{"id":2518,"date":"2020-12-22T10:31:14","date_gmt":"2020-12-22T05:01:14","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=2518"},"modified":"2020-12-22T12:58:00","modified_gmt":"2020-12-22T07:28:00","slug":"factorization-of-polynomials-using-factor-theorem","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/factorization-of-polynomials-using-factor-theorem\/","title":{"rendered":"Factorization Of Polynomials Using Factor Theorem"},"content":{"rendered":"
If p(x) is a polynomial of degree n \uf0b3 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x). People also ask<\/strong><\/p>\n Example 1: \u00a0 \u00a0<\/strong>Factorize x2<\/sup>\u00a0+4 + 9 z2<\/sup> + 4x \u2013 6 xz \u2013 12 z Example 2: \u00a0<\/strong> \u00a0Using factor theorem, factorize the polynomial x3<\/sup> \u2013 6x2<\/sup>\u00a0+ 11 x \u2013 6. Example 3: \u00a0 \u00a0<\/strong>Using factor theorem, factorize the polynomial x4<\/sup>\u00a0+ x3<\/sup> \u2013 7x2<\/sup> \u2013 x + 6. Example 4: \u00a0 \u00a0<\/strong>Factorize,\u00a0 2x4<\/sup>\u00a0+ x3<\/sup> \u2013 14x2<\/sup> \u2013 19x \u2013 6 Example 5: \u00a0 \u00a0<\/strong>Factorize,\u00a0 9z3<\/sup>\u00a0\u2013 27z2<\/sup> \u2013 100 z+ 300, if it is given that (3z+10) is a factor of it. Example 6: \u00a0 \u00a0<\/strong>Simplify Example 7: \u00a0 \u00a0<\/strong>Establish the identity Factorization Of Polynomials Using Factor Theorem Factor Theorem: If p(x) is a polynomial of degree n \uf0b3 1 and a is any real number, then (i) x \u2013 a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x \u2013 a is a factor of p(x). Proof: By the […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[5],"tags":[1072,1088,1089,1090,24],"yoast_head":"\n
\nProof:<\/strong> By the Remainder Theorem,
\np(x) = (x \u2013 a) q(x) + p(a).
\n(i) If p(a) = 0, then p(x) = (x \u2013 a) q(x), which shows that x \u2013 a is a factor of p(x).
\n(ii) Since x \u2013 a is a factor of p(x),
\np(x) = (x \u2013 a) g(x) for same polynomial g(x). In this case, p(a) = (a \u2013 a) g(a) = 0.<\/p>\n\n
\nx4<\/sup> + x3<\/sup> \u2013 7x2<\/sup> \u2013 x + 6 the constant term is 6 and its factors are \u00b1 1, \u00b1 2, \u00b1 3, \u00b1 6.<\/li>\n\n
Factorization Of Polynomials Using Factor Theorem Example Problems With Solutions<\/h2>\n
\nSolution:<\/strong>
\nThe presence of the three squares viz.x2<\/sup>, (2)2<\/sup>, and (3z)2<\/sup> gives a clue that identity (vii) could be used. So we write.
\nA = x2<\/sup> + (2)2<\/sup> + (3z)2<\/sup> + 4x \u2013 6 xz \u2013 12 z
\nWe note that the last two of the product terms are negative and that both of these contain z. Hence we write A as
\nA = x2<\/sup> + (2)2<\/sup> + (\u20133z)2<\/sup> + 2.2x \u2013 2.x.(\u20133z) + 2.2 (\u2013 3z)
\n= (x+2 \u2013 3z)2<\/sup>
\n= (x + 2 \u2013 3z) (x + 2 \u2013 3z)<\/p>\n
\nSolution:<\/strong>
\nLet f(x) = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6
\nThe constant term in f(x) is equal to \u2013 6 and factors of \u2013 6 are \u00b11, \u00b1 2, \u00b1 3, \u00b1 6.
\nPutting x = 1 in f(x), we have
\nf(1) = 13<\/sup> \u2013 6 \u00d712<\/sup> + 11\u00d7 1\u2013 6
\n= 1 \u2013 6 + 11\u2013 6 = 0
\n\u2234 (x\u2013 1) is a factor of f(x)
\nSimilarly, x \u2013 2 and x \u2013 3 are factors of f(x).
\nSince f(x) is a polynomial of degree 3. So, it can not have more than three linear factors.
