{"id":2839,"date":"2023-10-03T13:58:34","date_gmt":"2023-10-03T13:58:34","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2839"},"modified":"2023-10-03T14:03:22","modified_gmt":"2023-10-03T14:03:22","slug":"week-4-ss2-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-4-ss2-first-term-further-mathematics-notes\/","title":{"rendered":"Week 4 – SS2 First Term Further Mathematics Notes"},"content":{"rendered":"

WEEK 4
\n<\/strong>TOPIC: polynomial (continued)
\n<\/strong>CONTENT:
\n<\/strong>Factorization of polynomial
\n<\/strong>Show that x – 1 is a factor off(x) = 2x3<\/sup> + 3x2<\/sup> \u2013 5x \u2013 6<\/p>\n

\u00a0Solution
\n<\/strong>f(-1) \u00a0\u00a0\u00a0\u00a0= 2(-1)3<\/sup> + 3(-1)2<\/sup> \u2013 5(-1) \u2013 6
\n\u00a0\u00a0\u00a0\u00a0= -2 + 3 + 5 \u2013 6
\nHence x + 1 is a factor of f(x)<\/p>\n

\u00a0Show that x + 1 is a factor of f(x) = x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6
\nHence factorise f(x) completely.<\/p>\n

\u00a0Solution
\n<\/strong>f(-1)\u00a0\u00a0\u00a0\u00a0=(-1)3<\/sup> + 2(-1)2<\/sup> \u2013 5(-1) \u2013 6
\n\u00a0\u00a0\u00a0\u00a0= -1 + 2 + 5 \u2013 6 = 0
\n: x+1 is a factor of f(x)
\nUsing Long Division:
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + x – 5
\n\"\"\/(a)\u00a0\u00a0\u00a0\u00a0x + 1 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 2x2<\/sup> \u2013 5x – 6
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + x2<\/sup>
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 5x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-6x – 6
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-6x \u2013 6
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00
\nx2<\/sup> + x \u2013 6 \u00a0\u00a0\u00a0\u00a0= x2<\/sup> + 3x2<\/sup> \u2013 2x \u2013 6
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= x(x + 3) -2 (x + 3)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= (x + 3)(x \u2013 2)
\nHence f(x)\u00a0\u00a0\u00a0\u00a0= (x + 1)(x + 3) (x \u2013 2)<\/p>\n

\u00a0Factiorizef(x) = x3<\/sup> + 7x2<\/sup> \u2013 14x \u2013 8 completely.<\/p>\n

\u00a0Solution
\n<\/strong>In a complete factorised form, f(x) can be written in the form:
\nf(x) = (x \u00b1 p)(x \u00b1 r)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026(1)
\nThe first term of the expansion of (1) is x3<\/sup> while the last term of the expansion is \u00b1pqr. Hence, p. q and r must be factors of -8. We can therefore try:
\nx \u00b1 1
\nx \u00b1 2
\nx \u00b1 4
\nx \u00b1 8
\nf(-1) = -1 \u2013 7 \u2013 14 \u2013 8 = -30 \u2260 0
\n: x + r is not a factor of f(x)
\nf(1) = 1 \u2013 7 + 14 \u2013 8 = 0
\n: x \u2013 1 is a factor of f(x).
\nA similar procedure can be used for x \u00b1 2, x \u00b1 4, x \u00b1 8 to find the other two factors. But a less cumbersome procedure is to use long division method once a factor of f(x) is obtained.<\/p>\n

\u00a0x2<\/sup> \u2013 6x + 8
\n\"\"\/\u00a0\u00a0\u00a0\u00a0x \u2013 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 7x2<\/sup> + 14x \u2013 8
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-6x2<\/sup> + 14x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-6x2<\/sup>+ 6x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08x \u2013 8
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08x \u2013 8
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 0
\nNow, x2<\/sup> \u2013 6x + 8\u00a0\u00a0\u00a0\u00a0= x2<\/sup> \u2013 4x \u2013 2x + 8
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= (x \u2013 2)(x \u2013 4)
\nHence, f(x) = (x \u2013 1)(x \u2013 2)(x \u2013 4)<\/p>\n

\u00a0What must be subtracted from f(x) = x2<\/sup>+2x2<\/sup>-3x +5 so that it will be exactly divisible by x \u2013 2? Hence, find that polynomial H(x) which is exactly divisible by x \u2013 2 and factorise it completely.<\/p>\n

\u00a0Solution
\n<\/strong>f(x) = x2<\/sup> + 2x2<\/sup> – 3x + 5
\nf(2) = 8 + 8 \u2013 6 + 5
\n\u00a0\u00a0\u00a0\u00a0= 15
\n:When f(x) is divided by x \u2013 2 the remainder is 15. Hence 15 must be subtracted from f(x) to make it exactly divisible by x \u2013 2.
\nLet the new polynomial which is exactly divisible by x \u2013 2 be H(x), then:
\nH(x) \u00a0\u00a0\u00a0\u00a0= f(x) \u2013 15
\n\u00a0\u00a0\u00a0\u00a0= x2<\/sup> + 2x2<\/sup> – 3x + 5\u2013 15
\n\u00a0\u00a0\u00a0\u00a0= x2<\/sup> + 2x2<\/sup> – 3x \u201310<\/p>\n

