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.