\nThis page will look at further examples of factoring with synthetic division and solving with polynomials of degree 3 and higher.<\/p>\n
\n\n\n\n
Factoring with Synthetic Division and Solving,
Examples<\/span><\/h2>\n\n\n\n
\n(1.1) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nShow that (x + 1)<\/span> is a factor of f(x) = 3x^3 \\space {\\text{–}} \\space 7x^2 \\space {\\text{–}} \\space 18x \\space {\\text{–}} \\space 8<\/span>,\n
\nthen factorise fully and solve for f(x) = 0<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nx + 1 \\space = \\space 0 \\space => \\space x = {\\text{-}}1<\/span><\/font>\n
\n\n\n\n![\"Factoring](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/factoring-with-synthetic-division.png\")
<\/figure>\n\n\n\n
\nA remainder of 0 shows that (x + 1)<\/span> is a factor,\n
\nand we can now factorise fully and solve.\n
\n(3x^2 \\space {\\text{–}} \\space 10x \\space {\\text{–}} \\space 8)(x + 1)<\/span>\n
\nThe larger quadratic turns out to factorise nicely into two neat binomials.<\/font>\n
\n(3x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x + 1)<\/span> => (3x + 2)(x \\space {\\text{–}} \\space 4)(x + 1)<\/span>\n
\nSetting the factors equal to 0 and solving will give us the solutions.<\/font>\n
\n3x + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}\\frac{2}{3} \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 4<\/span>\n
\nx + 1 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}1<\/span>\n
\nThe solutions of 3x^3 \\space {\\text{–}} \\space 7x^2 \\space {\\text{–}} \\space 18x \\space {\\text{–}} \\space 8<\/span> are x = {\\text{-}}1 \\space , \\space {\\text{-}}\\frac{2}{3} \\space , \\space 4<\/span>.\n
\n(1.2) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nIf (x \\space {\\text{–}} \\space 2)<\/span> is a factor of f(x) = x^3 + 3x^2 \\space {\\text{–}} \\space 4x \\space {\\text{–}} \\space 12<\/span>.\n
\nUse synthetic division to find the other two factors, and subsequently all roots.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nWe’re given the information that (x {\\text{–}} \\space 2)<\/span> is a factor, so can procced straight away with synthetic division without having to make any guesses at values to try.<\/font>\n
\nx \\space {\\text{–}} \\space 2 \\space = \\space 0 \\space => \\space x = 2<\/span><\/font>\n
\n\n\n\n![\"Factoring](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/synth-div-factor.png\")
<\/figure>\n\n\n\n
\nA remainder of 0 is obtained as expected,\n
\nand we can now factorise fully and find the roots\/solutions.\n
\n(x^2 + 5x + 6)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nThe larger quadratic turns out to factorise nicely into two neat binomials.<\/font>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 2)<\/span> => (x + 3)(x + 2)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nSetting the factors equal to 0 and solving will give us the solutions.<\/font>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 2<\/span>\n
\nx + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}2<\/span>\n
\nThe solutions of f(x) = x^3 + 3x^2 \\space {\\text{–}} \\space 4x \\space {\\text{–}} \\space 12<\/span> are x = {\\text{-}}3 \\space , \\space {\\text{-}}2 \\space , \\space 2<\/span>.\n
\n(1.3) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nFind the roots\/zeros of h(x) = x^3 \\space {\\text{–}} \\space 3x^2 \\space {\\text{–}} \\space 10x + 24<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nSometimes we aren’t given any factors or roots to start with.\n
\nWe can apply the rational roots test, but we can also just proceed with a few basic initial guesses to find a first factor and root.<\/font>\n
