Question

FACTORING:
Can x^3+4x^2+24 be factored to find zeros? (fundamental theorem of algebra)

Answer 1

first find the factors of prime factors of 24

Answer 2

the factors of 24 are 1,2,3,4,6,8,12
put these values with +ve and -ve sign in the given expression and check for which particular values the expression vanishes

Answer 3

what's ve

Answer 4

positive (+ve) and negative (-ve)

Answer 5

by +ve i mean positive sign

Answer 6

Ohh!

Answer 7

so do I plug it in the x? and ?

Answer 8

yup

Answer 9

and check if the value of the expression vanishes

Answer 10

so if it gives me zero? would that number then be one of my "solutions"?

Answer 11

Yes, any value of x such that p(x) = 0 is a solution/root/zero.

Answer 12

then the value of x for which the above expession vanishes will be one of the roots of the equation

Answer 13

I understand now, both of you, thanks for your kind time (:

Answer 14

thanks
but if you ar not able to find any such specific x by hit and trial method
than opt for Cardans method for solving cubic equation

Answer 15

well, Cardan's method isn't what's being requested here — problem asks if it can be factored.

Answer 16

I found no zeros so I guess that means no real solution?

Answer 17

since its cubic it must have at least one real solution
as imaginary roots occur in conjugate pairs

Answer 18

can you plz check the sign of 24 its +ve or -ve

Answer 19

negative I suppose

Answer 20

but if its negative u have mentioned +ve

Answer 21

\(x^3+4x^2\color{red}{-} 24\)

Answer 22

like that ?

Answer 23

none of them work, negative or positive factors of 24

Answer 24

if its x^3+4x^2 - 24 then the expression vanishes for 2
if it vanishes for 2 then x-2 is afactor

Answer 25

the only thing I close to a 0 is -4 input but only with x^3+4x^2

Answer 26

@icetea
plz check if ur question is x^3+4x^2 + 24 or x^3+4x^2 - 24

Answer 27

I wrote down -24 but I typed +24 woops, yes it's +24

Answer 28

then the answer would be : It cannot be factored.

Answer 29

@matricked is right there are roots but they are all decimals so it can't be factored

Answer 30

I hate 314x^3+1256x^2-7536 then I factored it to 314(x^3+4x^2+24)

Answer 31

*had

Answer 32

hey u flipped the - sign in the end !

Answer 33

you can't do that

Answer 34

so 314x^3+1256x^2-7536
then 314(x^3+4x^2-24)

Answer 35

there are real roots to this equation and it can be factored keep trying!

Answer 36

=314(x^3+4x^2-24)
=314( x^3 -2x^2 +6x^2 -12x +12x -24 )
=314( x^2(x-2) + 6x(x-2) +12(x-2))
=314((x-2)

Answer 37

OH I see what you did hmm so then +2 would be my one real root?!

Answer 38

=314(x^3+4x^2-24)
=314( x^3 -2x^2 +6x^2 -12x +12x -24 )
=314( x^2(x-2) + 6x(x-2) +12(x-2))
=314(x-2)(x^2+6x+12)

Answer 39

i hope the rest u will be able to do...

Answer 40

yes, x= 2 is the real root, and you've got a pair of complex roots

Answer 41

-12 and +2 yess?

Answer 42

there are three roots so write the other one.

Answer 43

no.
\[x^2+6x+12 =0\] \(a=1,b=6,c=12\)
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-6\pm\sqrt{6^2-4(1)(12)}}{2} = \frac{-6\pm\sqrt{36-48}}{2}=\]

Answer 44

Your -12 is incorrect

Answer 45

I thought you only used the discriminant?

Answer 46

the discriminant only allows you to analyze the kinds of roots you'll encounter, not find them in the general case...

Answer 47

No... you see the \[\pm \] that whpalmer4 placed there.. You suppose to evaluate it at + and at -......

Answer 48

If 2 is a zero, why do you get x-2 as a ...factor? LCM = x or something? Or do we just apply this rule for long division standards?

Answer 49

factor theorem

Answer 50

ooh then the -6+/-sqrt -6 is my complex root yes?

Answer 51

if 2 is factor then the expresion is divisible by x-2

Answer 52

you solve for x so x = 2 which is a root according to what matricked stated.

Answer 53

I see... and so if a problem has multiple roots, they'd be written such as x - #?

Answer 54

if it can be factored yes

Answer 55

Then using long division with these factors, can see if it equals zero at the end?... trying to remember that root theorem.

Answer 56

once you have roots, you can factor it

Answer 57

@icetea

\[\frac{-6\pm\sqrt{-12}}{2} = \frac{-6\pm i2\sqrt{3}}{2} = -3\pm i\sqrt{3}\]

Answer 58

I understand that.

Answer 59

P(x)= (x-a)Q(x) +R(x)

Answer 60

that was an interesting question why (x-p) factors out when p is a root

Answer 61

@whpalmer that slipped my mind! thanks !!!!!

Answer 62

Thank you.

Answer 63

(x-p) factors out when p is a root because P(p) = 0 and P(x) can be written as (x-p)(x-q)... by the factor theorem

Answer 64

welcome

Answer 65

I'll have to look up the factor theorem.

Answer 66

have to lay back and think how factor theorem works

Answer 67

lol yea lets do that

Answer 68

well, the roots are the spots where f(x) = 0, right? if we write the polynomial as \[(x-r_1)(x-r_2)...(x-r_n) = 0 \text{ (at a zero)}\]then we see our remainder is identically 0 at a root and we can divide off a root as a factor

Answer 69

like when polynomial cuts the x-axis at p, then p is a zero.
when u factor out the polynomial :-
\(ax^n + bx^{n-1} + cx^{n-2} ... = a(x-p)(x-q).... \)

Answer 70

so we need to think reverse, when p is a zero, it factors into (x-p) i get it !!!