SPECIAL PRODUCTS
Square of Binomial
The product of a square of a binomial is the square of the first term plus or minus ( ± ) the product of the first and the second terms plus the square of the second term. In symbol,
( x ± y ) 2 = x2 ± 2xy + y2
Illustrative Examples:
1. Find the product of ( 4x2 + 5y2 ) 2
Solution: ( 4x2 + 5y2 ) 2 = 16x4 + 2 [ ( 4x2 ) ( 5y2 ) ] + 25y4
= 16x4 + 2 ( 20x2y2 ) + 25y4
= 16x4 + 40x2y2 + 25y4
2. Find the product of ( 2a – 3b ) 2
Solution: ( 2a – 3b ) 2 = 4a2 – 2 [ ( 2a ) ( 3b ) ] + 9b2
= 4a2 – 2 ( 6ab ) + 9b2
= 4a2 – 12ab + 9b2
3. Find the product of ( a3 – 5 ) 2
Solution: ( a3 – 5 ) 2 = a6 – 2 [ ( a3 ) ( 5 ) ] + 25
= a6 – 2 ( 5a3 ) + 25
= a6 – 10a3 + 25
Product of Two Binomials
A. In the form of ( x + a ) ( x + b )
The product of two binomials of the form ( x + a ) ( x + b ) is the square of x, plus x times the sum of a and b, plus the product of a and b.
( x + a ) ( x + b ) = x2 + ( a + b ) x + ab
Illustrative Examples:
1. ( x + 3 ) ( x + 8 ) = x2
+ ( 3 + 8 ) x + 3 ( 8 )
= x2 +
11x + 24
2. ( x + 5 ) ( x + 4 ) = x2
+ ( 5 + 4 ) x + 5 ( 4 )
= x2 +
9x + 20
3. ( x + 2 ) ( x + 6 ) = x2 +
( 2 + 6 ) x + 2 ( 6 )
= x2 +
8x + 12
The four terms are the product of firsts, outers, inners, and the lasts.
( x + a ) ( X + b ) = x2 + ax + bx +
ab
= x2 + ( a + b ) x + ab
Illustrative Examples: Find the product of the following by using
the FOIL method:
1. ( 3x + 4y ) ( x – 2y )
Solution:
(1) Find the product of the first
terms: 3x ( x )
(2) Find the product of the outside
terms: 3x ( -2y ) = -6xy
(3) Find the product of the inside
terms: 4y ( x ) = 4xy
(4) Multiply the last terms: 4y ( -2y
) = -8y2
(5) Combined the four terms obtained
above and simplify:
3x2 + ( -6xy + 4xy
) – 8y2 = 3x2 – 2xy – 8y2
This method is actually similar to the foil method. Make first a 2 by 2
table, thus,
Place each term of
the two binomials vertically in each square.
Multiply the terms horizontally and
diagonally, combine the result, then simplify: thus, ac + bd + ad + bc.
Illustrative Examples:
Use the foil method to find the product of the
following binomials:
( a + b ) ( a
– b ) = a2 – b2
In this product, special
characteristics are very evident and it may be obtained by: (i) squaring the
first term of the factors; and, (ii) subtracting the square of the second term
of the factors. And that is,
( first term – second term )2
= ( first term )2 – ( second term )2
Illustrative Examples:
1. ( x+ 4 ) - ( x – 4 ) = x2
– 42 = x2 - 16
2. ( X + 2y ) ( x – 2y ) = x2
– (2y)2 = x2 – 4y2
3. ( 2b + 9c ) ( 2b – 9c ) = ( 2b )2
– ( 9c )2 = 4b2 – 81c2
( a + b + c ) 2 =
a2 + b2 + c2 +2ab + 2ac + 2bc
That is, the square of polynomial is
equal to the sum of the squares of the terms and twice the product of each term
by every other term taken separately.
