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_________

+ 6 –2x+ 0–3xx2_________

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) = x2 – 5x + 6x(x) + x(–2) + (–3)(x) + (–3)(–2) =

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) =

Multiply using foil, then collect like terms.

x2 – 4x + 4x – 16 = x2 – 16

(x + 4)(x – 4) = x2 – 4x + 4x – 16 = x2 – 16

Here are more examples:

(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?

x2 – 16 x2 – 9 25 – y2

What do all of these have in common?

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) =

Now we FOIL and collect like terms.

x2 + 4x + 16 = + 4x x2 + 8x + 16

(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 16 = + 4x x2 + 8x + 16

Whenever we square a binomial like this, the same pattern always occurs.

See the first term?

In the final product it is squared…

…and it appears in the middle term.

(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 16 = + 4x x2 + 8x + 16

Whenever we square a binomial like this, the same pattern always occurs.

What about the second term?

…and the last term is 4 squared.

The middle number is 2 times 4…

(x + 4)2 = (x + 4)(x + 4) = x2 + 4x + 16 = + 4x x2 + 8x + 16

Whenever we square a binomial like this, the same pattern always occurs.

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 + 36 = – 6y 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 = Remember: (a – b)2 = a2 – 2ab + b2(z – 4)(z – 4)

= 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)2 = a2 – 2ab + b2

For two numbers a and b, (a + b)2 = a2 + 2ab + b2

For any two numbers a and b, (a + b)(a – b) = a2 – 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

(x + 7)2

x2 + 2(7)(x) + 72

x2 + 14x + 49

2. (x + 7)2

(x – 2)2

x2 – 2(2)(x) + 22

x2 + 4x + 4

3. (x – 2)2

(2x + 3y)2

22x2 – 2(2x)(3y) + 32y2

4x2 + 12x + 9y2

4. (2x + 3y)2