Question

Find all the real zeros of the function : g(x) = 4 (x-1) (x²+4) (x+6)

Answer 1

Answer:

g(x) = 4(x - 1)(x2 + 4)(x + 6)

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

The roots are:

1, -2, and -6

(x - 1), or 1,  has a multiplicity of 4,

(x + 2), or -2, has a multiplicity of 2,

and (x + 6), or -6, has a multiplicity of 1

To determine whether or not they're real zeros, substitute them into the equation.

g(1) = 4(1 - 1)2(1 + 2)(1 + 6)

= 4(0)(3)^2(7)

= 0(9)(7)

= 0

g(-2) = 4(-2 - 1)(-2 + 2)^2(-2 + 6)

= 4(-3)(0)^2(4)

= (-12)(0)(4)

= 0

g(-6) = 4(-6 - 1)(-6 +2)^2(-6 + 6)

= 4(-7)(-4)^2(0)

= (-28)(16)(0)

= 0

Since all of the roots, when substituted into the equation equal 0, they're all real zeros.

(Sorry for this being so long..I hope it helped!)