Find all real zeros of the polynomial $f ( x ) = x ^ { 4 } - 61 x ^ { 2 } + 900$ and determine the multiplicity of each.
A) $x = - 5 , \text { multiplicity } 2 ; x = - 6 , \text { multiplicity } 2$
B) $x = 25 , \text { multiplicity } 2 ; x = 36 \text {, multiplicity } 2$
C) $x = 5 , \text { multiplicity } 2 ; x = 6 , \text { multiplicity } 2$
D) $x = 25 , \text { multiplicity } 2 ; x = 6 , \text { multiplicity } 2$
E) $x = 5 \text {, multiplicity } 1 ; x = - 5 \text {, multiplicity } 1 ; x = 6 \text {, multiplicity } 1 ; x = - 6 \text {, multiplicity } 1$

Question

Quizplus · Accepted Answer

To find the real zeros of the polynomial $f(x) = x^4 - 61x^2 + 900$, we can treat it as a quadratic in form by substituting $y = x^2$, giving us $y^2 - 61y + 900$. Factoring this quadratic equation yields $(y - 25)(y - 36) = 0$, so $y = 25$ or $y = 36$. Substituting back $x^2$ for $y$, we get $x^2 = 25$ and $x^2 = 36$, which gives us $x = \pm5$ and $x = \pm6$, each with multiplicity 1, because when we revert the substitution, each $x^2$ term translates to two real roots for $x$, indicating they cross the x-axis at these points, each exactly once.

Find All Real Zeros of the Polynomial f(x)=x4−61x2+900f ( x ) = x ^ { 4 } - 61 x ^ { 2 } + 900f(x)=x4−61x2+900

Find All Real Zeros of the Polynomial $f ( x ) = x ^ { 4 } - 61 x ^ { 2 } + 900$