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

By dividing

x³ – px² + x + 6

And 2x³ – x² – (p + 3)x – 6

By x – 3, the remainder is same

Let x – 3 = 0, then x = 3

Now by Remainder Theorem,

Let p(x) = x³ – px² + x + 6

p(3) = (3)³ – p(3)² + 3 + 6

= 27 – 9p + 9

= 36 – 9p

And q(x) = 2x³ – x² – (p + 3)x – 6

q(3) = 2(3)² – (3)² – (3)² – (p + 3) × 3 – 6

= 2 × 27 – 9 – 3p – 9 – 6

= 54 – 24 – 3p

= 30 – 3p

∵ The remainder in each case is same

∴ 36 – 9p = 30 – 3p

36 – 30 = 9p – 3p

`\implies` 6 = 6p

`\implies p = (6)/(6) = 1`

∴ p = 1