Question

2. If P(x) = x^3 + x^2 - 9x -
9 , find P(0), P(3), P(-3) and
P(-1). What do you
conclude about the zeroes of
P(X)? Is O a zero of P(x)?
epresents raised to the​

Answer 1

Answer:

0, 0, 0 only p(0)= -9

Step-by-step explanation:

p(x)= x³+x²-9x-9

p(0)= 0³+0²-9×0-9= -9

p(3)= 3³+3²-9×3-9= 27+9-27-9= 0

p(-3) = (-3)³+(-3)²-9×(-3)-9

= -27+9+27-9 = 0

p(-1)= (-1)³+(-1)²-9×(-1)-9

= -1+1+9-9=0

Answer p(0)= -9 and other are zero

therefore ,3,-3 and -1 are zeroes of the polynomial

Answer 2

Answer:

p(0)=-9,p(-3),p(-1).p(3)=0

Step-by-step explanation:

p(x)=x^3+x^2-9x-9

p(0)=0^3+0^2-9(0)-9

=-9

p(3)=3^3+3^2-9(3)-9

      =27+9-27-9

       =0

p(-3)=(-3)^3+(-3)^2-9(-3)-9

       =-27+9+27-9

        =0

p(-1)=(-1)^3+(-1)^2-9(-1)-9

      =-1+-1+9-9

      =0

p(0) is not a zero of x^3+x^2-9x-9