There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/- 
x = +/- i
1824 the estimate is 1820
Answer:
The African Elephant is 2.27 times heavier than a African Lion.
Step-by-step explanation:
9 × 103 pounds = 927E.
4 × 102 pounds = 408L.
The answer is a.0.32 km.
The speed that a tsunami can travel is modeled by the equation is s = 356√d.
It is given:
s = 200 km/h
d = ?
Now, let's substitute s in the equation and find d:
s = 356√d
200 = 356√d
√d = 200 ÷ 356
√d = 0.562
Now, let's square both sides of the equation:
(√d)² = (0.562)²
d = (0.562)² = 0.316 ≈ 0.32
Therefore, <span> the approximate depth (d) of water for a tsunami traveling at 200 kilometers per hour is 0.32 km.</span>
Y = kx, so y/x is constant
you want y such that
y/12 = -2/-8