{"id":2837,"date":"2023-10-03T13:58:02","date_gmt":"2023-10-03T13:58:02","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=2837"},"modified":"2023-10-03T14:03:22","modified_gmt":"2023-10-03T14:03:22","slug":"week-3-ss2-first-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-3-ss2-first-term-further-mathematics-notes\/","title":{"rendered":"Week 3 – SS2 First Term Further Mathematics Notes"},"content":{"rendered":"

\u00a0WEEK 3
\n<\/strong>TOPIC: Polynomials
\n<\/strong>Consider the expressions formed from the sum of integral non negative powers of a variable x taken together with some numerical constants. Such expressions are called Polynomials.
\n<\/strong>The general polynomial takes the form an<\/sub>Xn<\/sup> + an1<\/sub>Xn1<\/sup> + \u2026 a2<\/sub>X2<\/sup> + a1<\/sub>X + a0<\/sub> where a11<\/sub> an-1<\/sub>\u2026a0<\/sub>(a0<\/sub> \u22600) are numerical constants.
\nThe numerical constants an’<\/sub> an1<\/sub>\u2026. a2<\/sub>, a1<\/sub> are called coefficients of Xn<\/sup>, Xn1<\/sup>, \u2026 X2<\/sup>, X respectively while a0<\/sub> is called the constant term of the polynomial.
\nThe highest power of the variable is n and is called the degree of the polynomial. Let us designate the general polynomial by p(x). Thus:
\nP(x) =an<\/sub>Xn<\/sup> + an-1<\/sub> Xn-1<\/sup> + \u2026 + a2<\/sub>X2<\/sup> + a1<\/sub>X + a0<\/sub>.
\nThe following are examples of polynomials:
\n(a) P1<\/sub>(x) = 3x2<\/sup> \u2013 2x + 4
\n(b) P2<\/sub>(x) = 3x4<\/sup> \u2013 2x2<\/sup> + x \u2013 1
\n(c) P3<\/sub>(x) = x + 1
\n(d) P1<\/sub>(x) = 2x3<\/sup> + x – 3
\nThe following are not polynomials:
\n(a) f1<\/sub>(x) = (x2<\/sup> + 2x \u2013 3)
\n(b) f1<\/sub>(x) = 3x2<\/sup> \u2013 4x2<\/sup> + 2x \u2013 1
\n\u00a0\u00a0\u00a0\u00a02x + 3
\n(c) f1<\/sub>(x) = (2x \u2013 3)1<\/sup><\/p>\n

\u00a0Equality of Polynomials
\n<\/strong>The polynomial p(x) = a11<\/sub>X11<\/sup> + an-1<\/sub> Xn-1<\/sup> + \u2026 + a2<\/sub>X2<\/sup> + a1<\/sub>X + a0<\/sub> is said to be equal to the polynomial.
\nQ(x) =bn<\/sub>Xn<\/sup> + bn-1<\/sub>X-n1<\/sup> + \u2026 b2<\/sub>X2<\/sup> + b1<\/sub> X + b0<\/sub> provided
\nan<\/sub> = bn’<\/sub> an 1<\/sub> = bn 1<\/sub> \u2026 a2<\/sub> = b2′<\/sub> a1<\/sub> = b1,<\/sub> a0<\/sub> = b0<\/sub><\/p>\n