Illustrative Examples:
1. Expand ( 3x – y – 2z )
Solution:
(1) Find the squares of each term of
the trinomial;
9x2, y2 and 4z2
(2) Find also twice the product of
each term by every other term taken separately;
-6xy, -12xz, and 4yz
(3) Combine the terms an simplify;
9x2 + y2 + 4z2
– 6xy – 12xz + 4yz
2. ( 2a + 3b – 5c )2 = 4a2
+ 9b2 – 25c2 + 12ab – 20ac – 30bc
3. ( 3a + 5b + 4c )2 = 9a2
+ 25b2 + 16c2 + 30ab + 24ac + 40bc
DEFINITION: If a, b, and c are integers,
and a • b = c, then a and b are factors of c and c is a multiple of a and b.
In the given 3 • 4 = 12, where 3 and 4 are factors of 12 and 12 is a
multiple of 3 and 4. Other factors of 12 are
( -3 ) and ( -4 ), 1 and 12, ( -1 ) and ( -12 ), 2 and 6, and ( -2 ) and
( -6 ).
DEFINITION: A PRIME NUMBER is a number that has no other factors except 1 and
itself. 1, 3, 5, and 7 are all examples of prime numbers. And also, 11, 13, 17,
19….. etc.
A PRIME FACTOR is an
expression that has no other factor(s) except 1 and itself.
DEFINITION: FACTORING is the process
of finding the factors of a given algebraic expression.
FORMULA: ax + ay = a ( x + y )
Illustrative Examples:
1. Factor 4ax3 – 8ay2
+12a2x
Solution: By inspection, the common
factor is 4a; thus,
4ax3 – 8ay2 +
12a2x = 4a ( x3 – 2y2 + 3ax )
2. Factor 2x ( a + b ) – 3y ( a + b )
Solution: The common factor is ( a +
b ); thus,
2x ( a + b ) – 3y ( a + b ) = ( 2x –
3y ) ( a + b )
3. Factor 20xyz + 20xyz2 – 60x2y2z2
Solution: The common factor is 10xyz;
thus,
20xyz + 30xyz2 – 60x2y2z2 = 10xyz ( 2
+ 3z – 6xyz )
FORMULA: a2 – b2 = ( a + b )
( a – b )
CONCEPT:
( 1st term )2 –
( 2nd term )2 = ( 1st term + 2nd
term ) • ( 1st term – 2nd term )
Procedures in Factoring a Difference
of Two Squares:
- Identify
that you have a perfect square minus another perfect square.
- Rewrite
the problem as a first base squared minus a second base squared.
- Factor
the problem into the first base plus the second base times the first base
minus the second base.
Illustrative Examples:
1. Write in completely factored form:
( 16a2 – 25b2 )
Solution:
A. Identify that you have a perfect
square minus another perfect square;
16a2, 25b2 are
perfect square
B. rewrite the problem as a first
base squared minus a second base squared;
( 4a )2 – ( 5b )2
C. Factor the problem into the first
base, plus the second base, times the first base minus the second base squared;
( 4a + 5b ) x ( 4a – 5b ) ® answer
2. 49x8y6 –
625z8
Solution:
( 7x4y3 )2
– ( 25z4 )2 = ( 7x4y3 + 25z4
) (7x4y3 – 25z4 )
3. Factor ( 16x4z8
– 81z12 )
Solution:
( 16x4z8 – 81z12
) = ( 4x2z4 )2 ( 9z6 )2
= ( 4x2z4
+ 9y6 ) ( 4x2z4 – 9y6 )
Steps in Factoring a trinomial in the
form ax2 + bxy + cy2:
- Write
all pairs of factors of the
coefficient of the first term
- Write
all pairs of factors of the last term
- Use
various combinations of this factor until the necessary middle term is
found.
- If
the necessary combination does not exist, the polynomial is a prime,
meaning, it has no factors except 1 and itself.
Illustrative Examples:
1. Factor completely the given ( x2 –
6x + 8 )
Solution:
(a) Factor the first term: The first
term is x2 and its coefficient is 1, so the factors of 1 are 1 and
1.
(b) The factors of the last term
which is 8 are 8 and 1, 4 and 2, ( -8 ) and ( -1 ), and ( -4 )
( -2 ).
(c) The sign of the middle term is
negative, and since the sign of the last term is positive, the sign of the
possible factors of the last term are both negative. The only combination of the factor of the last term to get the sum
of 6 is 2 and 4, thus, x2 – 6x + 8 is ( x – 4 ) ( x – 2 ).
2.
( x2 + 9x + 20 ) = ( x + 4 ) ( x + 5 )
3. ( x2 – 10x + 21 ) = ( x
– 7 ) ( x – 3 )
Illustrative Examples:
1. Factor x2 –
3xy – 10y2
Solution: The factor of x2
are x and x, while the factors of -10y2 are:
( -5y and 2y ), ( 5y and _2y ), ( -10y and y
), and ( 10y and –y ). Anyone of these sets of factors of -10y2
could be the correct ones. In our example, the correct factors are -5y and 2y;
thus,
x2 – 3xy 10y2
= ( x + 2y) ( x – 5y )
2. x2 – 8xy +
15y2 = ( x – 3y ) ( x – 5y )
3. x2 + 5xy +
6y2 = ( x + 2y ) ( x + 3y )
acx2 + ( ad +
bc ) x + bd (ax + b ) ( cx + d )
Not just like the other types of
factoring, this one involves trial and error
Illustrative Examples:
1. Factor 6x2 +
8xy – 30y2
Solution: Get the factors
of 6x2 and the factors of -30y2, then by trial and error,
get the correct factors. Thus, 6x2 + 8xy – 30y2 = ( 3x –
5y ) ( 2x + 6y ).
2. Factor 6x2 –
28xy + 16y2
Solution: 6x2 –
28xy + 16y2 = ( 6x – 4y ) ( x – 4y )
3. 64x2 + 96x
+35 = ( 8x + 5 ) ( 8x + 7 )
Factoring by grouping can be facilitated
by grouping together terms that have common factor, then factor.
Illustrative Examples:
1. Factor ax + bx - ay -by
Solution: Group the first
two terms together and also the last two terms, then get their common factors;
thus,
ax + bx – ay – by = ( ax +
bx ) – ( ay + by )
= x ( a + b ) – y ( a
+ b )
answer® = ( x – y ) ( a + b )
NOTE: The sign of ay + by has
changed from negative to positive when enclosed by a minus sign.
2. Factor 8x3 –
24x2 - 2x + 6
Solution: Group the first
two terms together, and the last two terms together, and get the common factors
for both group. Thus,
8x3 – 24x2 – 2x + 6 = ( 8x3
– 24x2 ) – 1 ( 2x – 6 )
=4x2 ( 2x – 6 )
– 1 ( 2x – 6 )
= ( 4x2
– 1 ) ( 2x – 6 )
answer ® = ( 2x + 1 ) ( 2x – 1 ) 2 ( x- 3 )
3. Factor 25x2
+ 30xy + 9y2 + 15x + 9y + 2
Solution: Group the first
three terms together, and also, the fourth and fifth terms together. Then,
factor both group of terms. The first group is actually a trinomial perfect
square. Add the last term and factor.
Thus, 25x2 +
30xy + 9y2 + 15x + 9y + 2
= ( 25x2 + 30xy
+ 9y2 ) + ( 15x + 9y ) + 2
= ( 5x + 3y )2
+ 3 ( 5x + 3y ) + 2
= [ ( 5x + 3y ) + 2 ] [ (
5x + 3y ) + 1 ]
= ( 5x + 3y + 2 ) ( 5x +
3y + 1 )
SUM: (a3 + b3
) = ( a + b ) ( a2 – ab + b2 )
 |
DIFFERENCE: ( a3
– b3 ) = ( a – b ) (a2 + ab + b2 )
NOTE:
The factors of the Sum or Difference of two cubes at binomial and trinomial.
STEPS IN FACTORING THE SUM
AND DIFFERENCE OF TWO CUBES:
A. Find the terms of the
binomial factor by getting the cube roots of the term of the given sum or
difference of two cubes.
B. Find the first term of
the trinomial factor by squaring the first term of the binomial factor
C. Find the second term of
the trinomial factor by getting the product of the first and second term of the
binomial factor.
D. Find the last term of
the trinomial factor by getting the square of the second term of the binomial
factor.
E. Check your answer by
multiplying the binomial and trinomial factors obtained. If your answer is
correct, you will get the original sum or difference.
