\n 1. Definition of Zero\\Root<\/a>\n
\n\t 2. Zeros on Graph<\/a>\n\t
\n\t 3. Zeros and Degree<\/a>\n
\n 4. Factor, Remainder Theorem<\/a><\/font>\n <\/div>\n
\n
The factoring polynomials<\/i><\/a> page introduced the concept of factors of polynomials.\n
\nWhich is a topic that leads into understanding zeros of polynomials, also referred to as roots.<\/p>\n
\n<\/a>\n
\n\n\n\nDefinition of a Zero\/Root<\/span><\/h2>\n\n\n\n
\nA value x = t<\/span> is a zero\/root of a polynomial p(x)<\/span>, if p(t) = 0<\/span>.\n
\nSo t<\/i><\/font> is a zero\/root, if it is a solution to p(x) = 0<\/span>.\n
\nFor example a polynomial we can look at is p(x) = x^2 + 5x + 6<\/span>.\n
\nA root\/zero will be an x<\/span> value where x^2 + 5x + 6 = 0<\/span>.\n
\nLooking at x = {\\text{-}}2<\/span>. p({\\text{-}}2) = ({\\text{-}}2)^2 + 5({\\text{-}}2) + 6 = 0<\/span>\n
\nSo x = {\\text{-}}2<\/span> is a zero\/root.\n
\nIt ‘s shown on the polynomial factors page how a polynomial,\n
\np(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
\ncan be factored to p(x) = \\space (x \\space {\\text{–}} \\space t)(x \\space {\\text{–}} \\space u)(x \\space {\\text{–}} \\space v)…..<\/span>.\n
\nWith the values of t<\/i><\/font>, u<\/i><\/font>, v<\/i><\/font> ….\n
\nwhere (x \\space {\\text{–}} \\space t) = 0 \\space\\space , \\space\\space (x \\space\\space {\\text{–}} \\space u) = 0 \\space\\space , \\space\\space (x \\space {\\text{–}} \\space v) = 0 \\space …..<\/span>\n
\nbeing the roots\\zeros of the polynomial.\n
\nSo we’re looking for a polynomials factors in order to obtain the roots\/zeros.\n
\n<\/a>\n\n\n\nZeros of Polynomials on a Graph:<\/span><\/h3>\n\n\n\n
\n\n\n\n![\"How](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/zeros-of-polynomials.png\")
<\/figure>\n\n\n\n
\n<\/a>\n
\n\n\n\nPolynomial Degree:<\/span><\/h3>\n\n\n\nThe number of factors a polynomial has being determined by the value of the ‘degree’ of the polynomial, means that the number of zeros\/roots also is indicated by the degree, which is highest power\/index of a variable that is present.\n
\n4x + 5<\/span> is a polynomial of degree 1, and has one solution.\n
\nx^2 + 3x \\space {\\text{–}} \\space 5<\/span> is a polynomial of degree 2 and has two solutions.\n
\n4x^3 + x^2 \\space {\\text{–}} \\space 6x + 7<\/span> is a polynomial of degree 3, so having three solutions, and so on.\n
\n
\nExamples <\/font><\/font><\/font><\/u><\/b>\n
\nWe can recap some quadratic polynomials where we have two factors and solutions.\n
\n(1.1) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nx^2 \\space {\\text{–}} \\space x \\space {\\text{–}} \\space 6 = 0<\/span> => (x + 2)(x \\space {\\text{–}} \\space 3) = 0<\/span>\n
\nx = {\\text{-}}2 \\space , \\space x = 3<\/span>\n
\n(1.2) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nx^2 + 2x + 1 = 0<\/span> => (x + 1)(x + 1) = 0<\/span>\n
\nx = {\\text{-}}1<\/span>\n
\nTechnically two solutions, they just happen to be the same.\n
\n<\/a>\n\n\n\n
\n\n\n\nZeros of Polynomials,
Factor and Remainder Theorem<\/span><\/h2>\n\n\n\n
\nWhen it comes to establishing and finding zeros of polynomials that are larger than a quadratic and of degree 3 or larger.\n
\nThe ‘factor theorem’ and the ‘remainder theorem’ can often be very helpful in assisting. \n
\nThe wording can vary slightly from website to website when stating the theorems, but we’ll summarise here.\n
\n\n\n\nFactor Theorem:<\/span><\/h3>\n\n\n\nFor a polynomial f(x)<\/span>, when f(c) = 0<\/span>.\n
\nThen (x \\space {\\text{–}} \\space c)<\/span> is a factor of f(x)<\/span>.\n
\n\n\n\nRemainder Theorem:<\/span><\/h3>\n\n\n\nWhen a polynomial f(x)<\/span> is divided by (x \\space {\\text{–}} \\space c)<\/span>.\n
\nThen the remainder is f(c)<\/span>.\n
\nExamples <\/font><\/font><\/font><\/u><\/b>\n
\n(2.1) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nDetermine if (x \\space {\\text{–}} \\space 3)<\/span> is a factor of f(x) = x^3 + 2x^2 \\space {\\text{–}} \\space 11x \\space {\\text{–}} \\space 12<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nIf (x \\space {\\text{–}} \\space 3)<\/span> is a factor, from the factor theorem we expect to see f(3) = 0<\/span>.\n
\nf(3) = 3^3 + 2(3)^2 \\space {\\text{–}} \\space 11(3) \\space {\\text{–}} \\space 12 \\space = \\space 27 + 18 {\\text{–}} \\space 33 \\space {\\text{–}} \\space 12 \\space = \\space 0<\/span>\n
\nf(3) = 0<\/span>, so (x \\space {\\text{–}} \\space 3)<\/span> is a factor of f(x)<\/span><\/font>.\n
\nSubsequently from the remainder theorem, x = 3<\/span> is a zero.<\/font>\n
\n(2.2) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nWhich of -2, -1, 1 are zeros of f(x) = 2x^3 + 7x^2 + x \\space {\\text{–}} \\space 10<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nWhen a value c<\/span> is a zero of f(x)<\/span>, then f(c)<\/span> will be zero.\n
\nf({\\text{-}}2) = 2({\\text{-}}2)^3 + 7({\\text{-}}2)^2 + {\\text{-}}2 \\space {\\text{–}} \\space 10 \\space = \\space {\\text{-}}16 + 28 \\space {\\text{–}} \\space 2 \\space {\\text{–}} \\space 10 \\space = \\space 0<\/span>\n
\nf({\\text{-}}1) = 2({\\text{-}}1)^3 + 7({\\text{-}}1)^2 + {\\text{-}}1 \\space {\\text{–}} \\space 10 \\space = \\space {\\text{-}}2 + 7 \\space {\\text{–}} \\space 1 \\space {\\text{–}} \\space 10 \\space = \\space {\\text{-}}6<\/span>\n
\nf(1) = 2(1)^3 + 7(1)^2 + 1 \\space {\\text{–}} \\space 10 \\space = \\space 2 + 7 + 1 \\space {\\text{–}} \\space 10 \\space = \\space 0<\/span>\n
\n-2 and 1 are zeros of f(x) = 2x^3 + 7x^2 + x \\space {\\text{–}} \\space 10<\/font>.\n
\n\n\n