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inn [45]
4 years ago
14

Find algebraically the zeros for p(x)=x^3+x^2-4x-4

Mathematics
1 answer:
raketka [301]4 years ago
3 0
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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