Question

12. It two zeroes of the polynomial x4 + x3 - 15x2 - 29x - 6 and 2 +- (V5). Find other zeroes.
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Answer 1

EXPLANATION.

Two zeroes of the polynomial

=> x⁴ + x³ - 15x² - 29x - 6

=> if two zeroes are = 2 ± √5

Find all zeroes.

products of zeroes of quadratic polynomial.

=> x = 2 + √5 and x = 2 - √5

=> x - ( 2 + √5 ) and x - ( 2 - √5 )

=> ( x - 2 ) ² - ( √5 ) ²

=> x² + 4 - 4x - 5

=> x² - 4x - 1

On dividing x⁴ + x³ - 15x² - 29x - 9 to

x² - 4x - 1

we get,

=> x² + 5x + 6

Now factories into middle term split.

we get,

=> x² + 3x + 2x + 6 = 0

=> x ( x + 3 ) + 2 ( x + 3 ) = 0

=> ( x + 2 ) ( x + 3 ) = 0

=> x = -2 and x = -3

Therefore,

All zeroes of polynomial are

=> 2 + √5 , 2 - √5 , -2 , -3