{"id":2484,"date":"2016-09-08T09:00:38","date_gmt":"2016-09-08T09:00:38","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=2484"},"modified":"2016-09-08T09:59:00","modified_gmt":"2016-09-08T09:59:00","slug":"factor-theorem","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/factor-theorem\/","title":{"rendered":"How Do You Use The Factor Theorem"},"content":{"rendered":"
Theorem:<\/strong> If p(x) is a polynomial of degree n \u2265\u00a01 and a is any real number, then To use factor theorem<\/strong><\/p>\n Example 1:<\/strong> \u00a0 \u00a0Examine whether x + 2 is a factor of x3<\/sup> + 3x2<\/sup> + 5x + 6 and of 2x + 4. Example 2: \u00a0 \u00a0<\/strong>Use the factor theorem to determine whether x \u2013 1 is a factor of Example 3: \u00a0 \u00a0<\/strong>Factorize each of the following expression, given that x3<\/sup>\u00a0+ 13 x2<\/sup>\u00a0+ 32 x + 20. (x+2) is a factor. Example 4: \u00a0 \u00a0<\/strong>Factorize x3\u00a0\u2013 23 x2<\/sup>\u00a0+ 142 x \u2013 120 Example 5: \u00a0 \u00a0<\/strong>Show that (x \u2013 3) is a factor of the polynomial x3<\/sup>\u00a0\u2013 3x2<\/sup> + 4x \u2013 12 Example 6: \u00a0 \u00a0<\/strong>Show that (x \u2013 1) is a factor of x10<\/sup>\u00a0\u2013 1 and also of x11<\/sup> \u2013 1. Example 7: \u00a0 \u00a0<\/strong>Show that x + 1 and 2x \u2013 3 are factors of 2x3<\/sup>\u00a0\u2013 9x2<\/sup>\u00a0+ x + 12. Example 8: \u00a0 \u00a0<\/strong>Find the value of k, if x + 3 is a factor of 3x2<\/sup> + kx + 6. Example 9: \u00a0 \u00a0<\/strong>If ax3<\/sup>\u00a0+ bx2<\/sup> + x \u2013 6 has x + 2 as a factor and leaves a remainder 4 when divided by (x \u2013 2), find the values of a and b. Example 10: \u00a0 \u00a0<\/strong>If both x \u2013 2 and x \u2013 1\/2 are factors of px2<\/sup> + 5x + r, show that p = r. Example 11: \u00a0 \u00a0<\/strong>If x2<\/sup>\u00a0\u2013 1 is a factor of ax4<\/sup>\u00a0+ bx3<\/sup>\u00a0+ cx2<\/sup>\u00a0+ dx + e, show that a + c + e = b + d = 0. Example 12: \u00a0 \u00a0<\/strong>Using factor theorem, show that a \u2013 b, b \u2013 c and c \u2013 a are the factors of\u00a0a(b2<\/sup>\u00a0\u2013 c2<\/sup>) + b(c2<\/sup>\u00a0\u2013 a2<\/sup>) + c(a2<\/sup>\u00a0\u2013 b2<\/sup>). <\/p>\n","protected":false},"excerpt":{"rendered":" Factor Theorem Theorem: If p(x) is a polynomial of degree n \u2265\u00a01 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 Remainder Theorem, p(x) = (x \u2013 … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[1072,1073,24],"yoast_head":"\n
\n(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).
\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),
\nwhich 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).
\nIn this case, p(a) = (a \u2013 a) g(a) = 0.<\/p>\n\n
Factor Theorem Example Problems With Solutions<\/strong><\/h2>\n
\nSolution:<\/strong> \u00a0 \u00a0The zero of x + 2 is \u20132.
\nLet p(x) = x3<\/sup> + 3x2<\/sup> + 5x + 6 and s(x) = 2x + 4
\nThen,\u00a0\u00a0\u00a0 p(\u20132) = (\u20132)3<\/sup> + 3(\u20132)2<\/sup> + 5(\u20132) + 6
\n= \u20138 + 12 \u2013 10 + 6
\n= 0
\nSo, by the Factor Theorem, x + 2 is a factor of x3<\/sup> + 3x2<\/sup> + 5x + 6.
\nAgain,\u00a0\u00a0 s(\u20132) = 2(\u20132) + 4 = 0
\nSo, x + 2 is a factor of 2x + 4.<\/p>\n
\n(a) x3<\/sup>\u00a0+ 8x2<\/sup> \u2013 7x \u2013 2
\n(b) 2x3<\/sup> + 5x2<\/sup> \u2013 7
\n(c) 8x4<\/sup> + 12x3<\/sup> \u2013 18x + 14
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\nSolution: \u00a0\u00a0<\/strong>
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