\u00a0Addition and Subtraction of Polynomials
\n<\/strong>Let P(x) = an<\/sub>Xn<\/sup> + an-1<\/sub> Xn-1<\/sup> + \u2026 + a2<\/sub>X2<\/sup> + a1<\/sub>X + a0
\n<\/sub>Q(x) = bn<\/sub>Xn<\/sup> + bn-1<\/sub>X-n1<\/sup> + \u2026 b2<\/sub>X2<\/sup> + b1<\/sub> X + b0<\/sub> then,
\nP(x) + Q(x) = (an<\/sub> + bn<\/sub>)Xn<\/sup> + (an-1<\/sub> + bn-1<\/sub>) Xn1<\/sup> + \u2026 + (a2<\/sub> + b2<\/sub>) X2<\/sup> + (a1<\/sub> + b1<\/sub>) X + a0<\/sub> + b0<\/sub>.
\nAlso,
\nP(x) – Q(x) = (an<\/sub>-bn<\/sub>) Xn<\/sup>– (an-1<\/sub>– bn-1<\/sub>) Xn1<\/sup>– \u2026 + (a2<\/sub>– b2<\/sub>) X2<\/sup>+ (a1<\/sub>– b1<\/sub>) X + a0<\/sub>– b0<\/sub>.
\nGiven that P1<\/sub>(x) = 7x3<\/sup> \u2013 4x2<\/sup> + 3x + 4;
\nP2<\/sub>(x) = 5x2<\/sup> + 6x + 1 and P3<\/sub>(x) = 4x3<\/sup> + 2x \u2013 3.
\nFind:
\n(a) P1<\/sub>(x) + P2<\/sub>(x)
\n(b) P1<\/sub>(x) + P3<\/sub>(x)
\n(c) P1<\/sub>(x) \u2013 P2<\/sub>(x)
\n(d) P3 <\/sub>(x) \u2013 P2<\/sub>(x)
\n(e) P1<\/sub>(x) + P2<\/sub>(x) + P3<\/sub>(x)<\/p>\n

\u00a0
\n\u00a0Solution
\n<\/strong>(a) P1<\/sub>(x) + P2<\/sub>(x)
\n= (7x3<\/sup> \u2013 4x2<\/sup> + 3x + 4) + (5x2<\/sup> + 6x + 1)
\n= 7x3<\/sup> + (-4x2<\/sup> + 5x2<\/sup>) + (3x + 6x) + (4 + 1)
\n= 7x3<\/sup> + x2<\/sup> + 9x + 5<\/p>\n

\u00a0(b) P1<\/sub> (x) + P3<\/sub>(x)
\n(7x3<\/sup> \u2013 4x2<\/sup> + 3x + 4) + (4x3<\/sup> + 2x \u2013 3)
\n= (7x3<\/sup> + 4x3<\/sup>) + (-4x2<\/sup>( + (3x + 2x) + (4-3)
\n= 11x3<\/sup> \u2013 4x2<\/sup> + 5x + 1
\n(c) P1<\/sub> (x) \u2013 P2<\/sub> (x)
\n= (7x3<\/sup> \u2013 4x2<\/sup> + 3x + 4) + (5x2<\/sup> + 6x + 1)
\n= 7x3<\/sup> + (-4x3<\/sup> \u2013 5x2<\/sup>) + (3x \u2013 6x) + (4 \u2013 1)
\n= 7x3<\/sup> \u2013 9x2<\/sup> \u2013 3x + 3
\n(d) P3 <\/sub>(x) \u2013 P2<\/sub> (x)
\n= (4x3<\/sup> + 2x \u2013 3) \u2013 (5x2<\/sup> + 6x + 1)
\n= (4x3<\/sup> \u2013 5x2<\/sup> + (2x \u2013 6x) + (-3 \u2013 1)
\n= 4x3<\/sup> \u2013 5x2<\/sup> \u2013 4x \u2013 4
\n(e)P1<\/sub> (x) + P2<\/sub> (x) + P3<\/sub>(x)
\n= (7x3<\/sup> \u2013 4x2<\/sup> + 3x + 4) + (5x2<\/sup> + 6x + 1) + (4x3<\/sup> + 2x \u2013 3)
\n= (7x3<\/sup> + 4x3<\/sup>) + (-4x2<\/sup> + 5x2<\/sup>) + (3x + 6x + 2x) + (4 + 1 \u2013 3)
\n= 11x3<\/sup> + x2<\/sup> + 11x + 2
\nGiven that P1<\/sub>(x) = 2x3<\/sup> + 4x2<\/sup> \u2013 x + 1
\nP2<\/sub>(x) = 3x4<\/sup> + x3<\/sup> \u2013 2x2<\/sup> + x \u2013 3 and
\nP3<\/sub>(x) = 4x3<\/sup> + 2x \u2013 4
\nFind:
\n(a) 2P1<\/sub>(x) + P2<\/sub>(x)
\n(b) 3P2<\/sub>(x) + 2P3<\/sub>(x)
\n(c) 3P1<\/sub>(x) \u2013 3P3<\/sub>(x)
\n(d) P3<\/sub>(x) + 2P1<\/sub>(x) \u2013 3P2<\/sub>(x)<\/p>\n

