For what value of a, the divisions of two polynomials (ax³ + 3x² – 3) and (2x³ – 5x + a) by (x – 4) give the same remainder —let us calculate and write it.

Let the two polynomials be,

P(x) = ax³ + 3x² – 3 …(i)

Q(x) = 2x³ – 5x + a …(ii)

Now, we understand by the question that,

P(x) and Q(x) divided by (x – 4) gives the same remainder.

We need to find zero of the linear polynomial, (x – 4).

To find zero, put (x – 4) = 0

⇒ x – 4 = 0

⇒ x = 4

By Remainder theorem that says, f(x) is a polynomial of degree n (n ≥ 1) and ‘a’ is any real number. If f(x) is divided by (x – a), then the remainder will be f(a).

Here, a = 4.

This means, remainder when P(x) is divided by (x – 4) is P(4).

⇒ Remainder = P(4)

⇒ Remainder = a(4)³ + 3(4)² – 3

⇒ Remainder = 64a + 48 – 3

⇒ Remainder = 64a + 45 …(iii)

And remainder when Q(x) is divided by (x – 4) is Q(4).

⇒ Remainder = Q(4)

⇒ Remainder = 2(4)³ – 5(4) + a

⇒ Remainder = 128 – 20 + a

⇒ Remainder = 108 + a …(iv)

When P(x) and Q(x) are divided (x – 4) , they leave same remainder.

Comparing equations (iii) and (iv), we have

64a + 45 = 108 + a

⇒ 64a – a = 108 – 45

⇒ 63a = 63

⇒

⇒ a = 1

Thus, a = 1.