\n 1. Descartes Rules of Signs Intro<\/b><\/a>\n
\n\t 2. Descartes Rule of Signs Examples<\/a><\/font>\n <\/div>\n
\n
The rational roots test page showed how to find all the possible values of real roots or solutions that can a specific polynomial can have.\n
\nDescartes rule of signs is something that can tell us at most how many possible real roots\/zeros we can expect a polynomial to have, along with what the nature of those roots will be, either positive or negative.\n
\nThis page will demonstrate a selection of Descartes rule of signs examples in order to show how it can be used when dealing with polynomials.<\/p>\n
\n<\/a>\n
\n\n\n\nDescartes Rules of Signs Intro<\/span><\/h2>\n\n\n\n
\nIf we have a standard polynomial.\n
\nf(x) = \\space 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
\nWhat we look to do is count how many sign changes between + and − there are throughout the polynomial.\n
\nWe do this for both f(x)<\/span> and f({\\text{-}}x)<\/span>.\n
\nBecause for the sign changes between + and −.\n
\nThe number of changes in f(x)<\/span> tells us the number of possible positive real roots.\n
\nThe number of changes in f({\\text{-}}x)<\/span> tells us the number of possible negative real roots.\n
\nExample <\/u><\/font><\/font><\/font><\/b>\n
\nWe can count how many sign changes there are in h(x) = x^3 + 4x^2 \\space {\\text{--}} \\space 2x + 1<\/span>.\n
\n\n\n\n![\"How](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/descartes-rule-of-signs.png\")
<\/figure>\n\n\n\n
\nBut there may be less than 2 real positive roots, why?\n
\nBecause there is a possibility of there being complex roots to the polynomial. Which always come in a conjugate pair.\n
\nIf there are any complex roots, there will be an even number of them. 2 , 4 ……. \n
\nThus we count down by a size of 2 until we get to 1 or 0.\n
\nWe can’t count down by 2 from 1, as we can’t have -1 roots\/zeros.\n
\nSo there are 0 or 2 real positive roots for h(x) = x^3 + 4x^2 \\space {\\text{–}} \\space 2x + 1<\/span>.\n
\n–<\/b><\/font> Now for the case of h({\\text{-}}x)<\/span>.\n
\nh({\\text{-}}x) = ({\\text{-}}x)^3 + 4({\\text{-}}x)^2 \\space {\\text{–}} \\space 2({\\text{-}}x) + 1<\/span>\n
\nOnly the signs of terms with an odd power will change sign.<\/font>\n
\nh({\\text{-}}x) = {\\text{-}}x^3 + 4x^2 + 2x + 1<\/span>\n
\nWe now have 1 sign change from − to +.<\/font> h({\\text{-}}x) = {\\underbrace{{\\text{-}}x^3}_{\\text{{\\color{red}1}}}} + 4x^2 + 2x + 1<\/span>\n
\nAs we can’t count down by 2 from 1.\n
\nThere is definitely 1 negative real root to the original polynomial h(x)<\/span>\n
\nWe can now look at the graph of h(x)<\/span>.\n
\nWhere we see that there is just the 1 negative real root\/zero, and in fact 0 positive real roots\/zeros. \n
\n\n\n\n![\"Polynomial](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/1-neative-real-root.png\")
<\/figure>\n\n\n\n
\n<\/a>\n
\n\n\n\n
\n\n\n\nDescartes Rule of Signs Examples<\/span><\/h2>\n\n\n\n
\n(1.1) <\/i><\/b><\/font><\/u><\/font><\/font>\n
\nUse Descartes rule of signs to establish the number possible real roots of f(x) = x^3 \\space {\\text{–}} \\space 21x \\space {\\text{–}} \\space 20<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nf(x) = {\\underbrace{x^3}_{\\text{{\\color{lightseagreen}1}}}} \\space {\\text{–}} \\space 21x \\space {\\text{–}} \\space 20<\/span>\n
\nThere is one positive real root for the polynomial.\n
\nf({\\text{-}}x) = ({\\text{-}}x)^3 \\space {\\text{–}} \\space 21({\\text{-}}x) \\space {\\text{–}} \\space 20<\/span>\n
\n = {\\underbrace{{\\text{-}}x^3}_{\\text{{\\color{red}1}}}} + {\\underbrace{21x}_{\\text{{\\color{red}2}}}} \\space {\\text{–}} \\space 20<\/span>\n
\nThere is possibly two negative real roots, or zero negative real roots.\n
\nThus there is either three real roots to the polynomial, or one real root in total.\n
\nNote that the maximum number of possible real roots was 3.\n
\nWhich happens to be the degree of the polynomial.\n
\nThis is always the case when dealing with Descartes rule of signs examples.\n
\nSo if you have a maximum number of possible roots that is larger or smaller after adding up the sign changes, then you know you have gone a bit wrong somewhere.\n
\nThis is a handy fact to know as a way of checking.<\/font>\n
\n(1.2) <\/i><\/b><\/font><\/u><\/font><\/font>\n
\nWith Descartes rule of signs to establish the number possible real roots of g(x) = x^5 \\space {\\text{–}} \\space 6x^4 + 3x^3 + 36x^2 \\space {\\text{–}} \\space 34x \\space {\\text{–}} \\space 60<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\ng(x) = {\\underbrace{x^5}_{\\text{{\\color{lightseagreen}1}}}} \\space {\\text{–}} \\space {\\underbrace{6x^4}_{\\text{{\\color{lightseagreen}2}}}} + 3x^3 + {\\underbrace{36x^2}_{\\text{{\\color{lightseagreen}3}}}} \\space {\\text{–}} \\space 34x \\space {\\text{–}} \\space 60<\/span>\n
\nThere is three real positive roots, or just one positive real root.\n
\ng({\\text{-}}x) = ({\\text{-}}x)^5 \\space {\\text{–}} \\space 6({\\text{-}}x)^4 + 3({\\text{-}}x)^3 + 36({\\text{-}}x)^2 \\space {\\text{–}} \\space 34({\\text{-}}x) \\space {\\text{–}} \\space 60<\/span>\n
\n = {\\text{-}}x^5 \\space {\\text{–}} \\space 6x^4 \\space {\\text{–}} \\space {\\underbrace{3x^3}_{\\text{{\\color{red}1}}}} + {\\underbrace{36x^2}_{\\text{{\\color{red}2}}}} \\space {\\text{–}} \\space 34x \\space {\\text{–}} \\space 60<\/span>\n
\nThere is possibly two negative real roots, or zero negative real roots.\n
\nThus there is either five real roots, three real roots or one real root to the polynomial g(x)<\/span>.\n
\n
\n\n\n\nDescartes Rule of Signs Summary<\/span><\/h3>\n\n\n\n–<\/b><\/font> Make sure the polynomial is in descending order of exponent value.\n
\n–<\/b><\/font> When switching to f({\\text{-}}x)<\/span>, only the terms where the variable has an odd index have the sign changed.\n
\n–<\/b><\/font> We may not end up with the exact number of real roots the polynomial has, unless the result is 0 or 1.\n
\n Due to the possible existence of complex roots.\n
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