{"id":1389,"date":"2015-11-12T18:35:29","date_gmt":"2015-11-12T18:35:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1389"},"modified":"2017-03-31T22:58:05","modified_gmt":"2017-03-31T22:58:05","slug":"use-the-linear-factorization-theorem-to-find-polynomials-with-given-zeros","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/use-the-linear-factorization-theorem-to-find-polynomials-with-given-zeros\/","title":{"raw":"Use the Linear Factorization Theorem to find polynomials with given zeros","rendered":"Use the Linear Factorization Theorem to find polynomials with given zeros"},"content":{"raw":"

A vital implication of the Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree n<\/em>\u00a0will have n<\/em>\u00a0zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n<\/em>\u00a0factors. The Linear Factorization Theorem<\/strong> tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x\u00a0\u2013\u00a0c<\/em>), where c<\/em>\u00a0is a complex number.<\/p>\r\n

Let f<\/em>\u00a0be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex],\u00a0is a zero of [latex]f\\left(x\\right)[\/latex].\u00a0Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].\u00a0For f<\/em>\u00a0to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex]\u00a0must also be a factor of [latex]f\\left(x\\right)[\/latex].\u00a0This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex],\u00a0when multiplied by [latex]x-\\left(a+bi\\right)[\/latex],\u00a0will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f<\/em>\u00a0with real coefficients has a complex zero [latex]a+bi[\/latex],\u00a0then the complex conjugate [latex]a-bi[\/latex]\u00a0must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the Complex Conjugate Theorem<\/strong>.<\/p>\r\n\r\n

\r\n

A General Note: Complex Conjugate Theorem<\/h3>\r\n

According to the Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where c<\/em>\u00a0is a complex number.<\/p>\r\n

If the polynomial function f<\/em>\u00a0has real coefficients and a complex zero in the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.<\/p>\r\n\r\n<\/div>\r\n

\r\n

How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex]\u00a0on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\r\n
  1. Use the zeros to construct the linear factors of the polynomial.<\/li>\r\n\t
  2. Multiply the linear factors to expand the polynomial.<\/li>\r\n\t
  3. Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\r\n\t
  4. Simplify.<\/li>\r\n<\/ol><\/div>\r\n
    \r\n
    \r\n
    \r\n

    Example 7: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\r\n

    Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Solution<\/h3>\r\n

    Because [latex]x=i[\/latex]\u00a0is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex]\u00a0is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\r\n\r\n

    [latex]\\begin{cases}f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{cases}[\/latex]<\/div>\r\n

    We need to find a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]\r\ninto [latex]f\\left(x\\right)[\/latex].<\/p>\r\n\r\n

    [latex]\\begin{cases}100=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right)\\hfill \\\\ 100=a\\left(-20\\right)\\hfill \\\\ -5=a\\hfill \\end{cases}[\/latex]<\/div>\r\n

    So the polynomial function is<\/p>\r\n\r\n

    [latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/div>\r\n

    or<\/p>\r\n\r\n

    [latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

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

    We found that both i<\/em>\u00a0and \u2013i<\/em> were zeros, but only one of these zeros needed to be given. If i<\/em>\u00a0is a zero of a polynomial with real coefficients, then \u2013i<\/em>\u00a0must also be a zero of the polynomial because \u2013i<\/em>\u00a0is the complex conjugate of i<\/em>.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n

    Q & A<\/h3>\r\n

    If 2 + 3i<\/em>\u00a0were given as a zero of a polynomial with real coefficients, would 2 \u2013\u00a03i<\/em>\u00a0also need to be a zero?<\/strong><\/p>\r\n

    Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\r\n\r\n<\/div>\r\n

    \r\n

    Try It 5<\/h3>\r\n

    Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132i<\/em>\u00a0such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    A vital implication of the Fundamental Theorem of Algebra<\/strong>, as we stated above, is that a polynomial function of degree n<\/em>\u00a0will have n<\/em>\u00a0zeros in the set of complex numbers, if we allow for multiplicities. This means that we can factor the polynomial function into n<\/em>\u00a0factors. The Linear Factorization Theorem<\/strong> tells us that a polynomial function will have the same number of factors as its degree, and that each factor will be in the form (x\u00a0\u2013\u00a0c<\/em>), where c<\/em>\u00a0is a complex number.<\/p>\n

    Let f<\/em>\u00a0be a polynomial function with real coefficients, and suppose [latex]a+bi\\text{, }b\\ne 0[\/latex],\u00a0is a zero of [latex]f\\left(x\\right)[\/latex].\u00a0Then, by the Factor Theorem, [latex]x-\\left(a+bi\\right)[\/latex]\u00a0is a factor of [latex]f\\left(x\\right)[\/latex].\u00a0For f<\/em>\u00a0to have real coefficients, [latex]x-\\left(a-bi\\right)[\/latex]\u00a0must also be a factor of [latex]f\\left(x\\right)[\/latex].\u00a0This is true because any factor other than [latex]x-\\left(a-bi\\right)[\/latex],\u00a0when multiplied by [latex]x-\\left(a+bi\\right)[\/latex],\u00a0will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f<\/em>\u00a0with real coefficients has a complex zero [latex]a+bi[\/latex],\u00a0then the complex conjugate [latex]a-bi[\/latex]\u00a0must also be a zero of [latex]f\\left(x\\right)[\/latex]. This is called the Complex Conjugate Theorem<\/strong>.<\/p>\n

