{"id":2841,"date":"2023-10-03T13:59:10","date_gmt":"2023-10-03T13:59:10","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2841"},"modified":"2023-10-03T14:03:22","modified_gmt":"2023-10-03T14:03:22","slug":"week-5-ss2-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-5-ss2-first-term-further-mathematics-notes\/","title":{"rendered":"Week 5 – SS2 First Term Further Mathematics Notes"},"content":{"rendered":"

WEEK 5
\n<\/strong>TOPIC: Cubic equations and their factorization , graphs of cubic equations
\n<\/strong>Polynomials of degree three have the general form y = ax3<\/sup> + bx2<\/sup> + cx + d(a \u2260 0). The curve is usually called a cubical parabola<\/strong>.
\nA cubical parabola has two shapes depending on whether a > 0 or a < 0.
\n\"\"\/<\/p>\n

\u00a0
\n\u00a0\"\"\/
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0a> 0<\/p>\n

\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0a< 0
\n\u00a0\u00a0\u00a0\u00a0
\n\u00a0
\n\u00a0
\n\u00a0Sketch each of the following curves represented by the following functions:
\n(a) y = x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6
\n(b) y = 12 + 4x \u2013 3x2<\/sup> \u2013 x3<\/sup><\/p>\n

\u00a0Solution
\n<\/strong>(a) y = x3<\/sup> + 2x2<\/sup> \u2013 5x \u2013 6
\nUsing the factor theorem and long division the expression can be factorized as:
\ny = (x \u2013 2)(x + 1)(x + 3)
\nThe zeros of the polynomial are therefore x = 2, x = -1 and x = 3, hence the x \u2013 intercepts are (2, 0), (-1, 0) and (-3, 0).
\nThe y \u2013 intercept is (0, -6).
\nNext, we shall consider the behaviour of the function at different intervals along the x \u2013 axis. This will enable us to see whether the curve is above or below the axis.
\nMark the x \u2013 intercepts on the x \u2013 axis.
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x\u2013axis
\n\u00a0\u00a0\u00a0\u00a0(-3, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(-1, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2, 0)
\nThe intervals we shall consider are:
\n(a) x< -3
\n(b) -3 < x < -1
\n(c) -1 < x < 2
\n(d) x> 2
\nWe shall examine the signs (+ve or \u2013ve) in each of the intervals.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
\nx = -4 is in the intervals x < -3
\nf(-4) = -18 < 0
\nHence the part of the graph in the interval x < -3 is below the axis.\u00a0\u00a0\u00a0\u00a0
\nx = -2 is in the interval -3 < x <-1
\nf(-2) = 4 > 0
\nHence the part of the graph in the interval -3 < x < 2 is above the x \u2013 axis x = 0 is in the interval -1 < x < 2
\nf(0) = -6 < 0.
\nHence the part of the graph in the interval -1 < x < 2 is below the x \u2013 axis.
\nx = 3 is in the interval x > 2
\nf(3) = 24 > 0
\nHence the part of the graph in the interval is above the x \u2013 axis.
\nThe intercept on the axis coupled with the behaviour of the function at different intervals on the x \u2013 axis will enable us to get the shape of the curve.<\/p>\n

\u00a0
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\n\u00a0\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y<\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (-3, 0)(-1, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2, 0)<\/p>\n

\u00a0
\n\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (0, 6)<\/p>\n

\u00a0(b) y = 12 + 4x \u2013 3x2<\/sup> \u2013 x3<\/sup>
\n\t\tUsing the factor theorem and long division method
\ny = (2 \u2013 x)(2 + x)(3 + x)
\nThe zeros of the polynomial are x = 2, x = -2 and x = -3, hence the x \u2013 intercepts are (2, 0), (-2, 0) and (-3, 0).
\nThe y \u2013 intercept is (0, 12).
\nMark the x \u2013 intercept on the axis.
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x \u2013 axis
\n\u00a0\u00a0\u00a0\u00a0(-3, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(-1, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(2, 0)
\nThe intervals we shall consider are:
\n(a) x< -3
\n(b) -3 < x < -1
\n(c) -1 < x < 2
\n(d) x> 2
\nx = -4 is in the interval x < -3
\nf(-4) = 12 < 0,
\nhence the part of the graph within this interval is above the x \u2013 axis.
\nx = -2.5 is in the interval -3 < x < -2
\nf(-2.5) =
\nhence the part of the graph within this interval -3 < x < -2 is above the axis.
\nx = 0 is in the interval -2 < x < 2
\nf(0) = 12 > 0,
\nhence the part of the graph within this interval -2< x <2 is above the axis.
\nx = 3 is in the interval x > 2
\nf(3) = -30< 0,
\nhence the part of the graph within the interval r > 2 is below the x \u2013 axis.
\n\"\"\/y\u00a0\u00a0\u00a0\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (0, 12)<\/p>\n

\u00a0
\n\u00a0
\n\u00a0
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (-3, 0)\u00a0\u00a0\u00a0\u00a0 (-2, 0)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2, 0)<\/p>\n

\u00a0
\n\u00a0
\n\u00a0Evaluation
\n<\/strong>Sketch the curve of these equations
\n(a) y = x3<\/sup> \u2013 6x2<\/sup> + 11x \u2013 6\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) y = x3<\/sup> + 3x2<\/sup> \u2013 6x \u2013 8 <\/p>\n

\u00a0General Evaluation
\n<\/strong>(1) Factorise the following completely
\n(a) x3<\/sup> + 10x2<\/sup> + 23x + 14\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) x4<\/sup> – 1 <\/p>\n

\u00a0Weekend Assignment
\n<\/strong>Given that a cubic equation x3<\/sup> + 2x2<\/sup> \u2013 19x \u2013 20 = 0 has 4 as one its roots, find the
\n(1) Second root\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a) -1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) 3
\n(2) Third root\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a) 5\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -8\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) 2(3) Sum of the second and third roots\u00a0\u00a0\u00a0\u00a0(a) -4\u00a0\u00a0\u00a0\u00a0 (b) 6\u00a0\u00a0\u00a0\u00a0 (c) 4\u00a0\u00a0\u00a0\u00a0(d) -6
\n(4) Product of the second and third roots\u00a0\u00a0\u00a0\u00a0 (a) 6 (b) 5 (c) -6 (d) -5
\n(5) Find the zeros of x2<\/sup> \u2013 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a) 2 or -2 (b) 1 or 2 (c) -1 or 1 (d) 1 or -2<\/p>\n

\u00a0Theory
\n<\/strong>(1) If (x + 1) is a factor of f(x) = x3<\/sup> + kx2<\/sup> + 3x + 10, find the value of the constant k.
\n(2) Factorise f(x) completely.<\/p>\n

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

WEEK 5 TOPIC: Cubic equations and their factorization , graphs of cubic equations Polynomials of…<\/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-2841","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\/2841","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=2841"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2841\/revisions"}],"predecessor-version":[{"id":2842,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2841\/revisions\/2842"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2841"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2841"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2841"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}