The original can be rewritten as 
. Because i^2 is equal to -1, we can replace the -1 in each radicand with i^2, like this: 
. Now, i-squared is a perfect square that can be pulled out of each radicand as a single i. 
. 24 has a perfect square hidden in it. 4 * 6 = 24 and 4 is a perfect square. So let's break this up, step by step. 
and then 
. We will now multiply the i and the 2i, and multiply the square root of 6 times the square root of 6: 
. 36 itself is a perfect square because 6 * 6 = 36. So we will do that simplification now. 
. Multiplying the 2 and the 6 gives us 
. But here we are back to the fact that i-squared is equal to -1, so 2(-1)(6) = -12. See how that works?
Which expression is equivalent to 8x + 14 + 9x - 5
1 answer:
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Answer:
- zeros: x = -3, -1, +2.
- end behavior: as x approaches -∞, f(x) approaches -∞.
Step-by-step explanation:
I like to use a graphing calculator for finding the zeros of higher order polynomials. The attachment shows them to be at x = -3, -1, +2.
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The zeros can also be found by trial and error, trying the choices offered by the rational root theorem: ±1, ±2, ±3, ±6. It is easiest to try ±1. Doing so shows that -1 is a root, and the residual quadratic is ...
x² +x -6
which factors as (x -2)(x +3), so telling you the remaining roots are -3 and +2.
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For any odd-degree polynomial with a positive leading coefficient, the sign of the function will match the sign of x when the magnitude of x gets large. Thus as x approaches negative infinity, so does f(x).
