Question

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Answer 1

Question :-

If both ax³ + 2x² - 3 and x² - ax + 4 leave the same remainder when divided by x - 2 . Find the value of a ?

Answer :-

Given :-

p ( x ) = ax³ + 2x² - 3

q ( x ) = x² - ax + 4

( x - 2 ) when divides p ( x ) & q ( x ) leaves same remainder

Required to find :-

• Find the value of a ?

Concept used :-

• Remainder theorem

• Factor theorem

Solution :-

p ( x ) = ax³ + 2x² - 3

q ( x ) = x² - ax + 4

( x - 2 ) when divides p ( x ) & q ( x ) leaves same remainder

We need to find the value of a

So,

p ( x ) = ax³ + 2x² - 3

( x - 2 ) when divides p ( x ) leaves remainder

Let

➜ x - 2 = 0

➜ x = 2

substitute the value of x in p ( x )

➜ p ( 2 ) = a ( 2 )³ + 2 ( 2 )² - 3

➜ a ( 8 ) + 2 ( 4 ) - 3

➜ 8a + 8 - 3

➜ 8a + 5

So,

when ( x - 2 ) divides p (x ) the remainder is 8a + 5

Similarly,

q ( x ) = x² - ax + 4

( x - 2 ) when divides q ( x ) leaves remainder 8a + 5

This is because,

In the question it is mentioned that when ( x - 2 ) divides p ( x ) and q ( x ) it leaves same remainder

So,

Let

x - 2 = 0

x = 2

substitute this value in place of x in q ( x )

So,

q ( 2 ) =

( 2 )² - a ( 2 ) + 4 = 8a + 5

4 - 2a + 4 = 8a + 5

4 - 2a + 4 - 8a - 5 = 0

8 - 10a - 5 = 0

3 - 10a = 0

- 10a = - 3

( - ) negative signs get cancelled on both sides

10a = 3

$\rm{a = \dfrac{3}{10}}$

a = 0.3

Verification :-

Substitute the value of a in p ( x )

p ( x ) = ax³ + 2x² - 3

p ( x ) = 0.3x³ + 2x² - 3

( x - 2 ) divides p ( x )

p ( 2 ) = 0.3 ( 2 )³ + 2 (2 )² - 3

=> 0.3 ( 8 ) + 8 - 3

=> 2.4 + 5

=> 7.4

Substitute the value of a in q ( x )

q ( x ) = x² - ax + 4

q ( x ) = x² - 0.3x + 4

q ( x ) when divided by ( x - 2 )

q ( 2 ) = ( 2 )² - 0.3 (2) + 4

=> 4 - 0.6 + 4

=> 8 - 0.6

=> 7.4

Since it is given that ( x - 2 ) when divides leaves same remainder so the remainder in the case of p ( x ) and q ( x ) is same .

So,

Hence verified .