Week 1 Factor & Remainder Past Paper Questions - SOLUTIONS

Transcription

1 Question Solution Marks C2 JUNE 2010 Q2 f(3) = 3(3) 3 5(3) 2 58(3) + 40 f(3) = 98 So, the remainder is 98 Substitute x = 3 into 98 (If it is not stated that 98 is the remainder, do not give this C2 JUNE 2010 Q2 3x 3 5x 2 58x + 40 (x 5)(3x x 8) [3x 3 ] [ 15x x 2 ] [ 50x 8x] [+40] = 0 (x 5)(3x x 8) = 0 (x 5)(3x 2)(x + 4) = 0 x = 5 or x = 2 or x = 4 3, using (x 5) 3x x 8 the quadratic (3x 2)(x + 4) x = 5, x = 2, x = 4 3 (Unless all three solutions are correct, do not give this

2 C2 JAN 2010 f ( 1 2 ) = 5 2 ( ) + a ( ) + b ( 1 2 ) 6 = a b = a + 2b = 4 a + 2b = 3 1 f( 2) = 0 2( 2) 3 + a( 2) 2 + b( 2) 6 = a 2b 6 = 0 4a 2b = 22 2a b = 11 2 a + 2b = a b = 11 2 Substitute x = 1 2 into and make f ( 1 2 ) equal to 5 a + 2b = 3 (An equation that is Substitute x = 2 into and make f( 2) equal to 0 2a b = 11 (An equation that is Attempt to solve both equations simultaneously, by eliminating either a or b a = 5, b = 1 2a + 4b = 6 1 2a b = 11 2 (If either answer is incorrect, do not award this 5b = 5 b = 1 Sub b = 1 into 1 : a + 2( 1) = 3 a = 5 C2 JAN 2010 = 2x 3 + 5x 2 x 6 2x 3 + 5x 2 x 6 (x + 2)(2x 2 + x 3) [2x 3 ] [+4x 2 + x 2 ] [+2x 3x] [ 6] 2x 3 + 5x 2 x 6 (x + 2)(2x + 3)(x 1), using (x + 2) 2x 2 + x 3 (2x + 3)(x 1)

3 C2 JUNE 2009 f(k) = 8 8 C2 JUNE 2009 f(2) = 4 (3(2) 2)((2) k) 8 = 4 4(2 k) 8 = 4 8 4k 8 = 4 4k = 4 k = 1 Substitute x = 2 into and make f(2) equal to 4 k = 1 C2 JUNE 2009 (c) = (3x 2)(x ( 1)) 8 = (3x 2)(x + 1) 8 = 3x 2 + 3x 2x 2 8 = 3x 2 + x 10 = (3x 5)(x + 2) Substitute k = 1 into and multiply out to get a three term quadratic the quadratic (3x 5)(x + 2)

4 C2 JAN 2009 Q6 f(2) = f( 1) (2) 4 + 5(2) 3 + a(2) + b = ( 1) 4 + 5( 1) 3 + a( 1) + b a + b = 1 5 a + b a + b = 4 a + b a = 4 a 3a = 60 a = 20 Substitute x = 2 into Substitute x = 1 into and make f( 1) equal to f(2) a + b (Anything that is equivalent to this will also get the mark) 1 5 a + b (Anything that is equivalent to this will also get the mark) a = 20 C2 JAN 2009 Q6 = x 4 + 5x 3 20x + b f( 3) = 0 ( 3) 4 + 5( 3) 3 20( 3) + b = b = b = 0 b = 6 Substitute a = 20 into and substitute x = 3 into and make f( 3) equal to b = 0 (Any equation that is b = 6 C2 JUNE 2008 f( 4) = 2( 4) 3 3( 4) 2 39( 4) + 20 f( 4) = 0 So, (x + 4) is a of Substitute x = 4 into Show f( 4) = 0 and state that (x + 4) is a C2 JUNE x 3 3x 2 39x + 20 (x + 4)(2x 2 11x + 5) [2x 3 ] [+8x 2 11x 2 ] [ 44x + 5x] [+20] 2x 3 + 5x 2 x 6 (x + 4)(2x 1)(x 5), using (x + 4) 2x 2 11x + 5 (2x 1)(x 5)

