Question

The polynomial ay3+3y3-3 and 2y3-5y-a when divided by (y-4) leaves remainder A and B respectively.Find the value of a if 2A-B=0

Answer 1

$(1) If\ given\ polynomial1\ is\ a*y^3+3y-3\ \ THEN$

$A = a*4^3+3*4^3-3 =64a+3*64-3 = 64a+189\\ \\ B = 2*4^3 - 5*4 - a = 108 - a \\ \\ 2A-B=0 \\ \\ 128a+378 = 108 - a \\ \\ 129a = -270 \\ a = \frac{-270}{129} = \frac{-90}{43}$

$(2) If\ given\ polynomial1\ is\ a*y^3+3y-3\ \ THEN \\ \\ A = 64a+9,\ \ B=108-a \\ \\ a = \frac{30}{43}$

Answer 2

$P(y) = ay^3 +3y^3 - 3$

P(y) is divided by (y-4), the remainder is P(4)

$P(4) = a(4)^3 +3(4)^3 - 3 = 64a + 189$

But the given remainder is A, 

So, 64a + 189 = A   .......(1)

$Q(y) = 2y^3 - 5y - a$

Q(y) is divided by (y-4), the remainder is Q(4)

$Q(4) = 2(4)^3 - 5(4) - a = 108 - a$

But the given remainder is B, 

So, 108 - a = B   .......(2)

Now, 2A - B = 0

2(64a + 189) - (108 - a) = 0

128a + 378 - 108 + a =0

129a + 270 = 0

129a = -270

a = $\frac{-270}{129} =\frac{-90}{43}$