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351. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and coefficients in each case:
x³ - 4x² + 5x - 2; 2, 1, 1

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Short Answer Type

352. Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

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Long Answer Type

353. If the zeroes of the polynomial x³ – 3x²+ x + 1 are (a – b), a, (a + b), find a and b.

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354. If two zeroes of the polynomial x⁴ – 26x³-26x² + 138x –35 are 2 ±

square root of 3

, find other two zeroes.

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355. If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² - 2x + k, the remainder comes out to be x + a, find ‘k’ and ‘a’.

By division algorithm, we have It is given that f(a) = x⁴ – 6x³ + 16x² – 25x + 10, when divided by x² – 2x + k leaves x + a as remainder.
:. f(x) – (x + a) = x⁴– 6x³ + 16x²–26x + 10 – a is is exactly divisible by x² – 2x + k.
Let us now divide x⁴ – 6x³ + 16x² – 26x + 10 – a by x² – 2x + k.
By division algorithm, we have It is given that f(a) = x4 – 6x3 +

By division algorithm, we have It is given that f(a) = x4 – 6x3 +

For f(x) – (x + a) = x⁴ – 6x³ + 16x² – 26x + 10 – a to be exactly divisible by x² – 2x + k, we must have
(–10 + 2k) x + (10 – a – 8k + k²) = 0 for all x
⇒ – 10 + 2k = 0, 10 – a – 8k + k² = 0
⇒ k = 5, 10 – a – 40 + 25 = 0
⇒ k = 5 and a = – 5.

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