    \n

    A General Note: Complex Conjugate Theorem<\/h3>\n

    According to the Linear Factorization Theorem<\/strong>, a polynomial function will have the same number of factors as its degree, and each factor will be in the form [latex]\\left(x-c\\right)[\/latex], where c<\/em>\u00a0is a complex number.<\/p>\n

    If the polynomial function f<\/em>\u00a0has real coefficients and a complex zero in the form [latex]a+bi[\/latex],\u00a0then the complex conjugate of the zero, [latex]a-bi[\/latex],\u00a0is also a zero.<\/p>\n<\/div>\n

    \n

    How To: Given the zeros of a polynomial function [latex]f[\/latex] and a point [latex]\\left(c\\text{, }f(c)\\right)[\/latex]\u00a0on the graph of [latex]f[\/latex], use the Linear Factorization Theorem to find the polynomial function.<\/h3>\n
      \n
    1. Use the zeros to construct the linear factors of the polynomial.<\/li>\n
    2. Multiply the linear factors to expand the polynomial.<\/li>\n
    3. Substitute [latex]\\left(c,f\\left(c\\right)\\right)[\/latex] into the function to determine the leading coefficient.<\/li>\n
    4. Simplify.<\/li>\n<\/ol>\n<\/div>\n
      \n
      \n
      \n

      Example 7: Using the Linear Factorization Theorem to Find a Polynomial with Given Zeros<\/h3>\n

      Find a fourth degree polynomial with real coefficients that has zeros of \u20133, 2, i<\/em>, such that [latex]f\\left(-2\\right)=100[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      Because [latex]x=i[\/latex]\u00a0is a zero, by the Complex Conjugate Theorem [latex]x=-i[\/latex]\u00a0is also a zero. The polynomial must have factors of [latex]\\left(x+3\\right),\\left(x - 2\\right),\\left(x-i\\right)[\/latex], and [latex]\\left(x+i\\right)[\/latex]. Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let\u2019s begin by multiplying these factors.<\/p>\n

      [latex]\\begin{cases}f\\left(x\\right)=a\\left(x+3\\right)\\left(x - 2\\right)\\left(x-i\\right)\\left(x+i\\right)\\\\ f\\left(x\\right)=a\\left({x}^{2}+x - 6\\right)\\left({x}^{2}+1\\right)\\\\ f\\left(x\\right)=a\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)\\end{cases}[\/latex]<\/div>\n

      We need to find a<\/em> to ensure [latex]f\\left(-2\\right)=100[\/latex]. Substitute [latex]x=-2[\/latex] and [latex]f\\left(2\\right)=100[\/latex]
      \ninto [latex]f\\left(x\\right)[\/latex].<\/p>\n

      [latex]\\begin{cases}100=a\\left({\\left(-2\\right)}^{4}+{\\left(-2\\right)}^{3}-5{\\left(-2\\right)}^{2}+\\left(-2\\right)-6\\right)\\hfill \\\\ 100=a\\left(-20\\right)\\hfill \\\\ -5=a\\hfill \\end{cases}[\/latex]<\/div>\n

      So the polynomial function is<\/p>\n

      [latex]f\\left(x\\right)=-5\\left({x}^{4}+{x}^{3}-5{x}^{2}+x - 6\\right)[\/latex]<\/div>\n

      or<\/p>\n

      [latex]f\\left(x\\right)=-5{x}^{4}-5{x}^{3}+25{x}^{2}-5x+30[\/latex]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\n

      We found that both i<\/em>\u00a0and \u2013i<\/em> were zeros, but only one of these zeros needed to be given. If i<\/em>\u00a0is a zero of a polynomial with real coefficients, then \u2013i<\/em>\u00a0must also be a zero of the polynomial because \u2013i<\/em>\u00a0is the complex conjugate of i<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

      \n

      Q & A<\/h3>\n

      If 2 + 3i<\/em>\u00a0were given as a zero of a polynomial with real coefficients, would 2 \u2013\u00a03i<\/em>\u00a0also need to be a zero?<\/strong><\/p>\n

      Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.<\/em><\/p>\n<\/div>\n

      \n

      Try It 5<\/h3>\n

      Find a third degree polynomial with real coefficients that has zeros of 5 and \u20132i<\/em>\u00a0such that [latex]f\\left(1\\right)=10[\/latex].<\/p>\n

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

      \n\t\t\t

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