Find all real zeros of the polynomial $f ( x ) = x ^ { 4 } - 41 x ^ { 2 } + 400$ and determine the mutiplicity of each.
A) $x = 25$ , multiplicity 2; $x = 16$ , multiplicity 2
B) $x = 5$ , multiplicity 2; $x = - 4$ , multiplicity 2
C) $x = 25$ , multiplicity 2; $x = - 4$ , multiplicity 1
D) $x = - 5$ , multiplicity 2; $x = 4$ , multiplicity 2
E) $x = 5$ , multiplicity 1; $x = - 5$ , multiplicity 1; $x = - 4$ , multiplicity 1; $x = 4$ , multiplicity 1

Question

Quizplus · Accepted Answer

To find the real zeros of the polynomial $f(x) = x^4 - 41x^2 + 400$, we can set it equal to zero and solve for $x$. This equation can be factored as a quadratic in terms of $x^2$, giving $(x^2 - 25)(x^2 - 16) = 0$. This further factors into $(x - 5)(x + 5)(x - 4)(x + 4) = 0$, yielding the real zeros $x = 5, -5, 4, -4$, each with multiplicity 1.

Find All Real Zeros of the Polynomial f(x)=x4−41x2+400f ( x ) = x ^ { 4 } - 41 x ^ { 2 } + 400f(x)=x4−41x2+400

Find All Real Zeros of the Polynomial $f ( x ) = x ^ { 4 } - 41 x ^ { 2 } + 400$