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.

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .