Question

If two polynomial ax³+4x²+3x-4 & x³-4x+a leave the same remainder when divided by (x-3), find the value of a.​

Answer 1

Given :-

p ( x ) = ax³ + 4x² + 3x - 4

q ( x ) = x³ - 4x + a

( x - 3 ) when divides p ( x ) and q ( x ) leaves same remainder

Required to find :-

• Value of " a " ?

Concept used :-

• Remainder Theorem

Solution :-

Given that ;

p ( x ) = ax³ + 4x² + 3x - 4

q ( x ) = x³ - 4x + a

( x - 3 ) when divides p ( x ) and q ( x ) leaves same remainder

we need to find the value of " a " .

So,

p ( x ) = ax³ + 4x² + 3x - 4

( x - 3 ) when divides leaves remainder

So,

Let

x - 3 = 0

x = 3

Substitute the value of x in p ( x )

p ( x ) = ax³ + 4x² + 3x - 4

p ( 3 ) = a ( 3 )³ + 4 ( 3 )² + 3 ( 3 ) - 4

p ( 3 ) = a ( 27 ) + 4 ( 9 ) + 9 - 4

p ( 3 ) = 27a + 36 + 9 - 4

p ( 3 ) = 27a + 45 - 4

p ( 3 ) = 27a + 41

Similarly,

q ( x ) = x³ - 4x + a

( x - 3 ) when divides q ( x ) leaves remainder

So,

Let

x - 3 = 0

x = 3

q ( 3 ) = ( 3 )³ - 4 ( 3 ) + a

Since it is given that ( x - 3 ) when divides p ( x ) and q ( x ) leaves same remainder

So,

(3)³ - 4 ( 3 ) + a = 27a + 41

27 - 12 + a = 27a + 41

27 - 12 - 41 = 27a - a

- 26 = 26a

interchange the terms on both sides

26a = - 26

$\text{ a =}{\dfrac{-26}{26}}$

$\implies{\tt{ a = - 1 }}$

$\large{\leadsto{\boxed{\rm{\therefore{Value \; of\;' a' = -1}}}}}$