Transcript Document

1. Recognize special polynomial product patterns.
2. Use special polynomial product patterns to multiply
two polynomials.
Multiplication of polynomials is an extension of the distributive
property. When you multiply two polynomials you distribute each
term of one polynomial to each term of the other polynomial.
We can multiply polynomials in a vertical format like we would
multiply two numbers.
(x – 3)
(x – 2)
x_________
–2x + 6
_________
x2 –3x + 0
x2 –5x + 6
Multiplication of polynomials is an application of the distributive
property. When you multiply two polynomials you distribute each
term of one polynomial to each term of the other polynomial.
We can also multiply polynomials by using the FOIL pattern.
(x – 3)(x – 2) = x(x) + x(–2) + (–3)(x) + (–3)(–2) = x2 – 5x + 6
Some pairs of binomials have special products.
When multiplied, these pairs of binomials always follow the
same pattern.
By learning to recognize these pairs of binomials, you can use
their multiplication patterns to find the product quicker and
easier.
One special pair of binomials is the sum of two numbers times
the difference of the same two numbers.
Let’s look at the numbers x and 4. The sum of x and 4 can be
written (x + 4). The difference of x and 4 can be written (x – 4).
Their product is
(x + 4)(x – 4) = x2 – 4x + 4x – 16 = x2 – 16
Multiply using foil, then collect like terms.
Here are more examples:
}
(x + 4)(x – 4) = x2 – 4x + 4x – 16 = x2 – 16
(x + 3)(x – 3) =
x2
– 3x + 3x – 9 =
x2
– 9
(5 – y)(5 + y) = 25 +5y – 5y – y2 = 25 – y2
What do all of these
have in common?
What do all of these
have in common?
x2 – 16
x2 – 9
25 – y2
They are all binomials.
They are all differences.
Both terms are perfect squares.
For any two numbers a and b, (a + b)(a – b) = a2 – b2.
In other words, the sum of two numbers times the difference of
those two numbers will always be the difference of the squares of
the two numbers.
Example: (x + 10)(x – 10) = x2 – 100
(5 – 2)(5 + 2) = 25 – 4 = 21
3
7
= 21
The other special products are formed by squaring a binomial.
(x + 4)2 and (x – 6)2 are two example of binomials that have been
squared.
Let’s look at the first example: (x + 4)2
(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 4x + 16 = x2 + 8x + 16
Now we FOIL and collect like terms.
Whenever we square a binomial like this, the same pattern always occurs.
(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 4x + 16 = x2 + 8x + 16
See the
first term?
In the final product
it is squared…
…and it appears in the middle term.
Whenever we square a binomial like this, the same pattern always occurs.
(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 4x + 16 = x2 + 8x + 16
What about the
second term?
The middle number is 2 times 4…
…and the last term is 4
squared.
Whenever we square a binomial like this, the same pattern always occurs.
(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 4x + 16 = x2 + 8x + 16
Squaring a binomial will always produce a trinomial whose first
and last terms are perfect squares and whose middle term is 2
times the numbers in the binomial, or…
For two numbers a and b, (a + b)2 = a2 + 2ab + b2
Is it the same pattern if we are subtracting, as in the expression
(y – 6)2?
(y – 6)2 = (y – 6)(y – 6) = y2 – 6y – 6y + 36 = y2 – 12y + 36
It is almost the same. The y is squared, the 6 is squared and the
middle term is 2 times 6 times y. However, in this product the
middle term is subtracted. This is because we were subtracting in the
original binomial. Therefore our rule has only one small change
when we subtract.
For any two numbers a and b, (a – b)2 = a2 – 2ab + b2
Examples:
(x + 3)2 = (x + 3)(x + 3) Remember: (a + b)2 = a2 + 2ab + b2
= x2 + 2(3)(x) + 32
= x2 + 6x + 9
(z – 4)2 = (z – 4)(z – 4) Remember: (a – b)2 = a2 – 2ab + b2
= z2 – 2(4)(z) + 42
= z2 – 8z + 16
You should copy these rules into your notes and try to remember them.
They will help you work faster and make many problems you solve
easier.
For any two numbers a and b, (a + b)(a – b) = a2 – b2.
For two numbers a and b, (a + b)2 = a2 + 2ab + b2
For any two numbers a and b, (a – b)2 = a2 – 2ab + b2
1. (2x – 5)(2x + 5)
2. (x + 7)2
3. (x – 2)2
4. (2x + 3y)2
1. (2x – 5)(2x + 5)
(2x – 5)(2x + 5)
22x2 – 52
4x2 – 25
2. (x + 7)2
(x + 7)2
x2 + 2(7)(x) + 72
x2 + 14x + 49
3. (x – 2)2
(x – 2)2
x2 – 2(2)(x) + 22
x2 + 4x + 4
4. (2x + 3y)2
(2x + 3y)2
22x2 – 2(2x)(3y) + 32y2
4x2 + 12x + 9y2