waecE-LEARNING
Further Mathematics Paper 2, May/June. 2014  
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Main
General Comments
Weakness/Remedies
Strength








Question 9
  1. Differentiate (x – 3) (x2+ 5) with respect to x.
  1. If (x + 1)2is a factor of f(x)=x3+ax2+bx+3,where aandb are constants, find the:
  2. values of aandb;
  3. zeros of f(x).

 

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Observation


                            x + (a – 2)
x2 + 2x + 1  x3 + ax2 + bx + 3
                    x3 + 2x2 + x
                          (a – 2)x2 + (b – 1)x + 3
                          (a – 2)x2 + 2(a – 2)x + (a – 2)
                                            (b – 2a + 3)x + (5 – a)
Since (x+1)2 was a factor, then it implied that (b -2a+3) =0...........equation 1
and(5 – a) =0......equation (2). From equation (2), a=5. Substituting 5 for a in equation (1) gave b-2(5) +3=0. i.e. b=7. Hence the polynomial was
f(x) = x3+5x2+7x+3. Factors of f(x) were (x+1)2 and x+ (5-2) i.e. f(x)=(x+1)(x+1)(x+3). Therefore the zeroes of f(x) were -1,-1and -3.


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