Question

When divided by (x-3) the polynomials x3+px2+x+6 and 2x3-x2(p+3)x-6 leave the same remainder.Find the value of P.?

Answer 1

Hey Friend ☺

Using remainder theorem

p ( 3 ) will be the remainder of both polynomials

The remainder for the first polynomial

p ( 3 ) = x^3 + px^2 + x + 6

= ( 3 )^3 + p ( 3 )^2 + ( 3 ) + 6 .

= 27 + 9p + 3 + 6

= 9p + 36

so the remainder is 9p + 36

The remainder for the second polynomial us given by

p ( 3 ) = 2x^3 - x^2 + ( P + 3 )x - 6

= 2 ( 3 )^3 - ( 3 )^2 + ( P + 3 )3 - 6

= 2 × 27 - 9 + 3p + 9 - 6

= 54 - 6 + 3p

= 3p + 48

The remainders are given same

so we get equation

9p + 36 = 3p + 48

9p - 3p = 48 - 36

6p = 12

p = 12/6

p = 2

so the value of p is 2

✌

Answer 2

Hey friend, Harish here.

Here is  your answer:

Let, f(x) =x³ +px²+x+6 

      g(x) = 2x³ - x²+(p+3)x -6

Now , when we divide by (x -3) both equations give same remainder.

Let the remainder be 'r'.

So, When we substitute x = 3 . Then they give 'r' as the value.

So, f(3) → 3³ + p(3²) + 3 + 6 = r .

             ⇒ 27 + 9p + 9 = r
 
             ⇒ 36 +9p = r   - (i)

      g(3) → 2(3³) - 3² + 3(p+3) -6 = r

             ⇒ 54 - 9 + 3p + 9 -6 =r

             ⇒ 48 +3p = r   - (ii)

Now equate (i) & (ii).

 Then, 48 + 3p = 36 + 9p  ( r = r)

           48 - 36 = 9p - 3p
  
            12 = 6p
 
Therefore p = 2.
______________________________________________________

Hope my answer is helpful to u.