Answer
The solutions to this equation are $x = -4, -2, 2, \text{ and }4$.
Work Step by Step
We want to simplify this equation by factoring. Factoring this equation is almost the same as factoring a quadratic equation.
To factor a polynomial in the form $x^4 + bx^2 + c = 0$, we look at factors of $c$ such that, when added together, equal $b$.
For the equation $x^4 - 20x^2 + 64 = 0$, $c=64$, so we have to look for factors of $64$ that when added together will equal $b$ or $-20$. We have a couple of possibilities:
$64 = (-8)(-8)$
sum of the factors is $-16$
$64=(-16)(-4)$
sum of the factors is $-20$
The second pair, $-16$ and $-4$, will work since the sum of the factors is equal to $-20$, the middle term's numerical coefficient.. Let us factor that polynomial:
$(x^2 - 16)(x^2 - 4)=0$
We can see that both binomials can be factored further because they are the difference of two squares. We use the formula to factor the difference of two squares, which is:
$a^2 - b^2 = (a + b)(a - b)$
In the equation $x^2 - 16 = 0$, $a$ is the $\sqrt {x^2}$ or $x$, and $b$ is $\sqrt {16}$ or $4$. In the equation $x^2 - 4 = 0$, $a$ is the $\sqrt {x^2}$ or $x$, and $b$ is $\sqrt {4}$ or $2$. We plug these values into the formula:
$(x + 4)(x - 4)(x + 2)(x - 2) = 0$
We use the Zero-Product Property by equating each factor to $0$, then solving each equation.
First factor:
$x + 4 = 0$
$x = -4$
Second factor:
$x - 4 = 0$
$x = 4$
Third factor:
$x + 2 = 0$
$x = -2$
Fourth factor:
$x - 2 = 0$
$x = 2$
The solutions to this equation are $x = 4, -4, 2, -2$.
