Question

If the polynomials (2x3 +ax2 +3x -5) and (x3 + x2 –4x + a) leave the same remainder when divided by (x – 2), find the value of ‘a’.

Answer 1

Hey there!

Let the given polynomials be f(x) and g(x).

ATQ, 
When f(x) and g(x) are divided by (x-2) they leave the same remainder.

I.e (x-2) is a factor of f(x) and g(x). It means 2 is the zero of f(x) and g(x)

So that,

f(2) = g(2)

2x³+ax²+3x-5 = x³+x²-4x+a

2(2)³+a(2)²+3(2)-5 = 2³+2²-4(2)+a

2(8)+a(4)+6-5 = 8+4-8+a

16+4a+1 = 4+a

17+4a = 4+a

4a-a = 4-17

3a = -13

a = -13/3.

Hope helped!

:)

Answer 2

$<b><i>$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

$\to\sf\:a=\dfrac{-13}{3}$

∴ The value of a is –13/3