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3:4 Factoring Cubic Squares

Page history last edited by Math in a Box - Susan Johnsey gm 3 years, 10 months ago

Lesson 3.4 Factoring Cubic Binomials

                                           and Differences of Squares.



Before you begin this lesson there is a worksheet you must complete.   It will make your work now and later easier.

It is called  Worksheet for 3:4.   You can type worksheet in the SEARCH box and it will appear.  But I may have placed one in your folder so look there first.  Click it and print it for yourself.  

Complete it.   Some items are repeated just so you learn them well.

                 If you do not see it  then let me know.



A cubic binomial has 2 terms and all numbers and variables in the binomial are cubes.

Learn your cubic numbers now. Copy and complete this.











Cube of number



























Cube of variable expression











        the rule for cubing numbers is write the number 3 times and multiply. 

           2 cubed is   2*2*2   = 8


AND the rule for cubing the variables is write the variable with its exponent down 3 times as multiplication

        and then what do we do with the exponents????      


             x2cubed is x2*x2*x2= x6.  Hope you did not think it was x8.      

                            We have xx*xx*xx and that becomes x6.    OR  (x2 )3  is  x2*3 = x6 . 


These are examples of special binomials:

sums of cubics   and   differences of cubics.


DO YOU see that all of these are cubics?  RECALL ( d2)3 = d6 .  So d6 is a cubic.

x3 - 27

s3+ 1

D6+ 27

c3+ 64

d3m3- 27

x3+ 8m3



Every item in the binomials in the above table is the result of "cubing".  


The -27  is from - 3 being cubed.

See the +8m3 is from +2m being cubed.

The +64  is from +4 being cubed.

The Dis from D2 being cubed.


Recognizing that you have a cubic binomial is imperative.


They require another type of factoring and

if you do not see that you have cubic binomials then you will not remember to follow these directions below.


As you can see from the lessons in this chapter, 

each type of polynomial requires its own style of factoring.

But the check to see if you have factored correctly is always to multiply.


Thus finding the answer to a factoring problem can be difficult , but the check is always easy -- just multiply!!


Multiply (x-5)(x2+5x +25)          Did you get 6 terms?

You must be able to do that in order to proceed.  

Please find them and then combine like terms.    


Did you get 6 terms but then after combining you have only x3- 125? 

This is the difference of two cubes.  


I will be giving you one of these and will be asking you to find

the expressions that were multiplied to get it.  


That is,   can you look at the x3- 125

                           and give me the (x-5)(x2+5x +25) ?  


Let us learn how:


A cubic binomial factors

                into a binomial times a quadratic trinomial.


Example:   x3 - 27 = (     )(                 ).        

We will find the binomial first. See the "short parenthesis"?         

Find the numbers or variables that must be cubed to yield the x3and - 27.

Hope you do not have to look back at the table of cubics; you do know them.           

You should find x and -3. These will make our binomial.


x3- 27 = (x - 3)(                 )

Now let''s find the 3 terms of the trinomial.  

See the "long parenthesis"   

 LOOK at your x and - 3.  


We use the x and - 3 to create the trinomial. 


Here's how:

First term of the trinomial is found by squaring the x, that is   x2

Second term: multiply x and - 3 then change the sign of your answer: +3x

Third term: square the - 3: + 9.      

There's your trinomial: x2 +3x +9


We have factored:   x3- 27 = (x - 3) (x2 +3x + 9).

We can check this by muliplying as we did in Lesson 2.2.

The check by multiplying:  We will have 6 terms but 4 of them will subtract away!!!!!!! 


(x   3) (x2 +3x + 9)=

                          x(x2 +3x + 9) - 3(x2 +3x + 9)=   

                              x3 +3x2 + 9x  - 3x2 -  9x  - 27= 

                                                                           x - 27.  


YEA,  that is it. PLEASE take the time to check one or two of your answers.



You try this one:    y3 - 8p3     

What  do you need to cube to get the y3and the - 8p3 ?  


 You need y and -2p. The binomial is (y -2p).

KEEP the ORDER.  NOW  finish by finding the trinomial.


What's the trinomial," the long parenthesis"??

     Follow the 3 purple steps above.


Also do these three problems and find the answers below by ******.

