Problem Statement: Find the value of 12a if x3−6x2+ax+b is exactly divisible by x2−3x+2.
Solution:Step 1: Finding the remainder
To find if x3−6x2+ax+b is exactly divisible by x2−3x+2, we need to find the remainder when it is divided by x2−3x+2.
We can use polynomial long division to find the remainder.
The division can be done as follows:
x - 9
_______________________
x2 - 3x + 2 | x3 - 6x2 + ax + b
- (x3 - 3x2 + 2x)
_________________
- 3x2 + ax
- (-3x2 + 9x - 6)
_____________
ax + 6
- (ax - 3x + 2)
____________
9x - 2
Therefore, the remainder is 9x - 2.
Step 2: Finding the conditions for exact division
For exact division, the remainder should be zero. Therefore, we need to find the values of a and b such that 9x - 2 is equal to zero for all values of x.
This means that 9x = 2, or x = 2/9.
Substituting x = 2/9 in x3−6x2+ax+b, we get:
(2/9)3 - 6(2/9)2 + a(2/9) + b = 0
Simplifying this expression, we get:
8a/27 + b - 8/3 = 0
8a/27 = 8/3 - b
a = 27(8/3 - b)/8
Step 3: Finding the value of 12a
We need to find the value of 12a.
Substituting the value of a in terms of b, we get:
a = 27(8/3 - b)/8
a = 27/3 - 27b/8
a = 9 - 3.375b
Therefore, 12a = 108 - 40.5b
We need more information to find the value of b and hence the value of 12a.