# 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.

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