If f(x) = x^3 – 10x^2 + 29x – 30 and f(6) = 0, then find all
of the zeros of f(x) algebraically.
1 answer:
Step 1: 1
(((x3) - (2•5x2)) + 29x) - 30 = 0
Step 2 2.1 x3-10x2+29x-30 is not a perfect cube
Step 3 Factoring: x3-10x2+29x-30
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 29x-30
Group 2: -10x2+x3
Pull out from each group separately :
Group 1: (29x-30) • (1)
Group 2: (x-10) • (x2)
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Answer:
B
Step-by-step explanation:
5.6 is bigger than 4.3
-3x+5y=-4
5y=3x-4
Y=3/5x-4/5
When it parallel to the line y=3/5x-4/5
It can be y=-3/5x+b
-5=6/5+b
B=-5-6/5=-25/5-6/5=-31/5
Y=-3/5x-31/5
3/5x+y=-31/5
Multiply 5
3x+5y=-31
The answer is h(5) = 75.
Where x=5, y=75
We have

so that
