Question

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

Answer 1

Answer:

x = 0

x = -4

x = -3

Step-by-step 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.