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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'.

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Solution

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.

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