\u00a0Solution
\n<\/strong>(a) 2P1<\/sub>(x) + P2<\/sub>(x)
\n= 2(2x3<\/sup> + 4x2<\/sup> \u2013 x + 1) + (3x4<\/sup> + x3<\/sup> \u2013 2x2<\/sup> + x \u2013 3)
\n= (4x3<\/sup> + 8x2<\/sup> \u2013 2x + 2) + (3x4<\/sup> + x3<\/sup> \u2013 2x2<\/sup> + x \u2013 3)
\n= 3x-1<\/sup> + 5x3<\/sup> + 6x2<\/sup> \u2013 x \u2013 1
\n(b)3P2<\/sub>(x) + 2P3<\/sub>(x)
\n= 3(3x4<\/sup> + x3<\/sup> \u2013 2x2<\/sup> + x \u2013 3) + 2 (4x3<\/sup> + 2x \u2013 4)
\n= 9x4<\/sup> + 3x3<\/sup> \u2013 6x2<\/sup> + 3x \u2013 9 + 8x3<\/sup> + 4x \u2013 8
\n= 9x4<\/sup> + 11x3<\/sup> \u2013 6x2<\/sup> + 7x \u2013 17
\n(c) 3P1<\/sub>(x) \u2013 3P3<\/sub>(x)
\n= 3(2x3<\/sup> + 4x2<\/sup> \u2013 x + 1) -3 (4x3<\/sup> + 2x \u2013 4)
\n= 6x3<\/sup> + 12x2<\/sup> \u2013 3x + 3 \u2013 12x3<\/sup> \u2013 6x + 12
\n= -6x3<\/sup> + 12x2<\/sup> \u2013 9x + 15
\n(d) P3<\/sub>(x) + 2P1<\/sub>(x) \u2013 3P2<\/sub>(x)
\n= (4x3<\/sup> + 2x \u2013 4) + 2 (2x3<\/sup> + 4x2<\/sup> \u2013 x + 1) \u2013 3(3x4<\/sup> + x3<\/sup> \u2013 2x2<\/sup> + x \u2013 3)
\n= 4x3<\/sup> + 2x \u2013 4 + 4x3<\/sup> + 8x2<\/sup> \u2013 2x + 2 \u2013 9x4<\/sup> \u2013 3x3<\/sup> + 6x2<\/sup> \u2013 3x + 9
\n: P3<\/sub>(x) + 2P1<\/sub>(x) \u2013 3P2<\/sub>(x) = -9x4<\/sup> + 5x3<\/sup> + 14x2<\/sup> \u2013 3x + 7
\nThe value of p(x) at x = a is denoted by p(a) and is obtained by substituting a for x in the polynomial.<\/p>\n

\u00a0Example
\n<\/strong>Giventhat p(x)= 2x3<\/sup> + 5x2<\/sup> \u2013 9x \u2013 18, find;
\n(a) P(1)
\n(b) P(-1)
\n(c) P(2)
\n(d) P(0)<\/p>\n

\u00a0Solution
\n<\/strong>(a) P(1) = 2(1)3<\/sup> + 5(1)2<\/sup> \u2013 9(1) \u2013 18
\n= 2 + 5 \u2013 9 \u2013 18
\n= -20
\n(b) P(-1) = 2(-1)3<\/sup> + 5 (-1)2<\/sup> -9(-1) \u2013 18
\n\u00a0\u00a0\u00a0\u00a0= -2 + 5 + 9 \u2013 18
\n\u00a0\u00a0\u00a0\u00a0= -6
\n(c) P(2) = 2(2)3<\/sup> + 5(2)2<\/sup> \u2013 9(2) \u2013 18
\n\u00a0\u00a0\u00a0\u00a0= 16 + 20 \u2013 18 \u2013 18
\n\u00a0\u00a0\u00a0\u00a0= 0
\n(d) P(0) = 2(0)3<\/sup> + 5(0)2<\/sup> \u2013 9(0) \u2013 18
\n\u00a0\u00a0\u00a0\u00a0= -18<\/p>\n

