Question Find all the real zeros of the function : g(x) = 4 (x-1) (x²+4) (x+6) if there is more than one answer , seperate them with commase

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!) Log in to Reply

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!)