Question

show that -1,1and 3 are zeroes of polynomial x3-3x2-x+3​

Answer 1

Step-by-step explanation:

let p(X)=x^3-3x^2-x+3

if -1,1,3 are zeroes of polynomial,

then p(-1)=p(1)=p(3)=0

p(-1)=(-1)^3-3(-1)^2-(-1)+3

=-1-3+1+3=0

p(1)=(1)^3-3(1)^2-(1)+3

=1-3-1+3=0

p(3)=(3)^3-3(3)^2-(3)+3

=27-27-3+3=0

Answer 2

Answer:

$x( - 11) = - 11 \times 3 - 3 \times 2 \times - 11 + 3 \\ = - 33 - 6 \times - 11 + 3 \\ = 30 \\ x(3) = 3 \times 3 - 3 \times 2 \times 3 + 3 \\ = 9 - 18 + 3 \\ = - 6$