Question

How do you factor and solve each polynomial completely `y=x^4-14x^2+40`?

Answer 1

`x^4-14x^2+40 = (x-sqrt(10))(x+sqrt(10))(x-2)(x+2)`

with zeros: `+-sqrt(10)` and `+-2`

Explanation:

The difference of squares identity can be written:

`a^2-b^2 = (a-b)(a+b)`

We use this a couple of times below.

Note that `10+4 = 14` and `10*4 = 40`

So we find:

`x^4-14x^2+40 = (x^2-10)(x^2-4)`

`color(white)(x^4-14x^2+40) = (x^2-(sqrt(10))^2)(x^2-2^2)`

`color(white)(x^4-14x^2+40) = (x-sqrt(10))(x+sqrt(10))(x-2)(x+2)`

So this quartic has zeros:

`x = +-sqrt(10)" "` and `" "x = +-2`