The zeroes of <span>f(x) = x^5-12x^2+32x can be found by factoring,
</span><span>f(x) = x^5-12x^2+32x=(x-8)(x-4)
By the zero product theorem, (x-8)=0 or (x-4)=0 which means
x=8 or x=4.
So the zeroes of f(x) are S={4,8}</span>
Pretty sure the answer is 8
Answer:
50
Step-by-step explanation:
- 8i = √-64
8i + 6
- -4i = -√-16
-4i + 3
[8i + 6][-4i + 3] → FOIL
>> -32i² + [24i - 24i] + 18
↑ ↑
-1 0
>>> 32 + 18 = 50
Information on Imaginary Numbers
√-1 = i
-1 = i²
-i = i³
1 = i⁴ [And every other exponent that is a multiple of 4; this cycle then repeats itself every time you go up one number at a time]
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Answer:
b=(2a-1)/(3a-1)
Step-by-step explanation:
a=(b-1)/(3b-2)
3ab-2a=b-1
3ab-b=2a-1
b(3a-1)=2a-1
b=(2a-1)/(3a-1)
