Hi,
As 9 is one of the 0's, we can write the function as:
f(x) = 2x³ -- 10x² -- 71x -- 9 = (2x² + nx + p)(x -- 9) (1).
Normally, I would have written mx² + nx + p, instead of 2x² + nx + p, but it is obvious that m = 2, because the only term on 3rd power is 2x³.
If we open the parentheses on (1), we get:
2x³ -- 10x² -- 71x -- 9 = (2x² + nx + p)(x -- 9) = 2x³ -- 18x² + nx² -- 9nx + px -- 9p, or
2x³ -- 10x² -- 71x -- 9 = 2x³ + (n -- 18)x² + (p -- 9n)x -- 9p.
Corresponding coefficients are equal:
n -- 18 = --10, therefore n = 8.
p -- 9n = --71, or p -- 9·8 = --71, therefore p = 1.
Therefore f(x) = (2x² + 8x + 1)(x -- 9).
Second and third 0s for f(x) (if they exist), need to be found from 2x² + 8x + 1 = 0.
Δ = b² -- 4·a·c = 8² -- 4·2·1 = 64 -- 8 = 56 = 2√14.
Second 0 would therefore be: x₂ = (--8 -- 2√14) / (2·2) = (--4 -- √14)/2.
Third 0 would therefore be: x₃ = (--8 + 2√14) / (2·2) = (--4 + √14)/2.
Green eyes.
