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.
Number

1

2

3

4

5

6

3

2

Cube of number

1

8

27

?

125

216

27

?










Variable

X

R^{2}

T

B^{2}

R^{4}

P^{2}

m^{3}

f^{4}

Cube of variable expression

X^{3}

R^{6}

?

?

R^{12}

?

?

?

BIG NOTE:
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????
x^{2}cubed is x^{2}*x^{2}*x^{2}= x^{6}. Hope you did not think it was x^{8}.
We have xx*xx*xx and that becomes x^{6}. OR (x^{2} )^{3} is x^{2*3} = x^{6} .
These are examples of special binomials:
sums of cubics and differences of cubics.
DO YOU see that all of these are cubics? RECALL ( d^{2})^{3 }= d^{6} . So d^{6} is a cubic.
x^{3}  27

s^{3}+ 1

D^{6}+ 27

c^{3}+ 64

d^{3}m^{3} 27

x^{3}+ 8m^{3}

Every item in the binomials in the above table is the result of "cubing".
The 27 is from  3 being cubed.

See the +8m^{3} is from +2m being cubed.

The +64 is from +4 being cubed.

The D^{6 }is from D^{2 }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 (x5)(x^{2}+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 x^{3} 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 x^{3} 125
and give me the (x5)(x^{2}+5x +25) ?
Let us learn how:
A cubic binomial factors
into a binomial times a quadratic trinomial.
Example: x^{3}  27 = ( )( ).
We will find the binomial first. See the "short parenthesis"?
Find the numbers or variables that must be cubed to yield the x^{3}and  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.
x^{3} 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 x^{2}
Second term: multiply x and  3 then change the sign of your answer: +3x
Third term: square the  3: + 9.
There's your trinomial: x^{2} +3x +9
We have factored: x^{3} 27 = (x  3) (x^{2} +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!!!!!!!
THE CHECK:
(x  3) (x^{2} +3x + 9)=
x(x^{2} +3x + 9)  3(x^{2} +3x + 9)=
x^{3 }+3x^{2} + 9x  3x^{2}^{ } 9x  27=
x^{3 }  27.
YEA, that is it. PLEASE take the time to check one or two of your answers.

You try this one: y^{3}  8p^{3 }^{ }^{ }
What do you need to cube to get the y^{3}and the  8p^{3} ?
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 ******.
54x^{3}+ 2y^{3 } First see the 2 as a factor in all terms: 2(27x^{3} + y^{3}) is 54x^{3}+ 2y^{3}. Now look inside the parenthesis at the cubic. Factor again.
2(27x^{3} + y^{3}) =?
x^{6}+ y^{3}
r^{3}t^{3}125
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.

DO NOT JUST READ THIS.
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.
y^{3} 8p^{3}

(y 2p)( y^{2} +2py +4p^{2} )

54x^{3}+ 2y^{3}

2( 27x^{3} +y^{3} )=
2(3x + y)( 9x^{2}  3xy + y^{2} )
Be sure you get the 3x +y correct before finding the second parenthesis

x^{6}+ y^{3}

(x^{2} + y)( x^{4}  x^{2} y + y^{2} )
Be sure you get the x^{2} +y from the x^{6}+ y^{3}

r^{3}t^{3}  125

(rt 5)( r^{2} t^{2} +5rt + 25)

Now let 's try some more challenging ones.
(x8)^{3}+ 27y^{3}
I know there is only one parenthesis in this problem,
but let''s temporarily call it , you guessed it, Big P.
(x8)^{3} + 27y^{3 }becomes P^{3}+ 27y^{3}.
These are ALL CUBES, did you notice?
P^{3 }+ 27y^{3} factors to (P + 3y) (P ^{2} 3P y +9y^{2}).
Now we will replace the P with the original (x8) and we have our answer.
In summary:
(x8)^{3}+ 27y^{3}=
P ^{3}+ 27y^{3}=
This is a cubic binomial so let's find the binomial and trinomial.
We will use P and +3y
(P + 3y)(P ^{2}3Py +9y^{2}) =
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 + 9y^{2}) and replaced each of
them with (x8)??
[ (x 8) + 3y] [ (x 8)^{2 } 3y(x 8) + 9y^{2}]
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: x^{2} + 9 does not factor!!)
Know your squares.
Number

1

2

3

4

5

6

3

2

square of number

1

4

9
, ,

16

25

36

9

4










Variable

X

R

T

B ^{2}

R^{2}

P^{3}

?

C^{5}

square of variable

X^{2}

R^{2}

T^{2}

B^{4}

R^{4}

P^{6}

Q^{8}

?

Examples : These are always subtraction with squares.
x^{2}  9

x^{2}m^{2}  t^{8}

25x^{2}  49z^{2}

121  x^{6}

Factor these then multiply to check. These too have a pattern.
It is much easier to memorize than the ones for the cubics. Watch.
x^{2} 9 = (x3)(x+3)
Check it. Multiply.
(x3)(x+3)=
x(x+3) 3(x+3)
= x^{2} +3x 3x 9= x^{2} 9
m^{2}  1 = (m 1)(m+1)
1 is a square, 1 times 1 = 1^{ }
^{.}
x^{2}m^{2}t^{8} = (xm t^{4})(xm+ t^{4})
25x^{2} 49z^{2} = (5x 7z)(5x +7z)
121  x^{6} = (11 x^{3} )(11 + x^{3} )
Do you see the pattern?
And these will always check by multiplying them out.
The terms are squares and they are SUBTRACTED.
z^{2} 9b^{2} = (z3b)(z+3b)
Tricky one I mentioned above: sums of squares: x^{2} + 9 does not factor!!
16y^{2} +25 is a sum of squares. It will not factor. Try guessing and then check it by multiplying.
(4y+5)(4y5) gives 16y^{2}  25.
(4y+5)(4y+5) gives 16y^{2} +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} (n8)^{2}
I will color code so it is easy to follow.
(y+6)^{2} (n8)^{2}
And now let (y+6) = P and (n  8)= Q
This becomes: P^{2}  Q^{2} =
now factor this to:
[P  Q ][P + Q ] = and then replace the P and Q.
[(y+6) (n8)][(y+6) +(n8)] OK? Now simplify the LIKE terms for the "best answer".
[(y+6) (n8)][(y+6) +(n8)] 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
Comments (0)
You don't have permission to comment on this page.