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)
(((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: 230cm²
Step-by-step explanation:
>We can do this by splitting the figure into 2 trapeziums.
>Additionally, you can see that both trapeziums bear the same lengths a and b. Hence, we can use this formula: .
>Thus,
Area =