The product of two binomial forms a trinomial.

Product of two binomials can be found using a method called **“FOIL”.**

F-front

O- outer

I-inner

L-lastConsider for example:

(2x+y) $\times$ (6y+x)

**F**-here the front terms of both the binomials are taken and multiplication is performed.

Here front terms are 2x and 6y.

Multiplying them=2x $\times$ 6y=12xy

**O**-here both the outer terms of the binomials are taken and multiplied.

Here outer terms are 2x and x.

Multiplying them=2x $\times$ x=2x^{2}

**I**-here both inner terms are taken and multiplied.

Here inner terms are y and 6y.

Multiplying them=y $\times$ 6y=6y^{2}

**L**-here the last terms of the binomials are taken and multiplied.

Here last terms are y and x.

Multiplying them=y $\times$ x=xy

So (2x+y) $\times$ (6y+x) =12xy+2x^{2}+6y^{2}+xy

Now like terms are combined so = 13xy+2x^{2}+6y^{2}

Example:

(y + 3)(y+4) = (y +…) (y +…) = y^{2}

= (y +…) (…+ 4) = 4y

= (…+ 3)(x +…) = 3y

= (…+ 3) (…+ 4) = 12

(y + 3)(y + 4) = y^{2}+4y+3y+12

= y^{2}+7y+12

When dealing with the negative terms in a binomial each step must be taken carefully.

Two methods can be employed for doing this.

**Method I**

Example: (x+5)(3x-2)

Here, first of all insert the parenthesis “[ ]”

So – [(x+5)(3x-2)]

Then use FOIL method

– [3x^{2}-2x+15x-10]

Simplify

- [3x^{2}+13x-10]

Remove [ ] and change the sign

-3x^{2}-13x+10

So, -(x+5) (3x-2) = -3x^{2}-13x+10

Method II

Example: -(x+5) (3x-2)

First distribute the sign for first binomial

(-x-5)(3x-2)

Expand using FOIL method

-3x^{2}+2x-15x+10

Simplify

-3x^{2}-13x+10