        54x3+ 2y First see the 2 as a factor in all terms: 2(27x3 + y3)  is 54x3+ 2y3. Now look inside the parenthesis at the cubic.  Factor again.

2(27x3 + y3) =?


         x6+ y3




You  need to know well the list of cubics.

Memorize the numbers before your test.

Study the worksheet that you completed at the beginning of this lesson.




BE SURE TO DO THESE examples by yourself.        

Copy them into you notes and then see if you can do them again without looking!!


FIND the binomial first

    then use the 3 steps above to find the quadratic trinomial.


y3- 8p3

(y -2p)( y2 +2py +4p2 )


54x3+ 2y3

2( 27x3 +y3 )=

2(3x + y)( 9x2 - 3xy + y2 )

Be sure you get the 3x +y correct before finding the second parenthesis


x6+ y3

(x2 + y)( x4 - x2 y + y2 )

Be sure you get the x2 +y from the x6+ y3

r3t3 - 125

(rt -5)( r2 t2 +5rt + 25)


Now let 's try some more challenging ones.

(x-8)3+ 27y3


I know there is only one parenthesis in this problem,

but let''s temporarily call it , you guessed it, Big P.  


(x-8)3 + 27ybecomes     P3+ 27y3.      


These are ALL CUBES, did you notice?

 P3 + 27y3   factors to (P + 3y) (2- 3P y +9y2).


Now we will replace the P with the original (x-8) and we have our answer


In summary:

(x-8)3+ 27y3=

3+ 27y3=

This is a cubic binomial so let's find the binomial and trinomial.


We will use P and +3y

(P  + 3y)(P 2-3Py +9y2) =  

Now carefully replace each P with (x- 8).   Look carefully at this.  


Do you see below that I removed all three P's

in  (P  + 3y)(P 2+ 3Py + 9y2)  and replaced each of

                                                       them with (x-8)??

(x -8) + 3y  [ (x -8)2 - 3y(x -8) + 9y2


Study it. Write it on paper and try it yourself.


TRY:            (r +t) 3- 64


Do Assignment 3.4A  


PART 2   Now the SQUARES (differences only!)


Do not continue until you understand the cubics.


Factoring Binomials that are Differences of Squares.

(note that I did not say sums of squares:  x2 + 9  does not factor!!)

Know your squares. 











square of number




, ,



















B 2





square of variable










Examples :  These are always subtraction with squares.

x2 - 9

x2m2 - t8

25x2 - 49z2

121 - x6


Factor these then multiply to check. These too have a pattern.

It is much easier to memorize than the ones for the cubics. Watch.


x2- 9 = (x-3)(x+3)


Check it. Multiply.



                   x(x+3) -3(x+3)

               = x2 +3x -3x -9= x2- 9


m2 - 1  = (m -1)(m+1) 

                         1 is a square, 1 times 1 = 1 


x2m2-t8 =  (xm -t4)(xm+ t4)


25x2 -49z2 = (5x -7z)(5x +7z)


121 - x6 = (11 -x3 )(11 + x3 )


Do you see the pattern?

And these will always check by multiplying them out.         

The terms are squares and they are SUBTRACTED.


z2- 9b2  = (z-3b)(z+3b)


Tricky one I mentioned above:  sums of squares:  x2 + 9  does not factor!!

16y2 +25  is a sum of squares.  It will not factor.  Try guessing and then check it by multiplying. 

  (4y+5)(4y-5)   gives   16y2 - 25. 

  (4y+5)(4y+5) gives 16y2 +40y+25  You know how to get that?  We learned it in earlier lesson.  I did not show all the steps, but you should know them.


LAST EXAMPLE  (important to copy into your notes)


(y+6)2- (n-8)2             


I will color code so it is easy to follow.

(y+6)2- (n-8)2

And  now let (y+6) = P   and (n - 8)= Q

This becomes: P2 - Q2 =     

 now factor this to:

                     [P  -  Q ][ + Q ]  =    and then replace the P and Q.


[(y+6) -(n-8)][(y+6) +(n-8)]    OK?    Now simplify  the LIKE terms for the "best answer".



[(y+6) -(n-8)][(y+6) +(n-8) is also [y - n + 14][(y + n  - 2].  Where did I get the 14?


Do Assignment 3.4B. 

By Susan Johnsey 2002 www.mathinabox.com



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