I forgot how to factor cubic equations?

  • Thread starter yurkler
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  • #1
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the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
 

Answers and Replies

  • #2
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Apply the rational root test to find a possible root.

The possible rational roots will have the form p/q where p divides 9 and where q divides 4. So what are the possible rational roots?? Is one of these possibilities an actual root?
 
  • #3
eumyang
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the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
Ah, this looks like the "factoring by grouping" method, which, unfortunately, doesn't work for ALL cubic polynomials. The rational root test, as micromass mentioned, is your best bet.
 
  • #4
epenguin
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the original equation is 4x^2-6x^2-12x+9=0

I got to this point: (2x^2)(2x-3)+(-3)(4x-3)=0
Do you mean 4x^3 ?
 
  • #5
98
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The equation you posted is not a cubic; it is a quadratic.

4 x^2 - 6 x^2 . . .

equals -2 x^2 . . .

Why didn't you combine the x^2 terms?

In any case, equations exist for solving a quadratic or cubic equation. Just do a search; I am sure you will find the appropriate formula.
 
  • #6
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You mean 4x3-6x2-12x+9=0?

The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really. Let's try x=4 for example. Then we set up the polynomial division like this:

2 | 4, -6, -12, 9
..|..................

First step. Drop the first term (from the left) to the right of the bar.

2 | 4, -6, -12, 9
..|..................
.....4

Next, multiply that number by 2, and put it below the next entry.

2 | 4, -6, -12, 9
..|......8...........
.....4

Add that column.

2 | 4, -6, -12, 9
..|......8...........
.....4...2

Repeat

2 | 4, -6, -12, 9
..|.......8....4.....
.....4....2...-8

2 | 4, -6, -12, 9
..|.......8....4..-16
.....4....2...-8..-7

The process I described reveals

4x3-6x2-12x+9 = 4x2-2x-8-7/(x-2)

The last term there, -7/(x-2) is called the remainder term. Since there is a remainder term, 2 is not a root. You just have to keep testing roots until you get one without a remainder term.
 
  • #7
I like Serena
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You mean 4x3-6x2-12x+9=0?

The roots of a polynomial must be a factor of the product of the first term and the last term. Therefore, we have to try all the factors for 4*9=36. There's this really convenient method for factoring polynomials like this. It's kind of like long division, but not really.

(Snip)
What about just trying to substitute a possible root in the original expression, instead of doing a polynomial division?
 
  • #8
209
0
You could do that and then use the long division to find the coefficients of the remaining quadratic.
 

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