Thursday, 17 January 2013

17 Jan Lesson summary

Explanation for Homework

-A01B Question4

Given that x^3+7x^2+ax-5=(x^2+x-2)f(x)+(x-1)(bx-1), find

 a)the value of a and of b

   if x=1, 
  1^3+7*1^2+1*a-5 = 0*(1+2)f(x)+0*(bx-1)
   so a=-3
   if x=-2, 
   so b=3
a=-3 b=3
   f(x)= x^3+7x^2+(-3)x-5(x-1)(3x-1)
A)  identity means it's true for all values of X
No matter what x is, LHS=RHS

when the main problem of Q4 is we don't know f(x), we can make x=1or -2, to make f(x)multiply zero and help us find the value of a and b.

x^3+7x^2+(-3)x-5 is like the Dividend
(x-1)(3x-1) is like the remainder
x^2+x-2 is like divisor
f(x) is like quotient

Remainder Theorem

3methods to find the remainder
M1:Traditional division(on P45)
M2:Synthetic division
M3: make the divisor(at the bottom) become zero and get that value of x
      put that value in the dividend
      if the result is zero, it means no remainder and this is FACTOR THEOREM
      if the result is other numbers, that is the remainder and this is REMAINDER THEOREM

No comments:

Post a Comment