\u00a0Multiplication of polynomials
\n<\/strong>The product of two polynomials of degrees m<\/em><\/strong> and n<\/em><\/strong> is another polynomial of degree m + n.<\/strong>
\n\t\t\t<\/em>
\n\u00a0Example
\n<\/em><\/strong>Given that\u00a0\u00a0\u00a0\u00a0P1<\/sub>(x) = 2x2<\/sup> + 5x + 6 and
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0P2<\/sub>(x) = 3x2<\/sup> \u2013 2x + 1, find P1<\/sub>P2<\/sub>
\n\t\tSolution
\n<\/strong>Method 1
\n<\/em><\/strong>P1<\/sub> x P2<\/sub>
\n\t\t= (2x2<\/sup> + 5x + 6)x (3x2<\/sup> \u2013 2x + 1)
\n= 2x2<\/sup>(3x2<\/sup> – 2x + 1) + 5x(3x2<\/sup> \u2013 2x + 1) + 6(3x2<\/sup> \u2013 2x + 1)
\n= 6x4<\/sup> \u2013 4x3<\/sup> + 2x2<\/sup> + 15x3 <\/sup>\u2013 10x2<\/sup> + 5x + 18x2<\/sup> \u2013 12x + 6
\n= 6x4<\/sup> +(4x3<\/sup> + 15x3<\/sup>) +(2x2<\/sup> – 10x2<\/sup> + 18x2<\/sup>)+ 5x – 12x + 6
\n= 6x4<\/sup> + 11x3<\/sup> + 10x2<\/sup> \u2013 7x + 6 <\/p>\n

\u00a0Method 2
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> + 5x + 6
\n3x2<\/sup> \u2013 2x + 1
\n\"\"\/2x2<\/sup> + 5x + 6
\n-4x3<\/sup> \u2013 10x2<\/sup> \u2013 12x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06x4<\/sup> + 15x3<\/sup> + 18x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06x4<\/sup> + 11x3<\/sup> + 10x2<\/sup> \u2013 7x + 6
\n\"\"\/
\n\t\tMethod 2 is usually called Long Multiplication method.
\n<\/strong>Given that <\/strong>P1<\/sub>(x) = 4x3<\/sup> \u2013 2x2<\/sup> + 3x \u2013 1 and P2<\/sub>(x) = 3x3<\/sup> \u2013 4 find P1<\/sub>(x) x P2<\/sub>(x)
\nSolution
\n<\/strong>Method 1
\n<\/strong>P1<\/sub>P2<\/sub>\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0(3x2<\/sup> \u2013 4) (4x3<\/sup> \u2013 2x2<\/sup> \u2013 3x \u2013 1)
\n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a03x2<\/sup> (4x3<\/sup> \u2013 2x3<\/sup> + 3x \u2013 1)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-4(4x3<\/sup> \u2013 2x2<\/sup> + 3x \u2013 1)
\n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a012x5<\/sup> \u2013 6x4<\/sup> + 9x3<\/sup> \u2013 3x2<\/sup> \u2013 16x3<\/sup> + 8x2<\/sup> \u2013 12x + 4
\n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a012x5<\/sup> \u2013 6x4<\/sup> \u2013 7x3<\/sup> + 5x2<\/sup> \u2013 12x + 4<\/p>\n

\u00a0Method 2
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x3<\/sup> \u2013 2x2<\/sup> \u2013 3x \u2013 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x2<\/sup>\u00a0\u00a0\u00a0\u00a0– 4
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-16×3+8×2 \u2013 12x + 4
\n12x5<\/sup> \u2013 6x4<\/sup> \u2013 9x3<\/sup> + 3x2<\/sup>\u00a0\u00a0\u00a0\u00a0
\n\"\"\/12x5<\/sup> \u2013 6x4<\/sup> \u2013 7x3<\/sup> + 5x2<\/sup> \u2013 12x + 4
\n\"\"\/
\n\t\tDivision of Polynomials
\n<\/strong>A polynomial of degree n <\/em>can be divided by another polynomial of degree m<\/em> if n \u2265 m.
\n<\/em>
\n\u00a0Divide the polynomial P(x) = 3x2<\/sup> -2x + 4 by the polynomial P(x) = x + 2
\nSolution
\n<\/strong>Since P1<\/sub> is being divided by P2<\/sub> it is called the dividend while P2<\/sub> is called the divisor. The result of division of P1<\/sub> by P2<\/sub> is called the quotient and whatever is left after division is called the remainder.
\nThe procedure of division of P1<\/sub> by P2<\/sub> can best be demonstrated step by step as follows:
\nStep 1
\n<\/strong>Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient.
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x + 2\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 2x + 4<\/p>\n

