71 !
use long division or synthetic division to factor each polynomial completely with the given factor then give the zeros of the function
f(x) = 6x^3+25x^2-29x-20; (2x+1)

f(x) = (2x+1)(3x−4)(x+5)

x = -5, x = -1/2, and x = 4/3

Step-by-step explanation:

Start by setting up the long division:

Look at the first terms.  2x goes into 6x³ a number of 3x² times.

2x+1 times 3x² is 6x³+3x².

Subtract.

Now drop down the next term (the -29x).

Repeat the process.  2x goes into 22x² a number of 11x times.

Multiply:

Subtract:

Drop down the next term (-20):

2x goes into -40x a number of -20 times.

Multiply:

Subtract:

The remainder is 0.  So f(x) can be factored as:

f(x) = (2x+1)(3x²+11x−20)

Which we can further factor using AC method:

f(x) = (2x+1)(3x−4)(x+5)

So the zeros are x = -5, x = -1/2, and x = 4/3.

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.