Can someone with these? will give brainliest.

Nine

Correct

Ten

The only way I have of solving any of these is either to graph them, or solve them as cubics.

This one is graphed. The choice is C. All are real and 2 are the same. See graph below.

Eleven

Using Pascal's triangle, the fifth line is 1    5   10   10    5    1

So the answer should be 5p^4q

Twelve

Use Synthetic Division

2  || 1     2   - 5     -6

            2     8      6

=====================

      1     4     3       0

So the quadratic that's left of this division is

x^2 +4x + 3 =0 for the roots.   Factor

(x +3)(x + 1)

These equal zero                

x + 3 = 0

x = - 3

x + 1 =0

x = - 1

===============

Roots (-3, 0)(-1 , 0)

-3,-1,2

Thirteen

Use synthetic division on all of them. Start out with trying to find -1 as a root. The one that works is C. Then you try -2 also using synthetic division. It also turns out to be a solution for the quartic. What is left over is x^ + 1 which does not factor in the reals.

- 1 || 1    3     3     3     2

           -1     -2    -1    -2

=======================

      1     2    1      2      0

-2 || 1    2    1     2

           -2    0   -2

===================

     1    +0   1

The left over is x^2 + 1 = 0 The roots of this are +/- i which is not in the real number system.    

C

idn

step-by-step explanation:

"a" determine whether sage found the correct quotient. show your work

step-by-step explanation: