Question

verify that -2,1,3 are zeroes of cubic polynomial p(x) = x^3 -2x^2 -5x+6

Answer 1

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Answer 2

Step-by-step explanation:

let p(x) = -2

therefore p(-2) = (-2)^3-2(-2)^2-5(-2)+6=0

= -8-8+19+6=0

= -16+16 =0

0=0 hence it is a zero of the polynomial

letp(x) = 1

p(1)=(1)^3-2(1)^2-5(1)+6=0

=1-2-5+6=0

=7-7=0

0=0 hence it is a zero of the polynomial

p(x) = 3

p(3)=(3)^3-2(3)^2-5(3)+6=0

=27-18-15+6=0

=33-33=0

0=0 hence it is a zero of the polynomial

Hence verified

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