Algebra (Period 1)

Factoring by Group 6-6

First a review:
Checklist of how to factor thus far-->

1. Look for a GCF of all terms

2. Binomials - look for difference of two squares

both perfect squares - double hug - one positive, one negative - square roots of both terms

3. Trinomials - look for Trinomial Square (factors as a binomial squared)

first and last must be perfect squares - middle must be double the product of the two square roots

SINGLE hug - square roots of both terms - sign is middle sign

4. Trinomials - last sign positive - double hug with same sign as middle term - factors that multiply to last and add to middle

5. Trinomials - last sign negative - double hug with different signs, putting middle sign in first hug - factors that multiply to last and subtract to middle - middle sign will always be with the bigger factor



REMEMBER: 
FACTORING WILL NEVER CHANGE THE ORIGINAL VALUE OF THE POLYNOMIAL SO YOU SHOULD ALWAYS CHECK BY MULTIPLYING BACK!!!!

(we're skipping 6-5 and then going back to it)
When you have 4 TERMS IN YOUR POLYNOMIAL!


You put the polynomial in 2 sets of 2 by using ( )

Then you factor out the GCF for each set of 2 terms individually


DOES THIS ALWAYS WORK FOR EVERY 4 TERM POLYNOMIAL?

Of course not!
But for this section of the math book, it will!

What happens if it doesn't work? The polynomial may just not be factorable!

MAKE SURE IT'S IN DESCENDING ORDER FIRST!!!!

EXAMPLE: 6x³ - 9x² + 4x - 6
First notice there is NO GCF of all the terms!!

Factoring by grouping says if there is no GCF of the 4 terms, look and see if there is a GCF of just 2 terms at a time!!

Put ( ) around the first 2 terms and another ( ) around the 2nd set of terms.
(6x³ - 9x²) + (4x - 6)
Factor out the GCF from each set of two terms
3x²(2x - 3) + 2(2x - 3)
Look for a COMMON factor to factor out between the two sets
In this case its (2x - 3)
Pull out
(2x - 3)(3x + 2)
and check to make sure you cannot continue to factor!!

Try these:

x³ + x² + 2x - 2
First... is there a GCF? No

okay
now set up in 2 groups of TWO

(x³ + x²) + (2x - 2)
x² (x + 1) + 2 (x-1)
wait... they are NOT the same...
cannot be factored.. not factorable!!

2x² - 4x + xz - 2z
Is there a GCF? NO

(2x² - 4x) + (xz - 2z)
2x(x -2) + z(x-2)
(x-2)(2x + z)

24x³ + 27x² - 8x - 9
Is there a GCF? NO

(24x³ + 27x²) + (-8x - 9)

3x²(8x +9) -1(8x + 9)
(8x + 9)(3x² - 1)

c⁶ -c⁴ - c² + 1
(c⁶ -c⁴) + (-c² + 1)
c⁴(c² -1) -1c² -1)

Look carefully at that results.. why did the second term become -1?

(c² -1)(c⁴ -1)
and ask yourself... are you finished factoring? ...
NO
I see The difference of Two Squares...


(c + 1)(c-1)(c²+1)(c² -1)
Now are you finished?
No.. I still see the difference of Two Squares... bring everything down...
(c + 1)(c-1)(c²+1)(c + 1)(c-1)
and now write it in the correct order
(c² + 1)(c + 1)(c + 1)(c-1)(c-1)


4y⁵ + 6y⁴ +6y³ +9y²
First thing-- Is there a GCF? YES

pull out a y² and you are left with
y²(4y³ + 6y² + 6y + 9)
Now put those 4 terms in 2 groups of two!!.. use brackets..

y²[(4y³ + 6y²) + 6y + 9)]
Look for a GCF in each of the hugs!!

y²[2y²(2y +3) + 3(2y+3)]

What do each of them have in common? What do they share? 2y + 3
when you put that in the first set of hugs... what's left?

y²(2y+3)(2y²+3)