\u00a0Step 2
\n<\/strong>Multiply each of the terms of the divisor by the quotient.
\n3x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x + 2\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 2x + 4
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 6x
\nStep 3
\n<\/strong>Subtract the product obtained in step 2 from the first two terms of the dividend and add the next term of the dividend.
\n3x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x + 2\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 2x + 4
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 6x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -8x + 4<\/p>\n

\u00a0Step 4
\n<\/strong>Using -8x + 4 as a new dividend repeat step 1, 2 and 3.
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x \u2013 8(c)
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x + 2\u00a0\u00a0\u00a0\u00a03x2<\/sup> \u2013 2x + 4 (b)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(a)\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 6x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -8x + 4
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -8x \u2013 16
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20 (d)
\nNote that:
\n<\/strong>(a) x + 2 is the divisor;\u00a0\u00a0\u00a0\u00a0
\n(b) 3x2<\/sup> \u2013 2x + 4 is the dividend;
\n(c) 3x – 8is the quotient
\n(d) 20 is the remainder
\nThe expression 3x2<\/sup> \u2013 2x + 4 can be written as:
\n3x2<\/sup> \u2013 2x + 4\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0(x + 2) (3x \u2013 8)\u00a0\u00a0\u00a0\u00a0+\u00a0\u00a0\u00a0\u00a020
\n\"\"\/\"\"\/\"\"\/\"\"\/<\/p>\n

\u00a0Dividend\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divisor Quotient\u00a0\u00a0\u00a0\u00a0Remainder
\nIn general P(x) = D(x) x Q(x) + R
\nP(x) = Dividend
\nD(x) = Divisor
\nQ(x) = Quotient
\nR = Remainder
\nDivide 4x3<\/sup> + 6x2<\/sup> \u2013 2x + y by 2x \u2013 3 and hence find the quotient and the remainder<\/p>\n

\u00a0Solution
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> + 6x + 8
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x \u2013 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x3<\/sup> + 6x2<\/sup> \u2013 2x + 7
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a04x3<\/sup> + 6x2<\/sup>
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a012x2<\/sup> \u2013 2x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 12x2<\/sup> \u2013 18x
\n\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16x + 7
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 16x \u2013 24
\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 31
\nQuotient = 2x2<\/sup> + 6x + 8
\nRemainder = 31
\nFind the quotient and remainder when 2x4<\/sup> \u2013 3x3<\/sup> + x2 \u2013 4x + 5 is divided by x2<\/sup> + 3x + 1
\nSolution
\n<\/strong>2x2<\/sup> \u2013 9x + 26
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> + 3x + 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x4<\/sup> \u2013 3x3<\/sup> + x2<\/sup> \u2013 4x + 5
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x4<\/sup> \u2013 6x3<\/sup> + 2x2<\/sup>
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-9x3<\/sup> \u2013 x2<\/sup> \u2013 4x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-9x3<\/sup> \u2013 27x2<\/sup> \u2013 9x
\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\u00a026x2<\/sup> + 5x + 5
\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\u00a026x2<\/sup> + 78x + 26
\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\u00a0-73x \u2013 21
\nHence,\u00a0\u00a0\u00a0\u00a0Quotient = 2x2<\/sup> \u2013 9x + 26
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0Remainder = -73x \u2013 21<\/p>\n

\u00a0Find the quotient and remainder when x3<\/sup> + 8 is divided by x2<\/sup> \u2013 2x + 4
\nSolution
\n<\/strong>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/strong>x + 2
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 2x + 4\u00a0\u00a0\u00a0\u00a0x3<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+8
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> \u2013 2x2<\/sup> + 4x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2x2<\/sup> + 4x + 8
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2x2<\/sup> + 4x + 8
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a00<\/p>\n

\u00a0
\n\u00a0Quotient = x + 2
\nRemainder = 0
\nHence, x3<\/sup> \u2013 2x2<\/sup> + 4x is a factor of x3<\/sup> + 8.<\/p>\n

\u00a0The Remainder Theorem
\n<\/strong>The Long Division Method.Although a little cumbersome enables us to find, not only the quotient, but the remainder as well. Suppose we are given the polynomial f(x) = 2x3<\/sup> \u2013 3x2<\/sup> + 4x \u2013 1 and we wish to find the remainders when f(x) is divided by:
\n(a) x \u2013 1
\n(b) x \u2013 2
\n(c) x \u2013 3
\nBy using the long division method we have:
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 x + 3
\n\"\"\/\u00a0\u00a0\u00a0\u00a0x\u2013 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 3x2<\/sup>+ 4x \u2013 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 2x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x2<\/sup> + 4x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x2<\/sup> + x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x \u2013 1
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x \u2013 3
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 2\u00a0\u00a0\u00a0\u00a0
\nSo when f(x) is divided by x \u2013 1, the remainder is 2.
\n2x2<\/sup> \u2013 x + 6
\n\"\"\/\u00a0\u00a0\u00a0\u00a0x \u2013 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x3<\/sup> \u2013 3x2<\/sup> + 4x \u2013 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x3<\/sup> \u2013 4x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x2<\/sup> + 4x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x2<\/sup> + 2x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06x \u2013 1
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a06x \u2013 12
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 11<\/p>\n

\u00a0So when f(x) is divided by x \u2013 2, the remainder is 11.<\/p>\n

\u00a02x2<\/sup> \u2013 3x + 13
\n\"\"\/\u00a0\u00a0\u00a0\u00a0x \u2013 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 3x2<\/sup> + 4x \u2013 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a02x2<\/sup> \u2013 6x2<\/sup>
\n\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a03x2<\/sup> + 4x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x2<\/sup> + 9x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a013x \u2013 1
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a013x \u2013 39
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 38
\nSo when f(x) is divided by x \u2013 3, the remainder is 38.<\/p>\n

\u00a0Now, find:
\na. f(1)
\nb. f(2)
\nc. f(3)
\nd. f(1) = 2 \u2013 3 + 4 \u2013 1 = 2
\ne. f(2) = 16 \u2013 12 + 8 \u2013 1 = 11
\nf. f(3) = 54 \u2013 27 + 12 \u2013 1 = 38
\nCompare the answers in each of the following pairs
\ni. (a) and (d)
\nii. (b) and (e)
\niii. (c) and (f)
\nWhat do you notice?
\nThis is a remarkable result and it is not a mere coincidence.
\nLet us try another one.
\nGiven that H(x) = x3<\/sup> + 2x3<\/sup> \u2013 4x + 3, find the remainder when H(x) is divided by
\n(a) x + 1
\n(b) x + 2
\n(c) x + 3<\/p>\n

\u00a0x2<\/sup> + x – 5
\n\"\"\/(a)\u00a0\u00a0\u00a0\u00a0x + 1\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 2x3<\/sup> \u2013 4x + 3
\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 4x
\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\u00a0-5x + 3
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-5x \u2013 5
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a08<\/p>\n

\u00a0When H(x) is divided by x + 1, the remainder is 8.
\nx2<\/sup>\u00a0\u00a0\u00a0\u00a0-4
\n\"\"\/(b)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x + 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 2x3<\/sup> \u2013 4x + 3
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 2x2<\/sup>
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-4x + 3
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-4 \u2013 8
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a011<\/p>\n

\u00a0So, when H(x) is divided by x + 2, the remainder is 11.
\nx2<\/sup> + x – 1
\n\"\"\/(c)\u00a0\u00a0\u00a0\u00a0x + 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 2x3<\/sup> \u2013 4x + 3
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x3<\/sup> + 3x2<\/sup>
\n\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x2<\/sup> \u2013 4x
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0– x2<\/sup> + 3x
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x + 3
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0-x \u2013 3
\n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 6<\/p>\n

