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
Well not all lines are lines of symmerty, because if you draw a let's say a rectangle on a piece of paper and draw a diagonal line through it, well the two sides don't really lie perfectly on one another!!
The missing reason to complete Hector's proof is
<span>Corresponding Parts of Congruent Triangles Are Congruent
It's been established in the previous statement that triangle LNO and triangle PNM are congruent by the AAS Postulate.
The proof
</span>Corresponding Parts of Congruent Triangles Are Congruent
is comprehensive.
The answer would be 8. hope this helps
Answer:
A) CUB
Step-by-step explanation:
Of the suggested planes, only CUB contains both points C and T.
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<em>Comments on the other answer choices</em>
BED contains point T, but not C
ACE contains point C, but not T
ABE contains neither C nor T