Multiply.
1 answer:
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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
===============
Answer: Roots (-3, 0)(-1 , 0)
Answer: -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.
Answer: C
