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5 votes
5 votes
Determine algebraically the zeros of f(x)=3x3+ 21x2+36x

User Jakewins
by
3.1k points

1 Answer

17 votes
17 votes

Answer:

x = 0

x = -4

x = -3

Explanation:

Hello!

First, lets factor out 3x (GCF) from the whole expression:


  • f(x) = 3x^3 + 21x^2 + 36x

  • f(x) = 3x(x^2 + 7x + 12)

Factor the polynomial by grouping:


  • f(x) = 3x(x^2 + 7x + 12)

  • f(x) = 3x(x^2 + 3x + 4x + 12)

  • f(x) = 3x(x(x+3) + 4(x + 3))

  • f(x) = 3x(x + 4)(x + 3)

Set the expression and each factor to 0, and solve for x:


  • 0 = 3x(x + 4)(x + 3)

3x = 0

  • x = 0

x + 4 = 0

  • x = -4

x + 3 = 0

  • x = -3

There are 3 roots for this equation: 0, -4, and -3.

User Tudor
by
2.9k points
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