\u00a0When H(x) is divided by x + 3, the remainder is 6. Now, find:
\n(d) H(-1)
\n(e) H(-2)
\n(f) H(-3)
\n(d) H(-1) = -1 + 2 + 4 + 3 = 8
\n(e) H(-2) = -8 + 8 + 8 + 3 = 11
\n(f) H(-3) = -27 + 18 + 12 + 3 = 6<\/p>\n

\u00a0Compare again the answers in each of the following pairs:
\n(i) (a) and (d)
\n(ii) (b) and (e)
\n(iii) (c) and (f)
\nWhat again do you notice?
\n(a) When H(x) is divided by x + 1 the remainder is H(-1)
\n(b) When H(x) is divided by x + 2 the remainder is H(-2)
\n(c) When H(x) is divided by x + 3 the remainder is H(-3)
\nSo we can safely conclude that if H(x) is divided by x \u2013 a the remainder is H(a) and this forms the basis of what is called the remainder theorem.
\nTheorem
\n<\/strong>If the polynomial f(x) is divided by x \u2013 a, the remainder is f(a).<\/p>\n

\u00a0Proof
\n<\/em><\/strong>The polynomial function f(x) can be written as:
\nf(x) = (x \u2013 a) Q(x) + R\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2026(1)
\nwhere x \u2013 a is the divisor, Q(x) the quotient and R the remainder.
\nPut x = a in (1)
\nf(a) = (a \u2013 a) Q(a) + R
\nf(a) = R
\nThe theorem whose proof we have just established, is called the RemainderTheorem<\/strong>. In general, if f(x) is divided by ax + b then the remainder is
\nf
\nA special case of the remainder theorem is when f(x) leaves no remainder when it is divided by x \u2013 a. We therefore say that x \u2013 a is a factor of f(x). The modified theorem is called, Factor Theorem<\/strong> and it states:
\nIf f(a) = 0 then x \u2013 a is a factor of f(x).<\/p>\n

\u00a0Example<\/strong>
\n\t\tGiven that f(x) = 3x3<\/sup> \u2013 4x2<\/sup> + 2x + 3, find the remainder when f(x) is divided by x \u2013 1.<\/p>\n

\u00a0Solution
\n<\/strong>Let R be the remainder. Then using the remainder theorem, R = f(1)
\nf(1) \u00a0\u00a0\u00a0\u00a0= 3(1)3<\/sup> \u2013 4(1)2<\/sup> + 2(1) +3
\n\u00a0\u00a0\u00a0\u00a0= 3 \u2013 4 + 2 + 3
\n\u00a0\u00a0\u00a0\u00a0= 4<\/p>\n

\u00a0Find the remainder when
\nf(x) = (x + 3)(x \u2013 2)(x + 2) is divided by x + 1.<\/p>\n

\u00a0Solution
\n<\/strong>Let R be the remainder when f(x) is divided by (x + 1) then R = f(-1)
\nf(-1) \u00a0\u00a0\u00a0\u00a0= (-1 + 3)(-1 -2)(-1 + 2)
\n\u00a0\u00a0\u00a0\u00a0= (2)(-3)(+1)
\n\u00a0\u00a0\u00a0\u00a0= -6
\nHence the remainder is -6
\nFind the remainder when f(x) = 2x3<\/sup> + 3x2<\/sup> \u2013 4x + 1 is divided by 2x \u2013 1. What conclusion can you draw?<\/p>\n

\u00a0Solution
\n<\/strong>Let R be the remainder when f(x) is divided by 2x \u2013 1 then R = f
\nf = 2 f3<\/sup> + 3 f3<\/sup>\u2013 4 f + 1
\n= \u00bc + \u00be – 2 + 1
\n= 0
\nHence2x \u2013 1 is a factor of f(x).<\/p>\n

\u00a0Evaluation
\n<\/strong><\/p>\n

    \n
  1. If g(x) is the quotient when f(x) = 2X3 <\/sup> + 3x2<\/sup> -2x +1 is divided by x+2 find the remainder when g(x) is divided by x-1
    \n\t\t\t\t<\/sup><\/li>\n<\/ol>\n

