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:



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.