Question

alg

Answer 1

\[x^2+x-30=0,x^2+6x-5x-30=0\]
make factors and find the values of x.

Answer 2

You didn't show the graph, so here is one:

Answer 3

The roots are the values of \(x\) at which the curve crosses the x-axis \( (y = 0)\).

Answer 4

No guessing allowed, you can get the answer exactly by examining the graph carefully, or by doing as @surjithayer suggested, though it is important to understand the graph technique.

Answer 5

At least on the image of the graph I'm looking at, it is very clear which of those two choices is correct, and which is not.

Answer 6

No need to guess. Look at the spot where it crosses on the left. Is that at x = -6, or x = -5?

Answer 7

Yes, B is correct. There shouldn't be any question that the curve crosses at x = -6, and x = 5.

Answer 8

Well, let's try it out:
For it to be a solution, plugging in the value of \(x\) will give us 0 for the result. If we get something other than 0, it is not a solution.

\[x^2-10x + 11\]\[(-11)^2 -10(-11) + 11 = 121 + 110 + 11 = \text{not 0!}\]

Answer 9

So \(x = -11\) is NOT a solution to that equation, and is therefore the correct choice.

Answer 10

Actually, are you sure you copied that problem correctly? it doesn't make a whole lot of sense as I reread it...

Answer 11

Okay, for end behavior, you look at the highest power term and see what it does if x is really big or really small. What's the highest power term here?

Answer 12

It's odd, because a quadratic only has 2 solutions, and there are 4 different answer choices, so at least 2 of them would have to be wrong...

Answer 13

By "4" do you mean "+4" or "-3x^4"?

Answer 14

Well, what is your answer if the polynomial is \[g(x) = 5x^4+2x^3+3x^2+77\]?

Answer 15

\(-3x^4\) is the highest power term in that equation. The size of the individual numbers doesn't matter, you just need to know which term has the highest exponent.

As \(x\rightarrow\infty, -3x^4\rightarrow -\infty\)
As \(x\rightarrow-\infty, -3x^4\rightarrow -\infty\)

Answer 16

The graph looks like this:

Answer 17

Hold on, do you understand why the answer to the last question is NOT D?

Answer 18

To factor \[x^3-4x^2-4x+16\]I would use factoring by grouping.

\[(x^3-4x^2) - (4x - 16)\] (notice that I changed the sign on the 16 when it went inside the -( ))

Yes, A describes the end behavior of that function.

Answer 19

Now factor each of those two groups:
\[x^2(x-4) - 4(x-4)\]Notice that the stuff in ( ) is the same in both parts!\[(x-4)(x^2-4)\]after "undistributing"
Now we can factor \(x^2-4\) because it is a difference of two squares:\[(x-a)(x+a) = x^2 -ax + ax - a^2 = x^2 -a^2\]
so that factors as \[(x-2)(x+2)\]and the whole thing is \[(x-4)(x+2)(x-2)\]

Answer 20

Can you find the values of \(x\) that make that produce = 0?

Answer 21

sorry, 'product' not 'produce' — I'm thinking about my dinner :-)

Answer 22

Okay, if \(a*b*c = 0\) what does that say about \(a, b, c\)?

Answer 23

one or more of them = 0, right?

Answer 24

So to find all the roots (values of x that make the polynomial = 0), we just find the values of x that make any of the product terms = 0

\[(x-4) = 0\]\[(x+2)=0\]\[(x-2) = 0\]
and the roots are clearly 4, -2, 2, right?

Answer 25

No, I wouldn't accept it as the answer, and I doubt your instructor would give full credit, either.

Answer 26

Those equations are trivially solved for the actual roots, so there's no reason not to do so.

Answer 27

\[x-4 = 0\]\[x = 4\]\[x+2 = 0\]\[x=-2\]\[x-2=0\]\[x=2\]
Roots are \(x = 4, x = \pm 2\)

Answer 28

You need all three roots, you only mentioned two in your earlier answer.

Answer 29

Yes.

Answer 30

okay well thanks for your help and patience. :) I've just turned in my assignment

Answer 31

You're welcome! I hope anything I taught you sticks with you :-)