My Math Forum (http://mymathforum.com/math-forums.php)
-   Algebra (http://mymathforum.com/algebra/)
-   -   Product of Linear Factors (http://mymathforum.com/algebra/311682-product-linear-factors.html)

 HowDoIMath January 3rd, 2016 06:54 PM

Product of Linear Factors

Problem:
Rewrite (4x^2+5x)^2 - 5(4x^2+5x) - 6 as a product of linear factors.

I have no idea where to start.

I tried using the substitution method to get (4x^2 + 5x - 6)(4x^2 + 5x + 1) and using the quadratic formula to get x = 3/4, -2, -1, and -1/4.

 greg1313 January 3rd, 2016 07:08 PM

That's a good approach. If you've got x = 3/4, then 4x - 3 = 0.

(4x - 3)(x + 2)(4x + 1)(x + 1) is equivalent to the original expression.

 HowDoIMath January 3rd, 2016 07:16 PM

Quote:
 Originally Posted by greg1313 (Post 503179) That's a good approach. If you've got x = 3/4, then 4x - 3 = 0. (4x - 3)(x + 2)(4x + 1)(x + 1) is equivalent to the original expression.
Thank you! It was the x = 3/4 that was messing with my answer.
I overlooked 4x - 3 = 0 and used x = 3/4 instead.

 All times are GMT -8. The time now is 05:12 PM.