If the polynomials (3x3 + ax2 + 3x + 5) and (4x3 + x2 – 2x + a)

1.	If the polynomials (3x3 + ax2 + 3x + 5) and (4x3 + x2 – 2x + a) leaves the same remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.
Answer» Let the given polynomials are p(x) = 3x³ + ax² + 3x + 5 and f(x) = 4x³ + x² – 2x + a According to question p(2) = f(2) 3(2)³ + a(2)² + 3(2) + 5 = 4(2)³ + (2)² – 2(2) + a ⇒ 3 x 8 + 4a + 6 + 5 = 32 + 4 – 4 + a ⇒ 24 + 4a + 11 = 32 + a ⇒ 4a – a = 32 – 35 ⇒ 3a = -3 ⇒ a = -1 As the remainder is same, so p(2) or f(2) would be same when a = – 1 p(2) = 3(2)³ + (-1)(2)² + 3 x 2 + 5 [∴ a = -1] = 24 – 4 + 6 + 5 = 35 – 4 = 31 Thus, a = – 1 and p(2) or f(2) = 31

If the polynomials (3x3 + ax2 + 3x + 5) and (4x3 + x2 – 2x + a) leaves the same remainder when divided by (x – 2), find the value of a. Also the find the remainder in each-case.

Answer»

Let the given polynomials are

p(x) = 3x³ + ax² + 3x + 5

and f(x) = 4x³ + x² – 2x + a

According to question p(2) = f(2)

3(2)³ + a(2)² + 3(2) + 5

= 4(2)³ + (2)² – 2(2) + a

⇒ 3 x 8 + 4a + 6 + 5

= 32 + 4 – 4 + a

⇒ 24 + 4a + 11 = 32 + a

⇒ 4a – a = 32 – 35

⇒ 3a = -3

⇒ a = -1

As the remainder is same,

so p(2) or f(2) would be same when a = – 1

p(2) = 3(2)³ + (-1)(2)² + 3 x 2 + 5 [∴ a = -1]

= 24 – 4 + 6 + 5

= 35 – 4 = 31

Thus, a = – 1 and p(2) or f(2) = 31