explain factor by grouping factoring by grouping

Question

Explain Factor by Grouping ?

Factoring by grouping is often a good way to factor polynomials of 4 terms or more. (Sometimes it isn't. It doesn't always work. But it's worth trying.)

Example with 4 terms

Take a look at this one:

2x³ -4x² + 3x -6

Before I work through the example, take a look at the first two coefficients (2 and -4) and the last two (3 and -6). Notice how the ratios are the same (2 : -4 = 3: -6)? That's a good clue that factoring by grouping might work. OK, now let's group the first two terms and the last two terms.

(2x³ -4x² ) + (3x -6 )
Now, in each of these groups, factor out any common monomial factors.

2x² (x -2 ) + 3(x -2)
See how you have the same factor, (x -2 ), left over in each term? That's how you know that this method really is going to work. ( Up to this point, one isn't really sure.) All you have to do is factor out the (x -2) using reverse distribution,

(2x² + 3)(x -2)
and you're done!
Nastiness with negative signs
This one is only slightly different from the previous one:

2x³ -4x² -3x + 6

Here's the first problem you encounter: it's easy to make the mistake of putting the minus sign outside the parentheses:

(2x³ -4x2 ) - (3x + 6) (Wrong!)

Be sure to put the minus sign inside the parentheses, because it belongs only to the 3x and not to the 6.

(2x³ -4x² ) + (-3x + 6)

The next step is to factor out, from each group, any common monomial factors:

2x² (x -2 ) + 3(-x + 2)

Now, ideally, the groups left, (x -2) and (-x + 2), should be the same. They're not. But notice that if you factor out a negative sign from the second group, then they will be the same.

2x² (x -2 ) -3(x -2).
At last you can factor out the (x -2 ).

(2x2 - 3)(x - 2)

Example involving more than 4 terms.
You sometimes have to experiment a little when you're grouping the terms. Often, one way of grouping the terms doesn't work, while another way does. Here are a couple of tips for grouping the terms:
• You must always have the same number of terms in each Group.

• The ratios of the coefficients in one group must be the same as the ratios in the other groups.
OK, here's the example.
2x⁹ + x⁸ + 6x⁷ + 3x⁶ - 3x² - 9
If you just try to group the three terms on the left and the three on the right, it won't work. Don't feel bad about this attempting to group it this way is not a "mistake". You don't know whether it will work until you try.
(2x⁹ + x⁸ + 6x⁷)+ (3x⁶ -3x² - 9)
x⁷ (2x² + x + 6) + 3(x⁶ - x² -3)
Doesn't work -the two groups aren't the same after removing common factors.
So, try it another way, rearranging some of the terms. Notice how the rations of coefficients are the same in each group!
(2x⁹ + x⁸ -3x² ) + (6x⁷ + 3x⁶ -9)
2 : 1 : -3 = 6 : 3 : -9
Now remove the common factors,
x² (2x⁷ + x⁶ -3) + 3(2x⁷ + x⁶ - 3)
And the two groups are the same! Finish it up with a reverse distribution,
(x² + 3)(2x⁷ + x⁶ -3)
and you're done.