Factoring Polynomials: 2x^5 - 7x^3 - 4x, x²(x + 2) - x(x + 2

[DEdesigns57]

Here are two problems that I cant seemt o figure out no matter what I do.
2x^5 - 7x^3 - 4x

and

x²(x + 2) - x(x + 2) - 12(x + 2)

[stapel_eliz]

Here are two problems that I cant seemt o figure out no matter what I do.

What have you done? (It's a lot easier to find errors in work that we can see....)

x²(x + 2) - x(x + 2) - 12(x + 2)

Take the common factor of x + 2 out front, and then factor the remaining quadratic. Hint: What factors of -12 add up to -1?

2x^5 - 7x^3 - 4x

Take the common factor out front. This will leave you with a polynomial of degree 4 but in the form of a quadratic:

. . . . .

Hint: What factors of (2)(-4) = -8 add up to -7?

One of the resulting factors (both of which will be genuine quadratics) will be a difference of squares. Apply the formula for that, and you're done!

[QM deFuturo]

You haven't shown what you have tried yet, so it's hard to tell exactly what you are "stuck" on. One suggestion, I see in both cases there are common factors that you can factor out of each equation and re-write them, so that the various factors will be lower powers, and easier to work with. Then you should be able to factor those simpler forms. I'm assuming you know how to work with radical roots and powers? I think you will need to know that for the first problem when you try to further factor it after first factoring out a common term.

QM

[DEdesigns57]

Thank you for you reply and sorry that I didnt show you what I tried. As for the problem you showed me, I got it and understand it. But you mentioned if I knew about Radicals and exponents. Yes I do but i dont understand how I would need that to factor out the first problem I posted. This is as far as I got for that prolem,

2x^5 - 7x^3 - 4x

x(2x^4 - 7x^2 - 4) By Factoring out X

But from here im stuck because im not sure if im allowed to rewrite the trinomial inside the () as a quadrtic by makeing the first term (2x^2)^2, but if I did what about the other 2 terms, what about desedning order of powers ect...

[stapel_eliz]

Yes, you are allowed to re-write the quadratic in the manner demonstrated and suggested. This will be helpful in doing the factorization.

[DEdesigns57]

I tryed to solve the problem 2x^5 - 7x^3 - 4x but only got so far plz help.

x(2x^4 - 7x^2 - 4x) BY FACORING X

X((2x^2)^2 - 7x^2 -4x) REWRITING

X(2x^2 + 1)(x - 4) FACTORING OUT THE QUADRATIC TRINOMIAL

BUT NOW WHAT! None of the terms have nothing in common and they are not in any forms that I can use a formaula for, can you help?

[stapel_eliz]

I tryed to solve the problem 2x^5 - 7x^3 - 4x but only got so far plz help.

x(2x^4 - 7x^2 - 4x) BY FACORING X

If you divided an x out of 4x, how did it remain as 4x?

To learn how to do this sort of factoring, try here.

X((2x^2)^2 - 7x^2 -4x) REWRITING

Since (2x²)² = 4x⁴, not 2x⁴, try instead re-writing in the manner which was provided to you earlier: 2(x²)² - 7(x²) - 4.

X(2x^2 + 1)(x - 4) FACTORING OUT THE QUADRATIC TRINOMIAL

Since (2x² + 1)(x - 4) = 2x³ - 8x² + x - 4, then this cannot be the correct factorization.

To learn how to factor quadratics, try here. Then try to do the factorization of 2y² - 7y - 4, and see how that relates to the quadratic-type quartic you were provided earlier.

[QM deFuturo]

x²(x + 2) - x(x + 2) - 12(x + 2)

If you are still confused about how to factor this the way it is now, how about this:

x²A - xA - 12A

Can you factor this? Notice in the first equation, you could let A = x +2, and maybe if you rewrite the polynomial temporarily in the second form you can simplify it in your mind? Then after you have factored it with the "A", you can change the "As" back to "(x+2)".

QM