Problem with Quadratic Factoring

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Jason76

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How would you factor this problem?

\(\displaystyle 3x^{3} + x^{2} + -12x - 4 = 0\)

If it was in the form

\(\displaystyle x^{2} + lx + mx + n\) (where l, m, and n represent different numbers)

then you would factor by grouping. But the problem equation is different. Well, actually you do use grouping on it, but not in the same way.
 
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Jason76 said:
How would you factor this problem?

\(\displaystyle 3x^{3} + x^{2} + -12x - 4 = 0\)

If it was in the form

\(\displaystyle x^{2} + lx + mx + n\)

then you would factor by grouping. But the problem equation is different.
Click to expand...

Factoring cubic equations is, at best, luck. Just try something. I'm suspicious that the first and third terms are 3x the second and fourth terms. Let's factor partially 1&2 and 3&4

\(\displaystyle x^{2}(3x + 1) -4(3x+1) = 0\)

That's it. Now what?
 
(EDITED)

Here is the full problem given in the book:

\(\displaystyle 3x^{3} + x^{2} - 12x - 4 = 0\)

Here is the method the book uses. But I don't understand it.

Grouping:

\(\displaystyle (3x^{3} + x^{2}) + (-12x - 4) = 0\)

\(\displaystyle x^{2}(x + 3) - 4(x + 3) = 0\)

\(\displaystyle (x + 3)(x^{2} - 4) = 0\)

\(\displaystyle x + 3 = 0\) or \(\displaystyle x^{2} - 4 = 0\)

Creating Linear Factors:

\(\displaystyle (x + 3) = 0, (x^{2} - 4) = 0\)

\(\displaystyle x = -3, x = 2, x = -2\) (the positive and negative 2 are derived from the linear factor with the exponent) (The \(\displaystyle x^{2}\) is removed by squaring 4.)

But I don't understand the factoring, though I do understand how the linear factors and roots came about.
 
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Jason76 said:
Here is the full problem given in the book

\(\displaystyle 3x^{3} + x^{2} - 12x - 4 = 0\)

Here is the [answer in] the book

\(\displaystyle x = -3\)\(\displaystyle , x = 2, x = -2\)
Click to expand...

The solution in red is not correct. (Some of the factoring shown is also not correct.)

Please proofread your post, and verify that everything matches your book. If you find no errors in your typing, and it matches the book, then the book is wrong.
 
mmm4444bot said:
The solution in red is not correct. (Some of the factoring shown is also not correct.)

Please proofread your post, and verify that everything matches your book. If you find no errors in your typing, and it matches the book, then the book is wrong.
Click to expand...

I went back and edited, because I left out one line. But it's now what the book said:

With Line Added:

Here is the full problem given in the book:

\(\displaystyle 3x^{3} + x^{2} - 12x - 4 = 0\)

Here is the method the book uses. But I don't understand it.

Grouping:

\(\displaystyle (3x^{3} + x^{2}) + (-12x - 4) = 0\)

\(\displaystyle x^{2}(x + 3) - 4(x + 3) = 0\)

\(\displaystyle (x + 3)(x^{2} - 4) = 0\)

\(\displaystyle x + 3 = 0\) or \(\displaystyle x^{2} - 4 = 0\)

Creating Linear Factors:

\(\displaystyle (x + 3) = 0, (x^{2} - 4) = 0\)

\(\displaystyle x = -3, x = 2, x = -2\) (the positive and negative 2 are derived from the linear factor with the exponent) (The \(\displaystyle x^{2}\) is removed by squaring 4.)

But I don't understand the factoring, though I do understand how the linear factors and roots came about.
Click to expand...
 
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If you multiply out the second line in the factorization, you will see the author's solution is incorrect. The author must have been factoring THIS polynomial: \(\displaystyle x^3+3x^2 -4x-12\)
 
daon2 said:
If you multiply out the second line in the factorization, you will see the author's solution is incorrect. The author must have been factoring THIS polynomial: \(\displaystyle x^3+3x^2 -4x-12\)
Click to expand...

Right, I agree. I wonder if I can point this out, then I can get some money :)

Especially looking at this line:

\(\displaystyle x^{2}(x + 3) - 4(x + 3) = 0\)

That actually de-factors to

\(\displaystyle x^{3} + 3x^{2} - 4x - 12\)

as you said.

Ok, let's finish this equation (the correct equation).

\(\displaystyle x^{3} + 3x^{2} - 4x - 12\)

Grouping since 4 terms:

\(\displaystyle (x^{3} + 3x^{2}) + (- 4x - 12)\)

\(\displaystyle x^{2}(x + 3) - 4(x + 3) = 0\)

Breaking it down to 2 linear factors:

\(\displaystyle (x^{2} - 4) (x + 3) = 0\)

Left Linear Factor:

\(\displaystyle (x^{2} - 4) = 0\) Yields: \(\displaystyle x^{2} = 4\)

Yields \(\displaystyle x = \sqrt{4}\) Yields \(\displaystyle x = 2, x = -2\)

Right Linear Factor:

\(\displaystyle x + 3 = 0\) Yields \(\displaystyle x = -3\)

Roots:

\(\displaystyle 2, -2, -3\) are the roots.
 
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Alternatively, using the book's given:

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

Group terms:

(3x^3 + x^2) + (-12x - 4) = 0

Factor each group:

x^2(3x + 1) - 4(3x + 1) = 0

Now factor out the common factor (3x + 1)

(3x + 1)(x^2 - 4) = 0

Apply the Zero Product Property

3x + 1 = 0

or

x^2 - 4 = 0

x = -1/3 or x = -2 or x = 2
 
mmm4444bot said:
Alternatively, using the book's given:

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

Group terms:

(3x^3 + x^2) + (-12x - 4) = 0

Factor each group:

x^2(3x + 1) - 4(3x + 1) = 0

Now factor out the common factor (3x + 1)

(3x + 1)(x^2 - 4) = 0

Apply the Zero Product Property

3x + 1 = 0

or

x^2 - 4 = 0

x = -1/3 or x = -2 or x = 2
Click to expand...

The book's given doesn't lead to the book's answer, so the book is wrong.
 
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