When divided by $x - 3$ the polynomials $x^{3} - p x^{2} + x + 6$ and $2 x^{3} - x^{2} - (p + 3) x - 6$ leave the same remainder. Find the value of $^{'} p^{'}$ .

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Solution

The Remainder is same whne $(x - 3)$ divides $(x^{3} - p x^{2} + x + 6)$ & $(2 x^{3} - x^{2} - (p + 3) x - 6)$
$∴$ Using Remainder Theorem
$R (3) = x^{3} - p x^{2} + x + 6$
$= 3^{3} - p (3^{2}) + 3 + 6$
$= 27 - 9 p + 3$
$= 36 - 9 p$

$R (3) = 2 x^{3} - x^{2} - (p + 3) x - 6$
$= 2 (3^{3}) - 3^{2} - (p + 3) 3 - 6$
$= 2 \times 27 - 9 - 3 p - 9 - 6$
$= 54 - 24 - 3 p$
$= 30 - 3 p$

Remainder are same
$∴ 36 - 9 p = 30 - 3 p$

$36 - 30 = - 3 p + 9 p$

$6 = 6 p$

$1 = p$

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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′.

When divided by $x - 3$ the polynomials $x^{3} - p x^{2} + x + 6$ and $2 x^{3} - x^{2} - (p + 3) x - 6$ leave the same remainder. Find the value of $^{'} p^{'}$ .