9514 1404 393
Answer:
Step-by-step explanation:
The remainder from division by (2x-3) is the same as the value of the function when that divisor is zero—at x = 3/2.
((2(3/2) +a)(3/2) +b)(3/2) -2 = 7
((3 +a)(3/2) +b)(3/2) = 9 . . . . add 2
9/2 +3/2a + b = 6 . . . . . . . . multiply by 2/3
3/2a +b = 3/2 . . . . . . . . . . . subtract 9/2
3a +2b = 3 . . . . . . . . . . . . . multiply by 2 [eq1]
The remainder from division by (x+3) is the same as the value of the function for x = -3.
((2(-3) +a)(-3) +b)(-3) -2 = -20
(-6 +a)(-3) +b = 6 . . . . . . add 2 and divide by -3
18 -3a +b = 6 . . . . . . . . . eliminate parentheses
3a -b = 12 . . . . . . . . . . . subtract 18, multiply by -1 [eq2]
Subtracting the [eq2] from [eq1], we have ...
(3a +2b) -(3a -b) = (3) - (12)
3b = -9 . . . . . simplify
b = -3 . . . . . . divide by 3
From [eq2], ...
3a -(-3) = 12 . . . . substitute for b
3a = 9 . . . . . . . . subtract 3
a = 3 . . . . . . . . . divide by 3
The values of 'a' and 'b' are 3 and -3, respectively.
