Can you find the zeroes of this function using synthetic division f(x)=2x^3-10x^2-71x-9 ; 9 (9 being the zero given) i can do the synthetic but i can do th ac/b chart because it ends up 2/8 and that doesn't work

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.

a

step-by-step explanation: