\n 1. Rational Roots Theorem<\/a>\n
\n\t 2. Examples<\/a><\/font>\n <\/div>\n
\n
The factors of polynomials<\/i><\/a> introduction page showed an example of working out the factors of a polynomial of degree 3.\n
\nWhere we often start out looking for the roots of a polynomial by trying a few initial guesses for values that would give a result of 0 when evaluated in the relevant polynomial.\n
\nBut to begin with how can we know for sure what values are suitable to try?\n
\nThis is where we can make use of what is known as the “rational roots theorem” or “rational roots test”. <\/p>\n
\n<\/a>\n\n\n\nRational Roots Theorem<\/span><\/h2>\n\n\n\n
\nIf we have a polynomial a_nx^n \\space + \\space a_{n \\space {\\text{–}} \\space 1}x^{n \\space {\\text{–}} \\space 1} \\space + ….. + \\space a_{\\tt{2}}x^{\\tt{2}} \\space \\space + a_1x \\space + \\space a_0<\/span> ,\n
\nwith integer coefficients a_n \\space ….. \\space a_1<\/span>,\n
\nand a non zero constant term a_0<\/span>.\n
\nThen every rational root of the polynomial is of the form, \\color{red}\\frac{\\pm \\space factor \\space of \\space a_0}{\\pm \\space factor \\space of \\space a_n}<\/span><\/font> .\n
\nThough we do have to note that not all values obtained with this fraction will be roots\/zeros, and sometimes none will be if a polynomial has only complex roots.\n
\nIn Practice <\/u><\/font><\/font><\/b>\n
\nWe can consider a simple quadratic polynomial, f(x) = x^2 + 3x + 2<\/span>.\n
\nThis actually factors nicely to (x + 2)(x + 1)<\/span>.\n
\nBut we’ll still show the rational roots theorem in action.\n
\nFrom the quadratic we can see that, a_n = 1 \\space , \\space a_0 = 2<\/span>.\n
\nFactors of 1 => 1. Factors of 2 => 1, 2.\n
\n\\frac{\\pm \\space 1,2}{\\pm \\space 1}<\/span><\/font> => ± \\frac{1}{1} \\space {\\tiny{,}} \\space \\frac{2}{1}<\/span><\/font> => {\\scriptsize{\\pm}} 1 \\space , \\space {\\scriptsize{\\pm}} 2<\/span><\/font>\n
\nAs we can see, -1 and -2 are among the possible values, which was expected as we knew beforehand that they are the factors.\n
\nEvaluating both values should give 0.\n
\nf({\\text{-}}2) = ({\\text{-}}2)^2 + 3({\\text{-}}2) + 2 \\space = \\space 4 \\space {\\text{–}} \\space 6 + 2 \\space = \\space 0<\/span>\n
\nf({\\text{-}}1) = ({\\text{-}}1)^2 + 3({\\text{-}}1) + 2 \\space = \\space 1 \\space {\\text{–}} \\space 3 + 2 \\space = \\space 0<\/span>\n
\n\n\n\n![\"The](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/rational-roots-theorem.png\")
<\/figure>\n\n\n\n
\nThis was a very basic example, it’s when we need to deal with polynomials of degree 3 and larger that using the rational roots theorem is most helpful.\n
\n<\/a>\n
\nExamples <\/font><\/font><\/font><\/u><\/b>\n
\n(1.1) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nFind all the possible rational roots of x^4 + 6x^3 + 7x^2 \\space {\\text{–}} \\space 6x \\space {\\text{–}} \\space 8<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nHere, a_n = 1 \\space , \\space a_0 = {\\text{-}}8<\/span>.\n
\nWe can just treat the -8 as 8, seeing as we look at plus and minus values of the factors.<\/font>\n
\nFactors of 1 => 1. Factors of 8 => 1, 2, 4, 8.\n
\n\\frac{\\pm \\space 1,2,4,8}{\\pm \\space 1}<\/span><\/font> => ± \\frac{1}{1} \\space {\\tiny{,}} \\space \\frac{2}{1} \\space {\\tiny{,}} \\space \\frac{4}{1} \\space {\\tiny{,}} \\space \\frac{8}{1}<\/span><\/font> => {\\scriptsize{\\pm}} 1 \\space , \\space {\\scriptsize{\\pm}} 2 \\space , \\space {\\scriptsize{\\pm}} 4\\space , \\space {\\scriptsize{\\pm}} 8<\/span><\/font>\n
\n(1.2) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nFind all the possible rational roots of 2x^3 \\space {\\text{–}} \\space 5x + 6<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nHere, a_n = 2 \\space , \\space a_0 = 6<\/span>.\n
\nFactors of 2 => 1, 2. Factors of 6 => 1, 2, 3, 6.\n
\n\\frac{\\pm \\space 1,2,3,6}{\\pm \\space 1}<\/span><\/font> => ± \\frac{1}{1} \\space {\\tiny{,}} \\space \\frac{2}{1} \\space {\\tiny{,}} \\space \\frac{3}{1} \\space {\\tiny{,}} \\space \\frac{6}{1}<\/span><\/font> => {\\scriptsize{\\pm}} 1 \\space , \\space {\\scriptsize{\\pm}} 2 \\space , \\space {\\scriptsize{\\pm}} 3\\space , \\space {\\scriptsize{\\pm}} 6<\/span><\/font>\n
\n\\frac{\\pm \\space 1,2,3,6}{\\pm \\space 2}<\/span><\/font> => ± \\frac{1}{2} \\space {\\tiny{,}} \\space \\frac{2}{2} \\space {\\tiny{,}} \\space \\frac{3}{2} \\space {\\tiny{,}} \\space \\frac{6}{2}<\/span><\/font> => {\\scriptsize{\\pm}} {\\Large{\\frac{1}{2}}} \\space , \\space {\\scriptsize{\\pm}} 1 \\space , \\space {\\scriptsize{\\pm}} {\\Large{\\frac{3}{2}}} \\space , \\space {\\scriptsize{\\pm}} 3<\/span><\/font>\n
\nThe possible rational roots are:<\/font>\n
\n{\\scriptsize{\\pm}} {\\Large{\\frac{1}{2}}} \\space , \\space {\\scriptsize{\\pm}} 1 \\space , \\space {\\scriptsize{\\pm}} {\\Large{\\frac{3}{2}}} \\space , \\space {\\scriptsize{\\pm}} 2 \\space , \\space {\\scriptsize{\\pm}} 3 \\space , \\space {\\scriptsize{\\pm}} 6<\/span><\/font>\n
\n\n\n