To show that the polynomial function [tex]\( f(x) = x^4 - 6x^3 - 30x^2 + 48x + 136 \)[/tex] has a real zero between the numbers [tex]\( -4 \)[/tex] and [tex]\( -3 \)[/tex], we will use the Intermediate Value Theorem. The Intermediate Value Theorem states that if a function [tex]\( f \)[/tex] is continuous on a closed interval [tex]\([a, b]\)[/tex] and if [tex]\( f(a) \)[/tex] and [tex]\( f(b) \)[/tex] have opposite signs, then there is at least one [tex]\( c \)[/tex] in the interval [tex]\((a, b)\)[/tex] such that [tex]\( f(c) = 0 \)[/tex].
First, let's calculate the value of the polynomial at [tex]\( x = -4 \)[/tex]:
[tex]\[
f(-4) = (-4)^4 - 6(-4)^3 - 30(-4)^2 + 48(-4) + 136
\][/tex]
Simplifying this step-by-step:
[tex]\[
(-4)^4 = 256
\][/tex]
[tex]\[
-6(-4)^3 = -6(-64) = 384
\][/tex]
[tex]\[
-30(-4)^2 = -30(16) = -480
\][/tex]
[tex]\[
48(-4) = -192
\][/tex]
Adding these results together:
[tex]\[
f(-4) = 256 + 384 - 480 - 192 + 136 = 104
\][/tex]
So, [tex]\( f(-4) = 104 \)[/tex].
Next, let's calculate the value of the polynomial at [tex]\( x = -3 \)[/tex]:
[tex]\[
f(-3) = (-3)^4 - 6(-3)^3 - 30(-3)^2 + 48(-3) + 136
\][/tex]
Simplifying this step-by-step:
[tex]\[
(-3)^4 = 81
\][/tex]
[tex]\[
-6(-3)^3 = -6(-27) = 162
\][/tex]
[tex]\[
-30(-3)^2 = -30(9) = -270
\][/tex]
[tex]\[
48(-3) = -144
\][/tex]
Adding these results together:
[tex]\[
f(-3) = 81 + 162 - 270 - 144 + 136 = -35
\][/tex]
So, [tex]\( f(-3) = -35 \)[/tex].
Since [tex]\( f(-4) = 104 \)[/tex] is positive and [tex]\( f(-3) = -35 \)[/tex] is negative, the signs change between [tex]\( f(-4) \)[/tex] and [tex]\( f(-3) \)[/tex]. According to the Intermediate Value Theorem, this indicates that there is at least one zero of the polynomial between [tex]\( -4 \)[/tex] and [tex]\( -3 \)[/tex].
Therefore, the correct answer is:
[tex]\[ \boxed{\text{B. A zero of the polynomial lies between -4 and -3.}} \][/tex]
