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प्रश्न
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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उत्तर
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
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