    General Evaluation<\/strong>
    \n\t\tIf f(x) = 6x3<\/sup> + 13x2<\/sup> + 2x \u2013 5,
    \n(a) show that f(-1) = 0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) find the factor of f(x)
    \n(1) When the polynomial f(x) = x4<\/sup> + px3<\/sup> + x2<\/sup> + qx + 1 is divided by x2<\/sup> + 3x + 2 quotient is x2 \u2013 1 and the remainder is 5x + 3. Find the value of constants p and q.
    \n(2) If (x + 1) is a factor of the polynomial f(x) = x3<\/sup> + kx2<\/sup> + 3x + 10. Find the value of the constant k, and factorise the polynomial completely.<\/p>\n

    \u00a0Reading assignment
    \n<\/strong>New Further Maths Project 1 pages71 \u2013 75 Exercise 6b Q1, 8, 22 and 26<\/p>\n

    \u00a0Weekend Assignment
    \n<\/strong>1. \u00a0\u00a0\u00a0\u00a0Given that f(x) = x5<\/sup> + 4x4<\/sup> \u2013 6x2<\/sup> + 2x + 2, find (-1)
    \n\u00a0\u00a0\u00a0\u00a0(a) 2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) -4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) -2
    \n(2) \u00a0\u00a0\u00a0\u00a0Determine the values of p and q if(x \u2013 1) and (x \u2013 2) are factors of 2x3 <\/sup>+ px \u2013 4 + q
    \n\u00a0\u00a0\u00a0\u00a0(a) p = 12, q = 13\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) p = -13, p = -12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) p = -13, q = 12\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
    \n\u00a0\u00a0\u00a0\u00a0(d) p = 10, q = 8
    \n(3) \u00a0\u00a0\u00a0\u00a0Find zero of the polynomial p(x) = x3<\/sup> + 4x2<\/sup> + x \u2013 6
    \n\u00a0\u00a0\u00a0\u00a0(a) x = -1, 2, 3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) x = -1, -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) x = 1, -2, -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) x = 1, 2, -3
    \n(4) \u00a0\u00a0\u00a0\u00a0If f(x) = 6x3<\/sup> + 13x2<\/sup> + 2x \u2013 5, find the factors of f(x)
    \n\u00a0\u00a0\u00a0\u00a0(a) (x \u2013 1) (2x \u2013 1) (3x \u2013 5)\u00a0\u00a0\u00a0\u00a0(b) (x + 1) (2x \u2013 1) (3x + 5)\u00a0\u00a0\u00a0\u00a0(c) (x + 1) (2x + 1) (3x \u2013 5)
    \n\u00a0\u00a0\u00a0\u00a0(d) (x + 1) (1 \u2013 2x) (5 \u2013 3x)
    \n(5) \u00a0\u00a0\u00a0\u00a0If (2x + 1) is a factor of the polynomial f(x) = 2x3<\/sup> \u2013 8x + x2<\/sup> + k, find the value of the \u00a0\u00a0\u00a0\u00a0constant k. (a) -2\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) -3\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(c) -4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) 3<\/p>\n

    \u00a0Theory
    \n<\/strong>(1) Find the values of the constant p, q and r such that, when the polynomial
    \nf(x) = x3<\/sup> + px2<\/sup> + qx + r is divided by (x + 2), (x \u2013 1) and (x \u2013 3), the remainder are respectively -48, 0 and 2. Hence factorise f(x) completely.
    \n(2) If x \u2013 2 is a factor of f(x) = x3<\/sup> \u2013 x2<\/sup> + px + q and f(x) leaves a remainder of 12 when it is divided by (x \u2013 3), find
    \n(a) the values of the constant p and q
    \n(b) the three values of c for which x3<\/sup> \u2013 x2<\/sup> + px + q = 0<\/p>\n

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

    \u00a0WEEK 3 TOPIC: Polynomials Consider the expressions formed from the sum of integral non negative…<\/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-2837","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\/2837","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=2837"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2837\/revisions"}],"predecessor-version":[{"id":2838,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2837\/revisions\/2838"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2837"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2837"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2837"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}