22 Jan lesson summary

Quiz B Answers
1)
The expression (3x^3 + x^2 + ax - 5) has the same remainder when divided by (x-1) and by (x+2). Evaluate a and hence find the remainder when the expression is divided by (x-2).

ANS: f(x) = 3x^3 + x^2 + ax -5
if (x-1) = 0, x = 1
if (x+2) = 0, x=-2

Let f(1) = a-1
Let f(-2) = -2a-25

hence (a-1) = (-2a-25)
a=-8

if (x-2) = 0, x=2

Let f(2) = 7

2)
The expression [f(x) = ax^3 - 9x^2 + 7x + b] leaves a remainder of [6(x+1)] when divided by (x^2 + 3x - 4). Find the values of a and b.

ANS: x^2 + 3x -4 can be factorised (cross method, formula, graphical method or remainder theorem) into (x+4)(x-1)
Thus x= -4 or 1

f(x) = ax^3 - 9x^2 + 7x + b
Let f(-4) = -64a + b -172
Let f(1) = a+b-2

if the remainder is 6(x+1)
-64a+b-172=-18

a+b-2=12
a=14-b
If we sub a with (14-b)

-64(14-b)+b = 154
=-896+65b=154
65b=1050
b= 16/2/13

a+16/2/3 -2 = 12
a= -2/2/13

Factorisation of quadratic equations
eg. x^2-6x+8

Cross method
x    -2  -2x
x    -4  -4x
_______
X^2 +8 -6x

(x-2)(x-4)

Formula [b+sqrt(-b^2-4(a)(c)]/2(a)
[-(-6)+sqrt((-6)^2 - 4(1)(8)] / 2(1)
[6+sqrt(36-32)]/2 = 2 or 4

Remainder Theorem
if (x-2)=0, x=2 [Guess and Check)
f(x) = x^2-6x+8
f(2) = 2^2-6(2)+8
     = 0
If there is no remainder it means that (x-2) is a factor.