- #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

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

- 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

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

- #2

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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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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.the original equation is 4x^2-6x^2-12x+9=0

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

- #4

epenguin

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Do you mean 4x^3 ?the original equation is 4x^2-6x^2-12x+9=0

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

- #5

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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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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

..|

.....4

Add that column.

2 | 4, -6, -12, 9

..|

.....4...2

Repeat

2 | 4, -6, -12, 9

..|

.....4....2...-8

2 | 4, -6, -12, 9

..|

.....4....2...-8..-7

The process I described reveals

4x

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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What about just trying to substitute a possible root in the original expression, instead of doing a polynomial division?You mean 4x-6x^{3}^{2}-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)

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