Question

If the polynomial f(x)=px^3+4x^2+3x-4 and g(x)=x^3-4+p are divided by (x-3) then the remainder is same in each case. Find the value of p.

Answer 1

Hey Mate :

Here is your solution :

Given,

f( x ) = px³ + 4x² + 3x - 4

g( x ) = x³ - 4 + p

r( x ) = ( x - 3 )

Now,

=> ( x - 3 ) = 0

=> x = 3

So, 3 is a zero of r( x ).

By Remainder Theorem,

=> f( x ) = px³ + 4x² + 3x - 4

By substituting x = 3.

=> f( 3 ) = p( 3 )³ + 4( 3 )²+ 3 × 3 - 4

=> f( 3 ) = p( 27 ) + 4( 9 ) + 9 - 4

=> f( 3 ) = 27p + 36 + 9 - 4

=> f( 3 ) = 27p + 45 - 4

=> f( 3 ) = 27p + 41 ------- ( 1 )

Now,

=> g( x ) = x³ - 4 + p

By substituting x = 3 ,

=> g( x ) = 3³ - 4 + p

=> g( x ) = 27 - 4 + p

=> g( x ) = 23 + p ------- ( 2 )

According to question ,

Remainders are equal.

So,

=> 23 + p = 27p + 41

=> 23 - 41 = 27p - p

=> -18 = 26p

=> p = ( -18 / 26 )

=> p = -9/13.

Hence, p = ( -9/13 ).

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Hope it helps !!

Answer 2

Answer:

p=(-9/13)

this is the answer