# Solving equations, but what type

• Sep 8th 2011, 11:31 AM
David Green
Solving equations, but what type
To solve an equation I would require first to be able to correctly identify it?

This is a typical equation I am looking at;

x^3 = 7x^2 + 12x = 0

At first I thought it was a quadractic but now I am not sure because the equation ax2 + bx + c = 0 does not contain a place for the x which is multiplied by the 12.

So if this type of equation is not a quadratic, would somebody please advise what type it is then I can have a go at solving it please.

Thank you
• Sep 8th 2011, 11:34 AM
TheChaz
Re: Solving equations, but what type
This is a third degree polynomial.
Initially, you will learn to solve these by
1) factor by grouping
2) rational roots theorem/synthetic division

Of course in this case you can factor:
x(x^2 + 7x + 12) = 0,
then factor the quadratic the usual way.
• Sep 8th 2011, 11:46 AM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by TheChaz
This is a third degree polynomial.
Initially, you will learn to solve these by
1) factor by grouping
2) rational roots theorem/synthetic division

Of course in this case you can factor:
x(x^2 + 7x + 12) = 0,
then factor the quadratic the usual way.

Thanks I'll start learning these types of equations now. Can't understand why they are in the course really as I can't find any information in the book(s) supplied except the quadratic formulas. These formulas give two routes or solutions, but the above I quote has three solutions?

Better get reading then.

Thanks

David
• Sep 8th 2011, 11:51 AM
TheChaz
Re: Solving equations, but what type
The degree of a polynomial is intimately linked with the number of solutions.
For our purposes, let's just say that the highest exponent is the maximum number of real solutions.

The methods I mentioned are more advanced. If you're only learning quadratics, they can still introduce higher-degree polynomials by multiplication with a monomial.
For instance, you can FOIL (2x + 3)(5x - 9) if needed.
Then if you multiply this by 11x^8 and set it to zero, you have a hard-looking problem that will reduce to factoring a quadratic after removing the GCF
• Sep 8th 2011, 12:08 PM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by TheChaz
This is a third degree polynomial.
Initially, you will learn to solve these by
1) factor by grouping
2) rational roots theorem/synthetic division

Of course in this case you can factor:
x(x^2 + 7x + 12) = 0,
then factor the quadratic the usual way.

So I have this equation which I now understand to be a cubic expression. If I am right I now require to find factor pairs for 7 and 12, these can be both positive and negative integers?

- 1, -7, 1,7. -2, 2, -3, 3, -4, 4

How am I doing so far?
• Sep 8th 2011, 12:39 PM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by David Green
So I have this equation which I now understand to be a cubic expression. If I am right I now require to find factor pairs for 7 and 12, these can be both positive and negative integers?

- 1, -7, 1,7. -2, 2, -3, 3, -4, 4

How am I doing so far?

The closest I have got so far is;

x^3 + 7x^2 = 12x = 0

x(x^2 - 4)(7x + 12x)

x^3 + 7x^2 + 12x - 28x - 48x

x^3 + 7x^2 + 8

I can't be far out now, I just need to find the correct factor so that my + 8 becomes + 12
• Sep 8th 2011, 12:48 PM
TheChaz
Re: Solving equations, but what type
Start with x^3 + 7x^2 + 12x = x(x^2 + 7x + 12) = 0

x^2 + 7x + 12 factors into (x + 3)(x + 4), separately from the x.

So you have x(x + 3)(x + 4) = 0
and set each factor to zero.
• Sep 8th 2011, 01:01 PM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by TheChaz
Start with x^3 + 7x^2 + 12x = x(x^2 + 7x + 12) = 0

x^2 + 7x + 12 factors into (x + 3)(x + 4), separately from the x.

So you have x(x + 3)(x + 4) = 0
and set each factor to zero.

In the above I worked that out to show; x^2 + 7x + 12. Is there something missing because I thought that when the brackets were multiplied out that the solution should be the same as the original equation?

