Question

verify that 1,-1,-3 are the zeroes of the cubic polynomial x cube+3x square - x-3 and check the relationship between zeros and coefficients​

Answer 1

Answer:

${1}^{3} + 3( {1}^{2} ) - 1 - 3 = 1 + 3 - 1 - 3 = 0$

${ - 1}^{3} + 3 ( { - 1}^{2} ) - ( - 1) - 3 = - 1 + 3 + 1 - 3 = 0$

${ - 3}^{3} + 3( { - 3}^{2} ) - ( - 3) - 3 = - 27 + 27 + 3 - 3 = 0$

sum of zeroes=-b/a=-3

sum of product of two zeroes=c/a=-1

product of zeroes=-d/a=3