Question

If p(x)= 2x3+ a x2+ 3x–3 and f(x) = x3 + x2-4x+ aleave the same remainder when divided by x-2, find the value of a.

Answer 1

Answer:

-13÷3

Step-by-step explanation:

Let p(x) = 2x3 + ax2 + 3x – 5 and q(x) = x3 + x2 – 4x + a

As we know by Remainder Theorem,

If a polynomial p(x) is divided by a linear polynomial (x – a) then, the remainder is p(a)

⇒ Remainder of p(x) when divided by x – 2 is p(2). Similarly, Remainder of q(x) when divided by x – 2 is q(2)

⇒ p(2) = 2(2)3 +a(2)2 + 3(2) – 5

⇒p(2) = 16 + 4a +6 – 5

⇒p(2) = 17 + 4a

Similarly, q(2) = (2)3 + (2)2 + –4(2) + a

⇒ q(2) = 8 + 4 –8 + a

⇒ q(2) = 4 + a

Since they both leave the same remainder, so p(2) = q(2)

⇒ 17 + 4a = 4 + a

⇒ 13 = -3a

\to\sf\:a=\dfrac{-13}{3}→a=3−13

∴ The value of a is –13/3