Also when I am looking for the factors, what do you mean by saying, set each factor to zero?

• Sep 8th 2011, 01:50 PM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by David Green
In the above I worked that out to show; x^2 + 7x + 12. Is there something missing because I thought that when the brackets were multiplied out that the solution should be the same as the original equation?

Also when I am looking for the factors, what do you mean by saying, set each factor to zero?

Just had a senior moment, I was working out the brackets incorrectly?

x(x+3)(x+4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12

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

It seems that I have only found two solutions though?

(x+3) and (x+4)

How do I go about finding the third solution?

Sorry Chaz you did say previously that the highest power is x^3, which I read afterwards in a maths book, but I am still unclear how to present it as three solutions in this example?
• Sep 9th 2011, 03:53 AM
HallsofIvy
Re: Solving equations, but what type
You have now written "x^3 = 7x^2 + 12x = 0", in your original post, "x^3 + 7x^2 = 12x = 0", in post number 6, and The Chaz wrote "x^3 + 7x^2 + 12x = 0" in post number 7. which is it?

If it is, indeed, x^3 + 7x^2 + 12x = x(x^2+ 7x+ 12)= x(x+ 3)(x+ 4)= 0, then the "zero property" gives you all three roots: the product of x(x+ 3)(x+ 4) is 0 only if at least one of the three factors is 0: x= 0, or x+ 3= 0, or x+ 4= 0.

You also seem to be very confused about exactly what you are trying to do. You say "It seems that I have only found two solutions though? (x+3) and (x+4)". Those are NOT "solutions". The three solutions you want are the three numbers that satisfy the equation. If x+ 3= 0 what is x? If x+ 4= 0, what is x? Now what was the third factor, above?
• Sep 9th 2011, 11:23 AM
David Green
Re: Solving equations, but what type
Quote:

Originally Posted by HallsofIvy
You have now written "x^3 = 7x^2 + 12x = 0", in your original post, "x^3 + 7x^2 = 12x = 0", in post number 6, and The Chaz wrote "x^3 + 7x^2 + 12x = 0" in post number 7. which is it?

If it is, indeed, x^3 + 7x^2 + 12x = x(x^2+ 7x+ 12)= x(x+ 3)(x+ 4)= 0, then the "zero property" gives you all three roots: the product of x(x+ 3)(x+ 4) is 0 only if at least one of the three factors is 0: x= 0, or x+ 3= 0, or x+ 4= 0.

You also seem to be very confused about exactly what you are trying to do. You say "It seems that I have only found two solutions though? (x+3) and (x+4)". Those are NOT "solutions". The three solutions you want are the three numbers that satisfy the equation. If x+ 3= 0 what is x? If x+ 4= 0, what is x? Now what was the third factor, above?

Please accept my applogies, the original equation should be;

x^3 + 7x^2 + 12x = 0

x(x + 3)(x + 4) = 0

x^2 + 4x + 3x + 12 = 0

x(x^2 + 7x + 12) = 0

x^3 + 7x^2 + 12x = 0

x = 0
x + 3 = 0
x + 4 = 0

to get that;

- 3 + 3 = 0
- 4 + 4 = 0

the three roots therefore must be;

x = 0
(-3 + 3) = 0
(-4 + 4) = 0

Please advise if I am now understanding this correctly, also remember I am learning from first principles.

Thanks

David
• Sep 10th 2011, 01:03 AM
Siron
Re: Solving equations, but what type
You seems confused about the solutions. You solve it correctly because the third degree equation can be factored as:
$x(x+3)(x+4)=0$

So you have the solutions:
$x=0$
$x+3=0 \Leftrightarrow x=-3$
$x+4=0 \Leftrightarrow x=-4$

Take a look at the post of HallsofIvy. A solution always looks like: $x=...$
But (-3+3)=0 doesn't make sense as a solutions of an equation.