hidden-answer a=\"981343\"]\r\n\r\n\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

    \"The<\/center>\r\n

     <\/p>\r\n

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

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

    [latex]\\begin{array}{l} f(x) = 6x^4 - x^3 - 15x^2 + 2x - 7 \\\\ f(2) = 6(2)^4 - (2)^3 - 15(2)^2 + 2(2) - 7 \\\\ = 25 \\end{array}[\/latex]<\/center>[\/hidden-answer]<\/section>\r\n
    \r\n

    [ohm2_question height=\"400\" hide_question_numbers=1]13806[\/ohm2_question]<\/p>\r\n<\/section>\r\n

    Using the Factor Theorem to Solve a Polynomial Equation<\/h3>\r\n

    The Factor Theorem<\/strong> is another theorem that helps us analyze polynomial equations. It tells us how the zeros of a polynomial are related to the factors.<\/p>\r\n

    \r\n
    \r\n

    the factor theorem<\/h3>\r\n

    According to the factor theorem,<\/strong> [latex]k[\/latex] is a zero of [latex]f(x)[\/latex] if and only if [latex](x\u2212k)[\/latex] is a factor of [latex]f(x)[\/latex].<\/p>\r\n<\/div>\r\n<\/section>\r\n

    Let's walk through the proof of the theorem.<\/p>\r\n

    \r\n

    Recall that the Division Algorithm.<\/p>\r\n

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

    If [latex]k[\/latex] is a zero, then the remainder [latex]r[\/latex] is [latex]f(k) = 0[\/latex] and [latex]f(x) = (x - k)q(x) + 0[\/latex] or [latex]f(x) = (x - k)q(x)[\/latex].<\/p>\r\n

    Notice, written in this form, [latex]x - k[\/latex] is a factor of [latex]f(x)[\/latex]. We can conclude if [latex]k[\/latex] is a zero of [latex]f(x)[\/latex], then [latex]x - k[\/latex] is a factor of [latex]f(x)[\/latex].<\/p>\r\n

    Similarly, if [latex]x - k[\/latex] is a factor of [latex]f(x)[\/latex], then the remainder of the Division Algorithm [latex]f(x) = (x - k)q(x) + r[\/latex] is [latex]0[\/latex]. This tells us that [latex]k[\/latex] is a zero.<\/p>\r\n<\/section>\r\n

    This pair of implications is the Factor Theorem. As we will soon see, a polynomial of degree [latex]n[\/latex] in the complex number system will have [latex]n[\/latex] zeros. We can use the Factor Theorem to completely factor a polynomial into the product of [latex]n[\/latex] factors. Once the polynomial has been completely factored, we can easily determine the zeros of the polynomial.<\/p>\r\n

    How to: Given a factor and a third-degree polynomial, use the Factor Theorem to factor the polynomial.<\/strong>\r\n
      \r\n\t
    1. Use synthetic division to divide the polynomial by [latex](x-k)[\/latex]<\/li>\r\n\t
    2. Confirm that the remainder is [latex]0[\/latex].<\/li>\r\n\t
    3. Write the polynomial as the product of [latex](x-k)[\/latex]\u00a0and the quadratic quotient.<\/li>\r\n\t
    4. If possible, factor the quadratic.<\/li>\r\n\t
    5. Write the polynomial as the product of factors.<\/li>\r\n<\/ol>\r\n<\/section>\r\n
      \r\n

      Show that [latex](x + 2)[\/latex] is a factor of [latex]x^3 - 6x^2 - x + 30[\/latex]. Find the remaining factors. Use the factors to determine the zeros of the polynomial.<\/p>\r\n\r\n\r\n[reveal-answer q=\"981344\"]Show Solution[\/reveal-answer] [hidden-answer a=\"981344\"]\r\n\r\n\r\n

      We can use synthetic division to show that [latex](x + 2)[\/latex] is a factor of the polynomial.<\/p>\r\n

      \"A<\/center>\r\n

       <\/p>\r\n

      The remainder is zero, so [latex](x + 2)[\/latex] is a factor of the polynomial. We can use the Division Algorithm to write the polynomial as the product of the divisor and the quotient:<\/p>\r\n

      [latex](x + 2)(x^2 - 8x + 15)[\/latex]<\/p>\r\n

      We can factor the quadratic factor to write the polynomial as<\/p>\r\n

      [latex](x + 2)(x - 3)(x - 5)[\/latex]<\/p>\r\n

      By the Factor Theorem, the zeros of [latex]x^3 - 6x^2 - x + 30[\/latex] are [latex]-2, 3,[\/latex] and [latex]5[\/latex].<\/p>\r\n\r\n\r\n[\/hidden-answer]<\/section>","rendered":"

      Evaluate Polynomials<\/h2>\n

      Evaluating a Polynomial Using the Remainder Theorem<\/h3>\n

      Previously, 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 [latex]x - k[\/latex], the remainder may be found quickly by evaluating the polynomial function at [latex]k[\/latex], that is, [latex]f(k)[\/latex].<\/p>\n

      \n
      \n

      the remainder theorem<\/h3>\n

      If a polynomial [latex]f(x)[\/latex] is divided by [latex]x\u2212k[\/latex], then the remainder is the value [latex]f(k)[\/latex].<\/p>\n<\/div>\n<\/section>\n

      Let’s walk through the proof of the theorem.<\/p>\n

      \n

      Recall that the Division Algorithm states that, given a polynomial dividend [latex]f(x)[\/latex] and a non-zero polynomial divisor [latex]d(x)[\/latex], there exist unique polynomials [latex]g(x)[\/latex] and [latex]r(x)[\/latex] such that<\/p>\n

      [latex]f(x) = d(x)g(x) + r(x)[\/latex]<\/p>\n

      and either [latex]r(x) = 0[\/latex] or the degree of [latex]r(x)[\/latex] is less than the degree of [latex]d(x)[\/latex]. In practice divisors, [latex]d(x)[\/latex] will have degrees less than or equal to the degree of [latex]f(x)[\/latex]. If the divisor, [latex]d(x)[\/latex], is [latex]x - k[\/latex], this takes the form<\/p>\n

      [latex]f(x) = (x - k)g(x) + r[\/latex]<\/p>\n

      Since the divisor [latex]x - k[\/latex] is linear, the remainder will be a constant, [latex]r[\/latex]. And, if we evaluate this for [latex]x = k[\/latex], we have<\/p>\n

      [latex]\\begin{array}{rl} f(k) & = (k - k)g(k) + r \\\\ & = 0 \\cdot g(k) + r \\\\ & = r \\end{array}[\/latex]<\/p>\n

      In other words, [latex]f(k)[\/latex] is the remainder obtained by dividing [latex]f(x)[\/latex] by [latex]x - k[\/latex].<\/p>\n<\/section>\n

      How to: Given a polynomial function [latex]f[\/latex], evaluate [latex]f(x)[\/latex] at [latex]x=k[\/latex] using the Remainder Theorem.<\/strong><\/p>\n
        \n
      1. Use synthetic division to divide the polynomial by [latex]x\u2212k[\/latex].<\/li>\n
      2. The remainder is the value [latex]f(k)[\/latex].<\/li>\n<\/ol>\n<\/section>\n
        \n

        Use the remainder theorem to evaluate [latex]f(x) = 6x^4 - x^3 - 15x^2 + 2x - 7[\/latex] at [latex]x = 2[\/latex].<\/p>\n