Question

Hiya, hlp plz?

-12x^3 - 65x^2 - 117x - 70 = 0
whats the easiest way to solve the cubic(by hand, not calc)?

Answer 1

you can factor out the negative 1 as a start

Answer 2

ok, so
\[-1 (12x^3 + 65x^2 + 117x + 70) = 0\] ?

Answer 3

Could use synthetic division to check for a factorization.

Answer 4

synthetic... how do i do that plz?

Answer 5

After two tries, I found a factor and was able to factor the remaining trinomial by inspection. There are three rational roots.

Answer 6

yep, but is there an easy way to do it rather than guess an check?

Answer 7

It isn't really guess and check. If there is an integer root, it has to be a factor of 70. So test to see if (x-a) divides the polynomial evenly, where a is a factor of 70. I recommend starting with the small ones first.

Answer 8

ok, so -2 is one of the roots
so to test: (x-a) = (-12 - 2) = -14
so divide polynomial by -14 to test...?

Answer 9

\[-12x^3 - 65x^2 - 117x - 70 = 0 \implies \]\[12x^3+65x^2 + 117x+ 70 = 0=(x+2)(4x+7)(3x+5) \implies\]\[x \in \left\{ -2,\frac{-7}{4},\frac{-5}{3} \right\}\]

Answer 10

12x3 + 65x2 + 117x + 70 = 0 =>> (x+2)(4x+7)(3x+5)

its this step here that gets me, how do u do this, pen and paper, in a simple way?

Answer 11

@AnimalAin ...? i realise it's pretty easy to do with a calculator, but how do u solve without...?

Answer 12

"rational root thm/decartes/synthetic" combo gives x = -2 as one root

Answer 13

fine wid that @Jack1 .. or u want to solve it in general ?

Answer 14

just googling rational root theorum,

Answer 15

ok, yep i remember that

Answer 16

oh no, its too easy for u to suck it all in one line :

rational root thm says this :

```
rational roots of a polynomial are of form : p/q
p = factors of constant term
q = factors of leading coefficient
```

Answer 17

so our guess and check limits to checking \(\pm 1, \pm 2, \pm 5, \pm 7 .... \)

Answer 18

descartes u lost me on
and synthetic ive nerver heard of, sorry

Answer 19

ok, yep, i agree with 1,2,5,and 7

Answer 20

decartes says this :

number of positive roots = number of sign changes in P(x)
number of negative roots = number of sign changes in P(-x)

Answer 21

^^above are upper limits, complex roots can kick in too

Answer 22

ur polynomial :

-12x^3 - 65x^2 - 117x - 70 = 0

Answer 23

ugh, h8 complex numbers >:(

Answer 24

there are no sign changes in P(x),
so there will not be any positive roots. so dont bother about them.

Answer 25

Having the knowledge that there will not be any positive roots,
out guess and check work reduces by half : \(-1, -2, -5, -7,....\)

Answer 26

ok, think i get descartes now too, cheers @ganeshie8 ;)

Answer 27

If you get lucky with any one root, then you're done.
u can divide it and solve the remaining depressed quadratic using quadratic formula

Answer 28

so always hit the easy possible numbers first then and hope to get lucky? (1,2,3 ? )

Answer 29

yes, thats the easy way if it works

Answer 30

if u dont get lucky we can solve it analytically

Answer 31

sweet... cheers man
while i've got u tho, could u give me a run down on synthetic division if u have a sec?

Answer 32

we can always solve cubic and quartic equations
but we cannot solve 5th degree and above always

Answer 33

yeah sure :)

Answer 34

synthetic is just a shortcut for long division

Answer 35

easy example first :

divide x^2 + 2x + 1 by x+1

Answer 36

\(
\begin{array}{}

-1~|&1&2&1&\\
&&&&\\
\hline
&&&&\\
&1&&&

\end{array}
\)

Answer 37

1, 2, 1 are the coefficients of polynomial

Answer 38

since you want to divide by x+1, put x = -1 on left side

Answer 39

and to start with, drop that "1" vertically down and put it in the last row

Answer 40

thats the start...

Answer 41

ok, follow what you did there, will have to try a few before it becomes muscle memory tho

cheers again for all this @ganeshie8 !! ur an awesome teacher / tutor hey

Answer 42

step 2 : 1(-1) = -1 :

\(
\begin{array}{}

-1~|&1&~~2&1&\\
&&-1&&\\
\hline
&&&&\\
&1&&&

\end{array}
\)

Answer 43

step3 : add them vertically and put the result down in the last row :

\(
\begin{array}{}

-1~|&1&~~2&1&\\
&&-1&&\\
\hline
&&&&\\
&1&~~1&&

\end{array}
\)

Answer 44

repeat step2 and step3 till u hit the end

Answer 45

\(
\begin{array}{}

-1~|&1&~~2&~~1&\\
&&-1&-1&\\
\hline
&&&&\\
&1&~~1&~~|0&

\end{array}
\)

Answer 46

look at last row :

1x + 1 is ur quotient
0 is ur remainder

Answer 47

ahhhhh!!!! gotcha now man

Answer 48

so,

(x^2 + 2x + 1 )/(x+1) = 1x + 1 with a remainder 0

Answer 49

:) you can try dividing ur polyomial by -2 now maybe

Answer 50

Since you know that "-2" is one root of cubic : -12x^3 - 65x^2 - 117x - 70 = 0


divide -2 and get the depressed quadratic equation

Answer 51

\(
\begin{array}{}

-2~|&-12&-65&-117&-70\\
&&&&\\
\hline
&&&&\\
&-12&&&

\end{array}
\)

Answer 52

multiply (-12)(-2) and put it in second row :

\(
\begin{array}{}

-2~|&-12&-65&-117&-70\\
&&~~~24&&\\
\hline
&&&&\\
&-12&&&

\end{array}
\)

Answer 53

-12x^3 - 65x^2 - 117x - 70 = 0
so:
-2 | -12 -65 -117 -70
24
-----------------------
umm..-12 -89 ???

Answer 54

add them vertically
repeat...

Answer 55

u need to add, not subtract

Answer 56

multiply (-12)(-2) and put it in second row :

\(
\begin{array}{}

-2~|&-12&-65&-117&-70\\
&&~~~24&&\\
\hline
&&&&\\
&-12&-41&&

\end{array}
\)

Answer 57

^^

Answer 58

-12x^3 - 65x^2 - 117x - 70 = 0
so:
-2 | -12 -65 -117 -70
24 82 -70
-----------------------
umm..-12 -41? 35 |0

is this right?

Answer 59

no wait

Answer 60

-12x^3 - 65x^2 - 117x - 70 = 0
so:
-2 | -12 -65 -117 -70
24 82 70
-----------------------
umm..-12 -41? -35 |0

is this right?

Answer 61

yep !

so 0 is the denominator, and the depressed quadratic is :

-12x^2 - 41x - 35

Answer 62

last row gives u `coefficients for the depressed polynomial` and the `remainder`

Answer 63

gotcha, so then we jam that into the quadratic equation an BAM, the other 2 roots?

Answer 64

that sit ! :) k gtg cya :)

Answer 65

cheers man, bye!