Please solve it. I really need it.
1 answer:
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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
Answer:
Yes
Step-by-step explanation:
In any triangle, the sum of any two sides must be larger than the third side. To test this, we only actually need to pick the two shorter sides. In this case, the following inequality is true: