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Consider this polynomial equation

6(x-3)(x^2+4)(x+1)=0

the equation has ( A. 2 B.3 C.4) solutions. it's real solutions are x = (-2 -1 and 2 or. -1 and 3)​

Consider this polynomial equation 6(x-3)(x^2+4)(x+1)=0 the equation has ( A. 2 B.3 C-example-1

1 Answer

4 votes

Answer:

2 solutions

-1 and 3 is the real solutions

User Bootsmaat
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