Question

Find all the zeroes of the equation. −2x^4−64x^2+288=0

Answer 1

-2x^4 - 64x^2 + 288 = -2(x^2-4)(x^2 + 36)

so the real roots are x = ±2

Answer 2

To find the zeroes of the equation -2x^4 - 64x^2 + 288 = 0, we can use factoring and the zero product property:

First, let's write the equation in standard form: -2x^4 - 64x^2 + 288 = 0.

Now we can see that there is a common factor of 2 in all the terms. Let's factor this out: 2(-x^4 - 32x^2 + 144) = 0.

Next, we can try factoring the expression within the parentheses:
-1(x^4 + 32x^2 - 144) = 0.

This further simplifies to:
-(x^4 + 32x^2 - 144) = 0.

Now, let's look for two numbers that multiply to -144 and add up to 32. It turns out that 18 and (-8) satisfy these conditions. Thus, we can factor the expression as:
-(x^2 - 8)(x^2 + 18) = 0.

Using the zero product property, we can set each factor equal to zero and solve for x:
x^2 - 8 = 0 OR x^2 + 18 = 0.

For the first equation, adding 8 to both sides gives us:
x^2 = 8.

Taking the square root of both sides, we get:
x = ±√8 = ±2√2.

For the second equation, subtracting 18 from both sides gives us:
x^2 = -18.

Since there is no real number that satisfies x^2 = -18 (as the square of a real number is always non-negative), there are no real zeroes for this equation.