1. Make sure your answers and work is SHOWN please and steps are in order. 2. List all of the possible rational roots.
3. Use synthetic division to test the possible rational roots and factor as far as possible.
4. Identify complex roots, if possible.
5. Sketch a graph of the function, please show:
a. the end behavior correctly
b. the shape of the near x-int
c. (if possible) anything you can learn by considering symmetry and transformations.

1a. p(x)- 2x^4-3x^3-6x^2+5x+6

2a. p(x)= x^4 + 4x^3 + 13x^2 + 36x + 36

3a. p(x)= 2x^5 - 9x^4 + 6x^3 + 22x^2 - 20x - 25

Been a while since I've done synthetic division.

1a. Let's assume that's supposed to be an equals sign

p(x) = 2x⁴ - 3x³ - 6x² + 5x + 6

possible rational roots have factors of 6 in the numerator and of 2 in the denominator.  We'll only worry about negative numerators.

Factors of six: 1,2,3,6, and we don't forget -1,-2,-3,-6

Factors of 2: 1,2

Possible rational roots:

(dividing by 1:) 1,-1,2,-2,3,-3,6,-6

(dividing by 2:) 1/2, -1/2  (2/2=1 is a duplicate, don't have to repeat it), 3/2, -3/2

Possible rational roots: 1,-1,2,-2,3,-3,6,-6, 1/2, -1/2, 3/2, -3/2

Synthetic division, trying x=1,

   1 | 2  -3  -6  5  6

             2  -1  -7  -2

       2    -1  -7  -2  4

Got a remainder of 4, so 1 isn't a root;

Trying x=-1  

-1 | 2  -3  -6  5  6

         -2   5  1  -6

     2  -5  -1  6  0

Zero remainder, found a root, x=-1.  This division says

(2x⁴ - 3x³ - 6x² + 5x + 6) / (x + 1) = 2x³ - 5x² - x + 6

Same set of rational roots on the cubic, we continue with x=2

2 | 2 -5 -1 6

         4 -2 -6

    2  -1 -3 0

Another zero remainder, x=2 is a root.  We're left with 2x² - x - 3 = 0, which factors as

(2x - 3)(x + 1) = 0

That's a second factor of x+1.  Our final factorization is

2x⁴ - 3x³ - 6x² + 5x + 6 = (x + 1)²(x-2)(2x - 3)

Fourth degree with a positive leading coefficient so goes to +infinity at both ends.  Double zero at x=-1, so it's tangent there, just touching the x axis, then down through x=3/2 and up through x=2.  

I'll leave the actually sketching and the other two polynomials to you -- that took some time.

step-by-step explanation:

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?