\nh(1) = 1^3 \\space {\\text{–}} \\space 3(1)^2 \\space {\\text{–}} \\space 10(1) + 24 \\space = \\space 1 \\space {\\text{–}} \\space 3 \\space {\\text{–}} \\space 10 + 24 \\space = \\space 13<\/span>\n
\nh({\\text{-}}1) = ({\\text{-}}1)^3 \\space {\\text{–}} \\space 3({\\text{-}}1)^2 \\space {\\text{–}} \\space 10({\\text{-}}1) + 24 \\space = \\space {\\text{-}}1 \\space {\\text{–}} \\space 3 + 10 + 24 \\space = \\space 13<\/span>\n
\nh(2) = 2^3 \\space {\\text{–}} \\space 3(2)^2 \\space {\\text{–}} \\space 10(2) + 24 \\space = \\space 8 \\space {\\text{–}} \\space 12 \\space {\\text{–}} \\space 20 + 24 \\space = \\space 0<\/span>\n
\n2 is a root, so with that information we can proceed with finding the other roots.<\/font>\n\n\n\n![\"Factoring](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/factors-ex-1.png\")
<\/figure>\n\n\n\n
\nA remainder of 0 is obtained as expected,\n
\nand we can now factorise fully and find the roots\/solutions.\n
\n(x^2 \\space {\\text{–}} \\space x \\space {\\text{–}} \\space 12)(x \\space {\\text{–}} \\space 2)<\/span>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 2)<\/span> => (x + 3)(x \\space {\\text{–}} \\space 4)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 4<\/span>\n
\nThe solutions of h(x) = x^3 \\space {\\text{–}} \\space 3x^2 \\space {\\text{–}} \\space 10x + 24<\/span> are x = {\\text{-}}3 \\space , \\space 2 \\space , \\space 4<\/span>.\n
\n(1.4) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nShow that (x + 2)<\/span> is a factor of f(x) = x^4 + 8x^3 + 17x^2 \\space {\\text{–}} \\space 2x \\space {\\text{–}} \\space 24<\/span>.\n
\nHence factorise fully and solve.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nf(x)<\/span> here is a polynomial of degree 4, so there are 4 factors and solutions.<\/span>\n
\nFirstly we deal with showing that (x + 2)<\/span> is a factor.\n
\nf({\\text{-}}2) = ({\\text{-}}2)^4 + 8({\\text{-}}2)^3 + 17({\\text{-}}2)^2 \\space {\\text{–}} \\space 2({\\text{-}}2) \\space {\\text{–}} \\space 24<\/span>\n
\n= 16 \\space {\\text{–}} \\space 64 + 68 + 4 \\space {\\text{–}} \\space 24 \\space = \\space 0<\/span>\n
\n(x + 2)<\/span> is a factor, with -2<\/b> being a zero.\n
\nWith that information we can proceed with finding the other roots.<\/font><\/font>\n
\n\n\n\n![\"Factoring](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/synth-degree-4.png\")
<\/figure>\n\n\n\n
\nA remainder of 0 is obtained as expected,\n
\nwe can now look to factorise further.\n
\n(x + 2)(x^3 + 6x^2 + 5 \\space {\\text{–}} \\space 12)<\/span>\n
\nWe can proceed with factoring with synthetic division for x^3 + 6x^2 + 5 \\space {\\text{–}} \\space 12<\/span>, but we don’t know any exact zeros\/roots to start with.\n
\nThe rational roots theorem<\/i><\/a> can help us find values to try.\n
\nThough sometimes we can try some basic values first, as we may get lucky and find a zero\/root.\n
\nA good value to try initially is usually 0 or 1.\n
\n\n\n\n![\"Factoring](\"http:\/\/www.learnermath.com\/wp-content\/uploads\/2023\/09\/further-roots.png\")
<\/figure>\n\n\n\n
\nFactoring further.\n
\nSuccess on this occasion, we have quickly found another zero\/root.\n
\nIt doesn’t always work out like this, but it’s nice when it does.\n
\nNow we can factor fully and solve.\n
\n(x^2 + 7x + 12)(x \\space {\\text{–}} \\space 1)(x + 2)<\/span>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 1)(x + 2)<\/span> => (x + 3)(x + 4)(x \\space {\\text{–}} \\space 1)(x + 2)<\/span>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x + 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}4<\/span>\n
