Question

1) How many zeros does the function f(x) = 2x^2- 3x^3 + x have?

2) Factor the expression 3x^3- 15x^2 + 18x.

Answer 1

And lastly, if you can see it. #10. @whpalmer4

Answer 2

Factor \[3x^3-15x^2+18x\]

\[3x^3-15x^2+18x = 3x(x^2 - 5x + 6) \]

Now we see that we have \[ax^2 + bx + c\]where a = 1, b = -5, c = 6

We also notice that 2*3 = 6 and 2+3 = 5

\[(x-2)(x-3)=x^2-3x-2x+(-2)(-3) = x^2 - 5x +6\]

So our factored equation is \[3x(x-2)(x-3)\]

Answer 3

That makes so much more sense! Thanks.

Answer 4

#10 we need to classify the function as linear, quadratic, cubic, or quartic. A linear equation in x will have the same change in y for a given change in x wherever we do it. If going from x = 0 to x = 1 makes y go from 3 to 6, going from x = 1 to x = 2 will cause the same change (adding 3), as will going from x = 1000 to x = 1001, etc. Our function isn't linear, because our table shows x increasing by 1 from 0 to 3 and y changes by 2, 4, 6 on those steps.

Answer 5

Sketching a graph of the values in the table, it looks like it could be a parabola up until the point we put in (5,19), which forces the line to bend away from the nice parabola shape. A quadratic would give us a parabola of some sort. A cubic gives you an s shape of some sort. I'm not sure what this one is, other than that it isn't linear or quadratic. Well, I think it's quartic, I just don't know how to get there without a lot of handwaving, sorry :-)

Answer 6

Now you need to count the zeros of f(x) = 2x^2- 3x^3 + x. That's easy — there's a zero for each power of x. We've got a cubic equation here (x^3 term) so there will be 3 zeros.
\[f(x) = 0 = x(2x-3x^2+1) \]

One zero is at x = 0, clearly.

We can use the quadratic formula to find the zeros of the other terms. Rearranging we get

\[-3x^2+2x + 1 = 0\]

so a = -3, b = 2, c = 1

and x = \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} = \frac{-2\pm\sqrt{2^2-4(-3)(1)}}{2(-3)}\]

\[= \frac{-2\pm 4}{-6} = 1,-\frac{1}{3}\]

Answer 7

So our zeros are f(x) = 0 for x = 0, x = 1, x = -1/3