Question

f(x) = x3 + 3x2 - x - 3

Part A: What are the factors of f(x)? Show your work. (3 points)

Part B: What are the zeros of f(x)? Show your work. (2 points)

Part C: What are the steps you would follow to graph f(x)? Describe the end behavior of the graph of f(x). (5 points)
a few moments ago

Answer 1

\[f(x) = x^3 + 3x^2 - x - 3\]

Answer 2

the numbers that possibly can be zeros are.
\[1,~~-1,~~~~~3,~~-3\]

Answer 3

Check them by plugging in,
x=1\[(1)^3+3(1)^2-(1)-3=0~~~~~~~~~1+3-1-3=0~~~~~~~~~~4-4=0\]WORKS.

Answer 4

\[checking~~~~~x=-1\]\[(-1)^3+3(-1)^2-1-3~~~~~~~~~=-1+3-(-1)-3~~~~~~~~~~~works.\]

Answer 5

\[(-3)^3+3(-3)^2-(-3)-3~~~~~~~~~~~~~-27+27+3-3~~~~~~works\]

Answer 6

zeros are 1, -1, -3

so the roots are
f(x)=(x-1)(x+1)(x+3)

Answer 7

Algebraic way to do it. First one was college algebra.\[f(x) = x^3 + 3x^2 - x - 3\]\[f(x) = x^3 -x+ 3x^2 - 3\]\[f(x) = x(x^2 -1) + 3(x^2 - 1)\]\[f(x)=(x^2-1)(3+x)\]\[f(x)=(x+1)(x-1)(x+3)\]

zeros.\[(x+1)(x-1)(x+3)=0~~~~~~~~~x=-1,~~~x=1,~~~~~x=-3\]

Answer 8

Back to college algebra.

y-intercept, looking at the\[f(x)=x^3+3x-x-3~~~~~~~~is~~-3.~~~or~~~~(0,-3)\]x-intercepts\[-1~~~1~~~3~~~~~~~OR.~~~~~~(-1,0)~~~~~(1,0)~~~~~~(3,0)\]
and the coefficient of the 3rd degree is positive, so it is going up to the right and down to the left.