Question

How do I turn x^3+6x^2+13x+10=0 into a Quadratic formula (ax^2+bx+c=0) so I can find the solutions. The question says to Solve for the shown equation and I'm TOTALLY stuck...

Answer 1

you familiar wid below goodies ?
1) Descartes rule of signs
2) rational root test

Answer 2

and,
3) synthetic division

Answer 3

if so, its a piece of cake for u. simply find one zero using above.
then apply quadratic formula for the depressed polynomial

Answer 4

I don't know what Descartes rule of signs is OR the rational root test :'O I kind of get synthetic division. Care to explain <3?

Answer 5

Sure :)
they're easy to *get*,
actually synthetic division is hard to get, compared to descartes/rational roots

Answer 6

Descartes rule of signs :-
1) Count number of sign changes in f(x)

Answer 7

x^3+6x^2+13x+10=0

how many sign changes do u see ?

all are positive, so 0 sign changes. that means there are NO POSIATIVE zeroes for this polynomial

Answer 8

So, there has to be atleast ONE negative zero for this cubic

Answer 9

we're done wid Descartes.
now we know that, there wont be any positive roots,
but, there has to be atleast ONE negative root.

Answer 10

you wid me still ?

Answer 11

Yeah, so far :3

Answer 12

Just taking notes as we go.

Answer 13

good, rational root theorem tells this :-
```
rational zeros for any polynomial will be of form : p/q
p are factors of constant term
q are factors of leading coefficient
```

Answer 14

x^3+6x^2+13x+10=0
^ ^

Answer 15

for this,
constant term = 10
leading coefficient = 1

Answer 16

Now tell me, wat are the factors of 10 ?
wat are the factors of 1 ?

Answer 17

For 10 its 1, 2, 5, and 10.

1 is just one right?

Answer 18

Awesome !
but u forgot negative factors

Answer 19

For 10 its 1, 2, 5, and 10
AND ALSO, -1, -2, -5 and -10

Answer 20

since Descartes already told us, there wont be any POSITIVE zeroes, dont wry about POSITIVE factors

Answer 21

lets test and see if -2 is a zero

Answer 22

use synthetic division

Answer 23

x^3+6x^2+13x+10=0


-2 | 1 6 13 10

Answer 24

can u divide, and see if it leaves 0 remainder ?

Answer 25

I did all the math and got 0 as the answer to the synthetic division. That means 0 remainder right?

Answer 26

yup! and wats the depressed quadratic ?

Answer 27

also, that means -2 is a zero/solution of given cubic

Answer 28

-2 | 1 6 13 10
| -2 -8 -10
------------------------
1 4 5 0

Answer 29

so the depressed quadratic is x^2 + 4x + 5

Answer 30

you can use quadratic formula on this, and find other two zeroes/solutions

Answer 31

Was doing the quadratic equation and got this on the top part of the division. How do I handle it?\[-4\pm \sqrt{-4}\]

Answer 32

pull out 4 from radical first

Answer 33

\(-4\pm \sqrt{-4}\)


\(-4\pm 2\sqrt{-1}\)

Answer 34

\(\sqrt{-1} = i\)

Answer 35

\(-4\pm \sqrt{-4}\)


\(-4\pm 2\sqrt{-1}\)

\(-4\pm 2i\)

Answer 36

leave it like that.
Done forget the denominator !

Answer 37

So now I'm at \[\frac{4\pm2i }{ 8 }\]

Would I be accurate in saying the answer/solution is

\[\frac{ 1 }{ 2 }\pm \frac{ 1 }{ 4 }i\]

Answer 38

Nope.

u should get :

\(\large \frac{-4\pm 2i}{2} \)

\(\large -2 \pm i\)