Can someone explain to me how you would completely factor 20x^2 – 28x – 48 by using the grouping method.

Question

anonymous · Answer

What's the "grouping method"? I'm kinda confused how you'd factor that by putting the terms into like groups.

anonymous · Answer

This new method of factoring is called factoring by grouping because pairs of terms are grouped together to factor the entire polynomial. The goal with this method is to find a common binomial that can be factored.

anonymous · Answer

There needs to be four terms (or, in general, an even amount of terms). There's 3.

You can't group pairs, since... well.. there is a pair and only one. Does it perhaps mean extracting the $4$?

$20x^2-28x-48=4(5x^2-7x-12)$?

anonymous · Answer

I don't know...

anonymous · Answer

Me neither. Hmm.

@Calcmathlete ???

anonymous · Answer

Factoring by grouping huh? Usually this is done for trinomials that have a leading coefficient greater than 1.
$$20x^{2} - 28x - 48$$$$4(5x^{2} - 7x - 12)$$Find factors of -60 that add up to -7
$$5x^{2} - 12x + 5x - 12$$$$(5x^{2} - 12x) + (5x - 12)$$$$x(5x - 12) + (5x - 12)$$Using Distribute property, combine.$$(x + 1)(5x - 12)$$

anonymous · Answer

Btw, that 4 I factored out at first is still there, but it bothered me putting it over and over again...

anonymous · Answer

Do you two get it!???? lol

anonymous · Answer

Uhh.. Kind of.

anonymous · Answer

Its really a process more than anything. What part are you stuck on?

anonymous · Answer

Well, I really don't understand the whole "grouping method". :/

anonymous · Answer

Well, let's take a look at something easier.

anonymous · Answer

$$x^{2} + 3x + 2$$Find factors of 2 that add up to 3.
$$x^{2} + 2x + x + 2$$Do you see how 2x and x add up to 3x?
$$(x^{2} + 2x) + (x + 2)$$$$x(x + 2) + (x + 2)$$$$(x + 1)(x + 2)$$Do you see how this is the same result when simply doing what you normally do?

anonymous · Answer

Oh, yes, I do.

anonymous · Answer

When you're finding factors of 2 that add up to 3, you're actually multiplying 2 and 1 because x^2 has a 1 in front right? I did the same thing with the previous problem. However, the factoring by grouping process works because when you have something like $$5x^{2} - 7x - 12$$,the 5 in front of the x^2 has to go somewhere right? You have to account for factors of 5 as well which is why you multiply.