Question

(4a^2)b–ab^2–(3a^2)b+ab^2–ab+6 for a=−3, b=2

Answer 1

Put the values of a = -3 and b = 2 to the expression

$4a^2b-ab^2-3a^2b+ab^2-ab+6$

$(4)(-3)^2(2)-(-3)(2)^2-(3)(-3)^2(2)+(-3)(2)^2-(-3)(2)+6\\\\=(4)(9)(2)+(3)(4)-(3)(9)(2)-(3)(4)-(-6)+6\\\\=72+12-54-12+6+6\\\\=\boxed{30}$

Answer 2

Answer:

30

Step-by-step explanation:

These are generally easier to evaluate by hand if they are simplified first.

... (4a^2)b -ab^2 -(3a^2)b +ab^2 -ab +6

... = (a^2)b(4 -3) + ab^2(-1 +1) -ab +6

... = a^2·b -ab +6

... = ab(a -1) +6

... = (-3)(2)(-3-1) +6

... = (-6)(-4)+6

... = 24 +6 = 30