Illustrative Examples:
1. Factor x3 +
64
Solution:
A. The cube root of x3 and 64 are x and 4, thus, ( x + 4 )
B To find the first term
of the trinomial factor, square the first term of the binomial factor, and that
is x2
c. Get the product of the
first and the second term of the binomial factor,
( 4 ) ( x ) = 4x
D. Get the square of the
second term of the binomial factors, thus, 42 = 16.
Therefore: x3 +
64 = ( x + 4 ) ( x2 + 4x + 16 ) ® answer
2. Factor 27x3
– y3
Solution: The given is the
difference of two cubes. Solving using the procedures, we will have the answer;
27x3 –y3
= ( 3x – y ) ( 9x2 + 3xy + y2 )
3.125a3 + 216b3
= ( 5a + 6b ) ( 25a2 –
30ab )
The sum or difference of two like odd
powers I always divisible by the sum or difference of the numbers respectively.
In factoring the sum or difference of two
like odd powers, it is important to remember the following:
1. The factors of the sum
or difference of two like odd powers are a binomial and a polynomial.
2. To find the terms of
the binomial factor get the odd power roots of the terms of the original binomials.
Original binomial refers to the sum or
difference of two like odd powers.
3. To find the terms of the polynomial factor,
divide the original binomial factor using either the long method or synthetic
division.
Illustrative Examples:
1. Factor a5 +
b5
Solution: The binomial
factor of a5 + b5 is ( a + b ). To get the polynomial
factor, divide a5 + b5 by ( a + b ), thus,
a4 –
a3b + a2b2 – ab3 + b4
a + b a5 + 0
+ 0 +
0 + b5
a5 + a4b
-a4b
-a4b
– a3b2
 |
a3b2
a3b2 + a2b3
 |
-a2b3
a2b3 – ab4
|
ab4 + b5
-ab4 + b5
 |
0
Thus, ( a5 + b5
) = ( a + b ) ( a4 – a3b a2b2 + ab3
+ b4 )
Alternative Way:
The factors of the sum or difference of
two like odd powers can also be found without using division and that is by
means of the following steps:
a. The binomial factor is
a sum or difference of the odd power roots of two variables of the original
binomial.
b. The terms of the
polynomial factors has a descending order of power of the first variable.
c. The power of the first
term and of the last term of the polynomial factor are one degree lower than
the powers of the first variable and of the second variable respectively of the
original binomial.
d. The signs of the terms
of the polynomial factor are all positive for difference of two like odd
powers, and has alternate positive an negative signs for sum of two like odd
powers.
Illustrative Examples:
1. Factor x7 +
y7
Solution:
X7 + y7 =
( x + y ) ( x6 – x5y + x4y2 + x3y3
+ x2y4 + xy5 + y6 )
2. Factor x5 –
y5
x5 – y5
= ( x + y) ( x4 + x3y + x2y2 + xy3
+ y4 )
Factoring Two – Like Even Powers
This one can be factored as a Difference
of Two Squares. Each factor should be factored repeatedly until the expression
is reduced to its prime factors.
Illustrative Examples:
1. Factor x6 –
y6
Solution:
X6 – y6
= [ ( x3 )2 – ( y3 )2 ]
= ( x3 + y3 )
( x3 – y3 )
= ( x + y ) ( x2 – xy + y2
) ( x – y) ( x2 + xy + y2 )
2. Factor x8 –
y8
Solution:
X8 – y8
= [ ( x4 )2 – ( y4 )2 ] = ( x4
+ y4 ) ( x4 – y4 )
= ( x4
+ y4 ) ( x2 + y2 ) ( x2 – y2
)
= ( x4
+ y4 ) ( x2 + y2 ) ( X + y ) ( x – y )
NOTE: ( x4 + y4
) and ( x2 + y2 ) are not factorable.
3. Factor a12 – b12
Solution:
a12 – b12
= [ ( a6 )2 – ( b6 )2 ] = ( a6
+ b6 ) ( a6 – b6 )
= ( a6 + b6 )
( a3 + b3 ) ( a3 – b3 )
= ( a6 + b6 )
( a + b ) (a2 – ab + b2 ) ( a – b ) ( a2 + ab
+ b2 )
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