\nx \\space {\\text{–}} \\space 1 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 1 \\space \\space \\space , \\space \\space \\space x + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}2<\/span>\n
\nThe solutions of f(x) = x^4 + 8x^3 + 17x^2 \\space {\\text{–}} \\space 2x \\space {\\text{–}} \\space 24<\/span> are x = {\\text{-}}4 \\space , \\space {\\text{-}}3 \\space , \\space {\\text{-}}2 \\space , \\space 1<\/span>.\n
\n\n\n
<\/figure>\n\n\n\n\nA remainder of 0 shows that (x + 1)<\/span> is a factor,\n
\nand we can now factorise fully and solve.\n
\n(3x^2 \\space {\\text{–}} \\space 10x \\space {\\text{–}} \\space 8)(x + 1)<\/span>\n
\nThe larger quadratic turns out to factorise nicely into two neat binomials.<\/font>\n
\n(3x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x + 1)<\/span> => (3x + 2)(x \\space {\\text{–}} \\space 4)(x + 1)<\/span>\n
\nSetting the factors equal to 0 and solving will give us the solutions.<\/font>\n
\n3x + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}\\frac{2}{3} \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 4<\/span>\n
\nx + 1 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}1<\/span>\n
\nThe solutions of 3x^3 \\space {\\text{–}} \\space 7x^2 \\space {\\text{–}} \\space 18x \\space {\\text{–}} \\space 8<\/span> are x = {\\text{-}}1 \\space , \\space {\\text{-}}\\frac{2}{3} \\space , \\space 4<\/span>.\n
\n(1.2) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nIf (x \\space {\\text{–}} \\space 2)<\/span> is a factor of f(x) = x^3 + 3x^2 \\space {\\text{–}} \\space 4x \\space {\\text{–}} \\space 12<\/span>.\n
\nUse synthetic division to find the other two factors, and subsequently all roots.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nWe’re given the information that (x {\\text{–}} \\space 2)<\/span> is a factor, so can procced straight away with synthetic division without having to make any guesses at values to try.<\/font>\n
\nx \\space {\\text{–}} \\space 2 \\space = \\space 0 \\space => \\space x = 2<\/span><\/font>\n
\n\n\n\n
<\/figure>\n\n\n\n\nA remainder of 0 is obtained as expected,\n
\nand we can now factorise fully and find the roots\/solutions.\n
\n(x^2 + 5x + 6)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nThe larger quadratic turns out to factorise nicely into two neat binomials.<\/font>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 2)<\/span> => (x + 3)(x + 2)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nSetting the factors equal to 0 and solving will give us the solutions.<\/font>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 2<\/span>\n
\nx + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}2<\/span>\n
\nThe solutions of f(x) = x^3 + 3x^2 \\space {\\text{–}} \\space 4x \\space {\\text{–}} \\space 12<\/span> are x = {\\text{-}}3 \\space , \\space {\\text{-}}2 \\space , \\space 2<\/span>.\n
\n(1.3) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nFind the roots\/zeros of h(x) = x^3 \\space {\\text{–}} \\space 3x^2 \\space {\\text{–}} \\space 10x + 24<\/span>.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nSometimes we aren’t given any factors or roots to start with.\n
\nWe can apply the rational roots test, but we can also just proceed with a few basic initial guesses to find a first factor and root.<\/font>\n
\nh(1) = 1^3 \\space {\\text{–}} \\space 3(1)^2 \\space {\\text{–}} \\space 10(1) + 24 \\space = \\space 1 \\space {\\text{–}} \\space 3 \\space {\\text{–}} \\space 10 + 24 \\space = \\space 13<\/span>\n
\nh({\\text{-}}1) = ({\\text{-}}1)^3 \\space {\\text{–}} \\space 3({\\text{-}}1)^2 \\space {\\text{–}} \\space 10({\\text{-}}1) + 24 \\space = \\space {\\text{-}}1 \\space {\\text{–}} \\space 3 + 10 + 24 \\space = \\space 13<\/span>\n
