You didn't give the fourth zero, but the answer is still false. If you have a root or an imaginary number as a zero, then its conjugate is also a zero. So if 8i is a zero, then -8i must also be a zero, and if 4i is a zero, then -4i must be a zero, with those zeros and -4, the number of zeroes exceeds the number of zeroes that a fourth degree polynomial can have.
1. (5+23)+65 = (5+65)+23 = 70+23 = 93 (D)
2. -(4x-7) = -4X+7 (C)
3. 2(6X+9) = 12X+18 (B)
4. 5(X-3) = 35
X-3 = 35/5
X-3 = 7
SO X= 7+3 = 10
THEN NO
9 IS NOT A SOLUTION
Answer:
dude you need to show the whole answer
Step-by-step explanation:
sorry I couldn't help
Righ, so negative times positive=negative
negative+positive=positive
thereofr
positive>negative
factor -84 where the negative is smaller ex -12 and 13, not -14 and 10 so
-2 and 42 nope
fast forward
-6 and 14 yes
-6+14=8
so the numbers are -6 and 14
Answer:
(X+3)
Step-by-step explanation:
