Please answer this...

a= 2, b=1 and c=-8

Step-by-step explanation:

+ f(x) is exactly dividsible by (2x-3), that means f(x)= (2x-3). g(x), then f(3/2)= 0

+ f(x) is divided by (x-2) and the remainder is 12, that means f(x)=(x-2). h(x) +12

So f(2)=12.

+ The last: f(x)=(x+3). k(x)+72, so f(-3)=72

Then we have 3 equations:

f(3/2)=0

f(2)=12

f(-3)=72

So we replace and we find:

81a+54b+36c=-72

16a+8b+4c=8

81a-27b+9c=63

From: 16a+8b+4c=8, the we have 4a+2b+c=2, or c=-4a-2b+2, we replace c for 2 equations:

9a+6b+4(-4a-2b+2)=-8

9a-3b+(-4a-2b+2)=7

Then

-7a-2b=-16

5a-5b=5, a= b+1

and -7(b+1) -2b =-16

or -9b=-9, b=1, a= b+1=1+1=2, c=-4a-2b+2 = -8-2+2= -8

So the final answer is a= 2, b=1 and c=-8

Hope you understand it.