5 C2 JAN 2008 (i) f(3) = (3) 3 2(3) 2 4(3) + 8 f(3) = 5 So, the remainder is 5 Substitute x = 3 into 5 (If it is not stated that 5 is the remainder, do not give this (ii) f( 2) = ( 2) 3 2( 2) 2 4( 2) + 8 f( 2) = 0 So, the remainder is 0 Substitute x = 2 into and put f( 2) equal to 0 (If it is not stated that 0 is the remainder, do not give this C2 JAN 2008 x 3 2x 2 4x + 8 (x + 2)(x 2 4x + 4) [x 3 ] [+2x 2 4x 2 ] [ 8x + 4x] [+8] x 3 2x 2 4x + 8 = 0 (x + 2)(x 2 4x + 4) = 0 (x + 2)(x 2)(x 2) = 0 x = 2 or x = 2, using (x + 2) x 2 4x + 4 (x 2)(x 2) x = 2, x = 2 (If either solution is incorrect, do not give this C2 JUNE 2007 Q2 f(2) = 3(2) 3 5(2) 2 16(2) + 12 f(2) = 16 So, the remainder is 16 Substitute x = 2 into 16 (If it is not stated that 16 is the remainder, do not give this C2 JUNE 2007 Q2 3x 3 5x 2 16x + 12 (x + 2)(3x 2 11x + 6) [3x 3 ] [+6x 2 11x 2 ] [ 22x + 6x] [+12] 3x 3 5x 2 16x + 12 (x + 2)(3x 2)(x 3), using (x + 2) 3x 2 11x + 6 (3x 2)(x 3)

6 C2 JAN 2007 f( 2) = ( 2) 3 + 4( 2) 2 + ( 2) 6 f( 2) = 0 So, (x + 2) is a Substitute x = 2 into Show f( 2) = 0 and state that (x + 2) is a C2 JAN 2007 x 3 + 4x 2 + x 6 (x + 2)(x 2 + 2x 3) [x 3 ] [+2x 2 + 2x 2 ] [4x 3x] [ 6] x 3 + 4x 2 + x 6 (x + 2)(x + 3)(x 1), using (x + 2) x 2 + 2x 3 (x + 3)(x 1) C2 JAN 2007 (c) x = 2 or x = 3 or x = 1 x = 2, x = 3, x = 1 (Unless all three solutions are correct, do not give this C2 JUNE 2006 Q4 f( 2) = 2( 2) 3 + 3( 2) 2 29( 2) 60 f( 2) = 6 So, the remainder is 6 Substitute x = 2 into 6 (If it is not stated that 6 is the remainder, do not give this C2 JUNE 2006 Q4 f( 3) = 2( 3) 3 + 3( 3) 2 29( 3) 60 f( 3) = 0 So, (x + 3) is a Substitute x = 3 into Show f( 3) = 0 and state that (x + 3) is a C2 JUNE 2006 Q4 (c) 2x 3 + 3x 2 29x 60 (x + 3)(2x 2 3x 20) [2x 3 ] [+6x 2 3x 2 ] [ 9x 20x] [ 60] 2x 3 + 3x 2 29x 60 (x + 3)(2x + 5)(x 4), using (x + 3) 2x 2 3x 20 (2x + 5)(x 4)

7 C2 JAN (1) 3 + (1) 2 5(1) + c = c = c = 0 c = 2 Substitute x = 1 into and put f(1) equal to 0 c = 2 C2 JAN 2006 = 2x 3 + x 2 5x + 2 As (x 1) is a : 2x 3 + x 2 5x + 2 (x 1)(2x 2 + 3x 2) [2x 3 ] [ 2x 2 + 3x 2 ] [ 3x 2x] [+2] 2x 3 + x 2 5x + 2 (x 1)(2x 1)(x + 2) Substitute c = 2 into and attempt to use algebraic division to, using (x 1) 2x 2 + 3x 2 (2x 1)(x + 2) C2 JAN 2006 (c) f ( ) = 2 (3 2 ) + ( ) 5 ( 3 2 ) + 2 f ( 3 2 ) = 7 2 So, the remainder is 7 2 Substitute x = 3 2 into 7 2 (If it is not stated that 7 2 is the remainder, do not give this C2 JUNE 2005 f( 4) = 2( 4) 3 + ( 4) 2 25( 4) + 12 f( 4) = 0 So, (x + 4) is a Substitute x = 4 into Show f( 4) = 0 and state that (x + 4) is a C2 JUNE x 3 + x 2 25x + 12 (x + 4)(2x 2 7x + 3) [2x 3 ] [+8x 2 7x 2 ] [ 28x + 3x] [+12] 2x 3 + x 2 25x + 12 (x + 4)(2x 1)(x 3), using (x + 4) 2x 2 7x + 3 (2x 1)(x 3)

8 C2 JAN 2005 f(2) = 1 (2) 3 2(2) 2 + a(2) + b = a + b = 1 2a + b = 1 1 f( 1) = 28 ( 1) 3 2( 1) 2 + a( 1) + b = a + b = 28 3 a + b = 28 a + b = a + b = 1 1 a + b = a = 30 a = 10 Sub a = 10 into 1 : 2( 10) + b = b = 1 b = 21 Substitute x = 2 into and make f(2) equal to 1 2a + b = 1 (An equation that is Substitute x = 1 into and make f( 1) equal to 28 a + b = 31 (An equation that is Attempt to solve both equations simultaneously, by eliminating either a or b a = 10, b = 21 (If either answer is incorrect, do not award this C2 JAN 2005 = x 3 2x 2 10x + 21 f(3) = (3) 3 2(3) 2 10(3) + 21 f(3) = 0 So, (x 3) is a Substitute x = 3 into Show f(3) = 0 and state that (x 3) is a