\u00a0Using long division
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 4x +5
\n\"\"\/x\u2013 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 2x2<\/sup> – 3x \u2013 10
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 2x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x2<\/sup> \u2013 3x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x2<\/sup> \u2013 8x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a05x \u2013 10
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a05x \u2013 10
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00
\nBut x2<\/sup> + 4x + 5 is not factorizable, hence H(x) = (x \u2013 2)(x2<\/sup> + 4x + 5).
\nGiven that f(x \u2013 1) = x2<\/sup> + 3x \u20131, find f(3).
\nPut\u00a0\u00a0\u00a0\u00a0x + 1 = 3
\n\u00a0\u00a0\u00a0\u00a0x = 2
\n: f(3)\u00a0\u00a0\u00a0\u00a0= (2)2<\/sup> + 3(2) \u2013 1
\n\u00a0\u00a0\u00a0\u00a0= 4 + 6 \u2013 1
\n\u00a0\u00a0\u00a0\u00a0= 9<\/p>\n

\u00a0(1) Given that p1<\/sub>(x) = 5x3<\/sup> + 3x2<\/sup> \u2013 2x + 6, p2<\/sub>(x) = x3<\/sup> + 4x2<\/sup> \u2013 3x + 1 and p3<\/sub>(x) 2x3<\/sup> \u2013 3x + 2, find
\n(i) (p2<\/sub> \u2013 p3<\/sub>) p1<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii)p2<\/sub>(p1<\/sub> + p3<\/sub>)<\/p>\n

\u00a0
\n\u00a0(2) Divide (i) 4x3<\/sup> + 3x2<\/sup> \u2013 2x + 1 by x2<\/sup> + 2x \u2013 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii) x4<\/sup> + 2x + 3 by x2<\/sup> \u2013 1 to obtain the
\nremainder and quotient.<\/p>\n

\u00a0Evaluation<\/strong>
\n\t\tFind the quadratic equation whose roots are (i) 3 & -2\u00a0\u00a0\u00a0\u00a0(ii) -1 & 8\u00a0\u00a0\u00a0\u00a0(iii) \u00be & \u00bd <\/p>\n

\u00a0General Evaluation
\n<\/strong>(1) When x2<\/sup> + bx + 2 is divided by x + 3 the remainder is 5. Find the value of b.
\n(2) If 2x2<\/sup> \u2013 (b \u2013 4) x \u2013 4 (b + 2) = 0, has equal roots, find the possible values of b.
\n(3) Factorise completely x3<\/sup> + 5x2<\/sup> \u2013 3x + 1
\n(4) Solve the following pair of equation simultaneously 4x \u2013 3y = 17, 3x2<\/sup> – 2y2<\/sup> + x \u2013 4y = 73<\/p>\n

\u00a0Reading Assignment
\n<\/strong>Further Maths 1 pages 66 \u2013 69 Exercise 6a Q5, 9 and 10<\/p>\n

\u00a0Weekend Assignment
\n<\/strong>(1) Find the remainder when 2x3<\/sup> \u2013 4x2<\/sup> + x \u2013 3 is divided by x + 3
\n(a) 84\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -86\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) -64\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) 76
\n(2) Factorise x4<\/sup> \u2013 1 completely\u00a0\u00a0\u00a0\u00a0(a) (x + 2) (x \u2013 3)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) (x + 1) (x \u2013 1) (x2<\/sup> + 1)
\n(c) (x + 1) (x + 2) (x \u2013 3)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) (x2<\/sup> + 1) (x + 1) (x \u2013 2)
\n(3) Given that p1<\/sub>(x) = 2x4<\/sup> + 3x3<\/sup> \u2013 x2<\/sup> + 2x \u2013 3 and p1<\/sub>(x) = 3x3<\/sup> + 2x + 2. Find 3p1<\/sub>(x) \u2013 3p1<\/sub>(x)
\n(a) 6x4<\/sup> \u2013 3x2<\/sup> \u2013 15\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 5x4<\/sup> \u2013 3x4<\/sup> + 2x2<\/sup> \u2013 x + 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 6x4<\/sup> = 9x3<\/sup> \u2013 15\u00a0\u00a0\u00a0\u00a0
\n(d) 6x4<\/sup> + 18x2<\/sup> + 6x2<\/sup>
\n\t\t(4) Given that p(x) = x3<\/sup> + 4x2<\/sup> \u2013 3x + 1, find p( \u00bd )
\n(a)
\n(5) Given that p(x) = ax2<\/sup> + bx + 1, p( \u00bd ) = \u00bd and p(-2) = 23, determine the values of a and b
\n(a) a = -3, b = 4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) a = 2, b = -3\u00a0\u00a0\u00a0\u00a0(c) a = 4, b = -3\u00a0\u00a0\u00a0\u00a0(d) a = 3, b = -3<\/p>\n

\u00a0Theory
\n<\/strong>(1) Find the quotient and remainder when 2x4<\/sup> \u2013 3x3<\/sup> + x2<\/sup> \u2013 4x + 5 is divided by x2<\/sup> + 3x + 1
\n(2) If p1<\/sub> = 3x3<\/sup> + 2x2<\/sup> \u2013 x + 2, p2<\/sub> = 2x2<\/sup> + x \u2013 6 and p3<\/sub> = x3<\/sup> + 3x2<\/sup> + 2x \u2013 4, find
\n(i) p3<\/sub>(p1<\/sub> + p2<\/sub>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(ii) p2<\/sub> + p3<\/sub> \u2013 3p1<\/sub>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(iii) p2<\/sub> x (p3<\/sub> + p1<\/sub>)<\/p>\n

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\n\t\t\t<\/strong>\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"

WEEK 4 TOPIC: polynomial (continued) CONTENT: Factorization of polynomial Show that x – 1 is…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,230],"tags":[],"class_list":["post-2839","post","type-post","status-publish","format-standard","hentry","category-posts","category-first-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2839","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2839"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2839\/revisions"}],"predecessor-version":[{"id":2840,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2839\/revisions\/2840"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2839"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2839"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2839"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}