\nLet f(x) = k (x\u20131) (x\u2013 2) (x \u2013 3). Then,
\nx3<\/sup>\u2013 6x2<\/sup> + 11x \u2013 6 = k(x\u20131) (x\u2013 2) (x\u2013 3)
\nPutting x = 0 on both sides, we get
\n\u2013 6 = k (0 \u2013 1) (0 \u2013 2) (0 \u2013 3)
\n\u21d2 \u2013 6 = \u2013 6 k \u21d2 k = 1
\nPutting k = 1 in f(x) = k (x\u2013 1) (x\u2013 2) (x\u20133), we get
\nf(x) = (x\u20131) (x\u2013 2) (x \u2013 3)
\nHence, x3<\/sup>\u20136x2<\/sup> + 11x \u2013 6 = (x\u2013 1) (x \u2013 2) (x\u20133)<\/p>\n
\nSolution:<\/strong>
\nLet f(x) = x4<\/sup> + x3<\/sup>\u2013 7x2<\/sup> \u2013x + 6
\nthe factors of constant term in f(x) are \u00b11, \u00b12, \u00b13 and \u00b1 6
\nNow,
\n
\nSince f(x) is a polynomial of degree 4. So, it cannot have more than 4 linear factors
\nThus, the factors of f (x) are (x\u20131), (x+1),
\n(x\u20132) and (x+3).
\nLet f(x) = k (x\u20131) (x+1) (x\u20132) (x + 3)
\n\u21d2 x4<\/sup> + x3<\/sup> \u2013 7x2<\/sup> \u2013 x + 6
\n= k (x\u20131) (x +1) (x \u2013 2) (x + 3)
\nPutting x = 0 on both sides, we get
\n6 = k (\u20131) (1) (\u20132) (3) \u21d2 6 = 6 k \u21d2 k = 1
\nSubstituting k = 1 in (i), we get
\nx4<\/sup> + x3<\/sup> \u2013 7x2<\/sup> \u2013 x + 6 = (x\u20131) (x +1) (x\u20132) (x+3)<\/p>\n
\nSolution:<\/strong>
\nLet f(x) = 2x4<\/sup> + x3<\/sup> \u2013 14x2<\/sup> \u2013 19x \u2013 6 be the given polynomial. The factors of the constant term \u2013 6 are \u00b11, \u00b12, \u00b13 and \u00b16, we have,
\nf(\u20131) = 2(\u20131)4<\/sup> + (\u20131)3<\/sup> \u2013 14(\u20131)2<\/sup> \u2013 19(\u20131)\u2013 6
\n= 2 \u2013 1 \u2013 14 + 19 \u2013 6 = 21 \u2013 21 = 0
\nand,
\nf(\u20132) = 2(\u20132)4<\/sup> + (\u20132)3<\/sup> \u2013 14(\u20132)2<\/sup> \u2013 19(\u20132)\u2013 6
\n= 32 \u2013 8 \u2013 56 + 38 \u2013 6 = 0
\nSo, x + 1 and x + 2 are factors of f(x).
\n\u21d2 (x + 1) (x + 2) is also a factor of f(x)
\n\u21d2 x2<\/sup> + 3x + 2 is a factor of f(x)
\nNow, we divide
\nf(x) = 2x4<\/sup> +x3<\/sup> \u2013 14x2<\/sup>\u201319x \u2013 6 by
\nx2<\/sup> + 3x + 2 to get the other factors.
\n<\/p>\n
\nSolution:<\/strong>
\nLet us divide 9z3<\/sup> \u2013 27z2<\/sup> \u2013 100 z+ 300 by
\n3z + 10 to get the other factors
\n
\n\u2234 9z3<\/sup> \u2013 27z2<\/sup> \u2013 100 z+ 300
\n= (3z + 10) (3z2<\/sup>\u201319z + 30)
\n= (3z + 10) (3z2<\/sup>\u201310z \u2013 9z + 30)
\n= (3z + 10) {(3z2<\/sup>\u201310z) \u2013 (9z \u2013 30)}
\n= (3z + 10) {z(3z\u201310) \u2013 3(3z\u201310)}
\n= (3z + 10) (3z\u201310) (z\u20133)
\nHence, 9z3<\/sup>\u201327z2<\/sup>\u2013100z+ 300
\n= (3z + 10) (3z\u201310) (z\u20133)<\/p>\n
\n\\(\\frac{4x-2}{{{x}^{2}}-x-2}+\\frac{3}{2{{x}^{2}}-7x+6}-\\frac{8x+3}{2{{x}^{2}}-x-3}\\)
\nSolution:<\/strong>
\n<\/p>\n
\n\\(\\frac{6{{x}^{2}}+11x-8}{3x-2}=\\left( 2x+5 \\right)+\\frac{2}{3x-2}\\)
\nSolution:<\/strong>
\n<\/p>\n","protected":false},"excerpt":{"rendered":"