\nh(2) = 2^3 \\space {\\text{–}} \\space 3(2)^2 \\space {\\text{–}} \\space 10(2) + 24 \\space = \\space 8 \\space {\\text{–}} \\space 12 \\space {\\text{–}} \\space 20 + 24 \\space = \\space 0<\/span>\n
\n2 is a root, so with that information we can proceed with finding the other roots.<\/font>\n\n\n\n
<\/figure>\n\n\n\n\nA remainder of 0 is obtained as expected,\n
\nand we can now factorise fully and find the roots\/solutions.\n
\n(x^2 \\space {\\text{–}} \\space x \\space {\\text{–}} \\space 12)(x \\space {\\text{–}} \\space 2)<\/span>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 2)<\/span> => (x + 3)(x \\space {\\text{–}} \\space 4)(x \\space {\\text{–}} \\space 2)<\/span>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x \\space {\\text{–}} \\space 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 4<\/span>\n
\nThe solutions of h(x) = x^3 \\space {\\text{–}} \\space 3x^2 \\space {\\text{–}} \\space 10x + 24<\/span> are x = {\\text{-}}3 \\space , \\space 2 \\space , \\space 4<\/span>.\n
\n(1.4) <\/i><\/b><\/font><\/font><\/font><\/u>\n
\nShow that (x + 2)<\/span> is a factor of f(x) = x^4 + 8x^3 + 17x^2 \\space {\\text{–}} \\space 2x \\space {\\text{–}} \\space 24<\/span>.\n
\nHence factorise fully and solve.\n
\nSolution<\/i> <\/b><\/font><\/font><\/u><\/font>\n
\nf(x)<\/span> here is a polynomial of degree 4, so there are 4 factors and solutions.<\/span>\n
\nFirstly we deal with showing that (x + 2)<\/span> is a factor.\n
\nf({\\text{-}}2) = ({\\text{-}}2)^4 + 8({\\text{-}}2)^3 + 17({\\text{-}}2)^2 \\space {\\text{–}} \\space 2({\\text{-}}2) \\space {\\text{–}} \\space 24<\/span>\n
\n= 16 \\space {\\text{–}} \\space 64 + 68 + 4 \\space {\\text{–}} \\space 24 \\space = \\space 0<\/span>\n
\n(x + 2)<\/span> is a factor, with -2<\/b> being a zero.\n
\nWith that information we can proceed with finding the other roots.<\/font><\/font>\n
\n\n\n\n
<\/figure>\n\n\n\n\nA remainder of 0 is obtained as expected,\n
\nwe can now look to factorise further.\n
\n(x + 2)(x^3 + 6x^2 + 5 \\space {\\text{–}} \\space 12)<\/span>\n
\nWe can proceed with factoring with synthetic division for x^3 + 6x^2 + 5 \\space {\\text{–}} \\space 12<\/span>, but we don’t know any exact zeros\/roots to start with.\n
\nThe rational roots theorem<\/i><\/a> can help us find values to try.\n
\nThough sometimes we can try some basic values first, as we may get lucky and find a zero\/root.\n
\nA good value to try initially is usually 0 or 1.\n
\n\n\n\n
<\/figure>\n\n\n\n\nFactoring further.\n
\nSuccess on this occasion, we have quickly found another zero\/root.\n
\nIt doesn’t always work out like this, but it’s nice when it does.\n
\nNow we can factor fully and solve.\n
\n(x^2 + 7x + 12)(x \\space {\\text{–}} \\space 1)(x + 2)<\/span>\n
\n(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space \\space \\space \\space )(x \\space {\\text{–}} \\space 1)(x + 2)<\/span> => (x + 3)(x + 4)(x \\space {\\text{–}} \\space 1)(x + 2)<\/span>\n
\nx + 3 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}3 \\space \\space \\space , \\space \\space \\space x + 4 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}4<\/span>\n
\nx \\space {\\text{–}} \\space 1 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = 1 \\space \\space \\space , \\space \\space \\space x + 2 = 0 \\space \\space {\\scriptsize{=>}} \\space \\space x = {\\text{-}}2<\/span>\n
\nThe solutions of f(x) = x^4 + 8x^3 + 17x^2 \\space {\\text{–}} \\space 2x \\space {\\text{–}} \\space 24<\/span> are x = {\\text{-}}4 \\space , \\space {\\text{-}}3 \\space , \\space {\\text{-}}2 \\space , \\space 1<\/span>.\n
\n\n\n