Quadratic (Most efficient)

3x^3+21x^2+36x=0

What's the quickest most efficient way to tackle this problem?

Is it possible without the use of forulas?

EDIT: wrote question incorrectly

(edited 9 years ago)

Reply 1

ghostwalker

Original post by Abdul-Karim

3x^2+21x^2+36x=0

You have two terms in x squared - did you mean that?

Reply 2

brianeverit

Original post by Abdul-Karim

3x^2+21x^2+36x=0

What's the quickest most efficient way to tackle this problem?

Is it possible without the use of forulas?

combine the two

x^2

terms and take out a common factor of x.
That is assuming you have written the question correctly.

Reply 3

gr8wizard10

Original post by brianeverit

combine the two

x^2

terms and take out a common factor of x.
That is assuming you have written the question correctly.

My bad, I re-edited OP.

Original post by ghostwalker

You have two terms in x squared - did you mean that?

As above.

Reply 4

Manitude

3x^{3} + 21x^{2} + 36x=0

Take a factor of 3x and then be inspection you can see one root is x=0.
You should be able to do the rest by factorising, I shouldn't really give the full solution. Think about what numbers add to give 7 and multiply to give 12.

Obviously this method doesn't always work as you don't always get a combination of number that facilitates a nice factorisation and there is no coefficient of x^{0} in this example. The quadratic formula always works and is quick to use once you've got a lot of practice with it. There is a cubic formula too, but I wouldn't both learning it as it is very long from what I remember. I don't remember too much of what I was taught about solving cubics, but often finding a factor of the form (x+a) where a is any real number (usually an integer) is a great start as it splits the cubic into a linear equation multiplied by a quadratic equation which is much more manageable. Another approach is to test small numbers such as x=+/-1, x=+/-2 and see if they work. If they "x=a" is a solution then you can pull a factor of (x-a) out of the cubic. You'll have to do polynomial division for this, but like most computations in maths the more you do them the quicker they get.

(edited 9 years ago)

Reply 5

TenOfThem

Original post by Abdul-Karim

3x^3+21x^2+36x=0

What's the quickest most efficient way to tackle this problem?

Is it possible without the use of forulas?

EDIT: wrote question incorrectly

3x^3 + 21x^2 + 36x = 3x(x^2 + 7x + 12)

The quadratic factorises into 2 brackets quite easily

Reply 6

Biryani007

Take 3x out and then you have a simple equation.

Reply 7

gr8wizard10

Original post by TenOfThem

3x^3 + 21x^2 + 36x = 3x(x^2 + 7x + 12)

The quadratic factorises into 2 brackets quite easily

From this:

3x(x^2+3x+4x+12)

Within the brackets:

x(x+3)+4(x+3)
= (x+4)(x+3)

So altogher according to your initial input: 3x((x+4)(x+3)) = 0

Am I heading in the right direction

Reply 8

TenOfThem

Original post by Abdul-Karim

From this:

3x(x^2+3x+4x+12)

Within the brackets:

x(x+3)+4(x+3)
= (x+4)(x+3)

So altogher according to your initial input: 3x((x+4)(x+3)) = 0

Am I heading in the right direction

yes that is fine

Reply 9

gr8wizard10

Original post by TenOfThem

yes that is fine

What can I do from that to get my values?

I get that without the 3x, the values would've be x=-4 and x=-3

Reply 10

TenOfThem

Original post by Abdul-Karim

What can I do from that to get my values?

I get that without the 3x, the values would've be x=-4 and x=-3

Do you know why -4 and -3 are possible answers?

The purpose of factorising is because ab=0 means that a or b = 0

Similarly abc = 0 means that a, b, or c must = 0

Reply 11

TenOfThem

Original post by Manitude

3x^{3} + 21x^{2} + 36x=0

The forum guidelines ask us not to give full solutions

Reply 12

gr8wizard10

Original post by TenOfThem

Do you know why -4 and -3 are possible answers?

The purpose of factorising is because ab=0 means that a or b = 0

Similarly abc = 0 means that a, b, or c must = 0