This is a polynomial of degree 4, because you have 4 roots real or imaginary:1st) x= 5 →(x-5)=0
2nd) x = - 3→(x+3) =0
3rd) x = -1+3i →(x+1-3i) = 0
and the 4th one that is not mentioned which is the conjugate of
-1+3i, that is -1-3i (in any polynomial if a root has the form of a+bi, there is always a conjugate root = a-bi)
4th) x= -1-3i→(x+1+3i) = 0
Hence the polynomial =(x-5)(x+3)((x+1-3i)(x+1+3i)
Solving the above will give you (unless I am mistaken, pls recalculate):
x⁴-9x²-50x-150<u />
Answer:
the value is 8
Step-by-step explanation:
-2s + 14 > 4s + 8
6 > 6s
s < 1
So it's {-<span>∞, . . . . ., -2, -1, 0}
Hope this helps.</span>
Squaring both sides of the given equation: w^2=7w, or w^2-7w=0.
w^2-7w=0 can be factored: w(w-7)=0. Then, w=7 or w=0. Both 0 and 7 are solutions of this equation.