Question

2​x^​3+​7​x^​2-​8*x-​21 = 0 .. the answer is x=-3/2, please show the working.

Answer 1

\[2x^3+7x^2-8x-21=0\]The show the working part is for you, we are here to help you show that for yourself. What you need to do to see whats going on in here is to change the equation to a form where you can interpret it. I'll start as an example for you:\[2x^3+7x^2-8x=21\]\[x(2x^2+7x-8)=21\]Try to make whats inside the parenthesis simpler.

Answer 2

I made it this far already. I don't really know what to do next

Answer 3

i thought it was 8 times x

Answer 4

do I use the quadratic equation?

Answer 5

\[(2​x^​3+​7​x^​2-​8)(x-​21) = 0\] ?

Answer 6

I wrote the way he said it, its \[2x^3+7x^2-8x-21=0\]
Yes, now you need to work on the quadratic equation, do you have any idea on how to do that?

Answer 7

yes i do.. \[ \frac{ -b +or- \sqrt{b^2-4ac} }{ 2a }\]

Answer 8

I got x = -4.41 and x= 0.91

Answer 9

is that correct? what's next?

Answer 10

Probably, the formula is correct.

Answer 11

ok, how do I put that in the parentheses to continue working?

Answer 12

since those are the roots of the quadratic equation, it can be written as follows:\[2x^2+7x-8=2(x+4.41)(x-0.91)\]

Answer 13

Oh, wait, now I cant see how we continue, just a sec

Answer 14

thanks ok.

Answer 15

where do I start? I don't know

Answer 16

We have an equation with an exponent 3, that means that the two functions that we need to rey to find that when multiplied will give this one must be exponent 1 and the other exponent 2. The general form is the following:\[(ax+b)(cx^2+dx+c)=acx^3+x^2(bc+ad)+x(bd+ac)+bc=0\]Now we have: ac=2, bc+ad=7, bd+ac=-8, and bc=-21

Answer 17

I have never studied that before..

Answer 18

I guess I'll need to do some more work before trying to solve it

Answer 19

Did you understand the part of the exponents?

Answer 20

we need to equations, one with exponent 1, and another with exponent 2, yeh?

Answer 21

the general form of the one with the exponent 1 ia ax+b right?
and the general form of the one with exponent 2 ia cx^2+dx+e
now its correct, I messed up with the letters before
Now that you have the general form of both you just need to multiply them to get something similar to what you have (exponent 3)\[(ax+b)(cx^2+dx+e)=(ac)x^3+(bc+ad)x^2+(ae+bd)x+(be)=0\]The only way that your equation with exponent 3 is the same as the general one, is when the coeficients are the same.

Answer 22

\[ac=2\]\[bc+ad=7\]\[ae+bd=-8\]\[be=-21\]

Answer 23

ok .. so far so good. I'm following

Answer 24

Now you need to solve this so that you can write the two equations that represent your problem and not a general situation.

Answer 25

so I put this: 2​x^​3+​7​x^​2-​8*x-​21 in the form of (ax+b)(cx2+dx+e)

Answer 26

yes, and now you have two equations you know how to solve, because at least one of them must be zero

Answer 27

I get what we're trying to do

Answer 28

since we've already got 2x+3, we can simply divide the cubic function to get the equation with exponent 2

Answer 29

?

Answer 30

Well, we can do that, but you need to get to the 2x+3 first, from where did you get it?

Answer 31

exactly.. i didn't get it. It was in the question. how do I get it anyways?

Answer 32

Well, those equations we got with a, b, c, d, and e should give it to us, but there is a little bit of trial and error since we have 4 equations and 5 variables we need to assume a value to one of them, in this case c so that we can solve the system

Answer 33

But you are correct, since the problem gives that to you

Answer 34

thank you. I appreciate your time and help :)

Answer 35

youre welcome