Question

Obtain all the zeroes of the polynomial x⁴ + 6x³ + x² - 24x - 20 if two of its zeroes are 2 and -5​

Answer 1

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Note :- if a & b are zeros of Any Polynomial f(x) , than (x - a)(x - b) is completely divide the polynomial f(x).

Solution :-

Given that, 2 & (-5) are the two zeros of polynomial.

So,

→ (x-2)(x+5)

→ x² + 5x - 2x - 10

→ x² + 3x - 10 is completely divide the given Polynomial .

So,

x² + 3x - 10 )x⁴ + 6x³ + x² -24x - 20( x²+3x + 2

x⁴+ 3x³- 10x²

(-ve) 3x³ + 11x² - 24 x

3x³ + 9x² - 30x

(-ve) 2x² + 6x - 20

2x² + 6x - 20

(-ve) 0.

So,

→ x² + 3x + 2 = 0

→ x² + 2x + x + 2 = 0

→ x(x + 2)+1(x+2) = 0

→ (x + 2)(x + 1) = 0

Putting Both Equal to Zero we get,

→ x = (-2) & (-1). (Ans).

Hence, All 4 Zeros of Given Polynomial are [(-5),(-2),(-1) & 2 ]..

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