peterblessd6588 peterblessd6588
  • 09-01-2018
  • Mathematics
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Find algebraically the zeros for p(x)=x^3+x^2-4x-4

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Аноним Аноним
  • 09-01-2018
One  zero is - 1
because p(-1) =  (-1)^3 + (-1)^2 -4(-1)- 4 = 0
So one factor of p(x) is x + 1
Dividing  p(x) by  x + 1  gives the quotient
x^2 - 4
x - 2)(x + 2) = 0

x = 2, -2


The zeroes are -2,-1 and 2
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