MULTIPLYING BINOMIALS USING THE "FOIL" METHOD

So FOIL stands for First, Outer, Inner, and Last respectively. This is how the structure of the operation goes when multiplying binomials. We will have a discussion on this. But first, let us have a short recall of the vocabulary that we will be using here.

Let's say we have a term x and y. The 'x' and 'y' are called terms. Based on a Google search "Term in math is defined as the values on which mathematical operations occur in an algebraic expression". Also, a Term in math is defined as the values on which mathematical operations occur in an algebraic expression. In our simple words here, it is simply a "letter(variable) or a number, may it be single or compound, sometimes with a positive or a negative number preceding before the letter, and called an expression". See the illustration below.

Illustration:

x, xy, 2xy, -3ab

It is said to be a monomial term or expression if there's no "+" or "-" symbol separating these terms. Otherwise, it can be called an equation. Now let's check the sequence of how to multiply binomials using the FOIL method.

Illustration:

Notice that "a and b", and "c and d" was separated by the "+" symbol. The a, b, c, and d are monomial. The "+" symbol separates them making binomials.

Photo source: study.com

The red indicates the first operation, and the result can be found on the right side of the photo. Followed by blue, yellow, and purple respectively. This is for us easy to visualize the operation as well.

Now Let's have a few examples.

Example 1

Multiply (x+1) and (x+2)

So by the FOIL method, we have

${\color{Red} F=x*x=x^2}$

${\color{Blue} O=x*2=2*x=2x}$

${\color{Orange} I=1*x=1x=x}$

${\color{Purple} L=1*2=2}$

Now, combining all results. We will simplify it by its order. The color was made to guide you in following the sequence order.

$(x+1)(x+2)={\color{Red} x^{2}}+{\color{Blue} 2x}+{\color{Orange} x}+{\color{Purple} 2}=\underline{x^2+3x+2}$

Example 2

Multiply (x-1) and (x-2). We just changed signs from a "+" into a "-", other elements are just the same.

So by the FOIL method, we have

${\color{Red} F=x*x=x^2}$

${\color{Blue} O=x*-2=-2*x=-2x}$

${\color{Orange} I=-1*x=-1x}$

${\color{Purple} L=-1*-2=2}$

Finalizing the result, we have

$(x-1)(x-2)={\color{Red} x^{2}}-{\color{Blue} 2x}-{\color{Orange} x}+{\color{Purple} 2}=\underline{x^2-3x+2}$

In the third example, we will be combining the "+ and -" signs. In this way, we can see how the result goes.

Example 3

Multiply (x+1) and (x-2)

${\color{Red} F=x^2}$

${\color{Blue} O=x*-2=-2*x=-2x}$

${\color{Orange} I=1*x=x}$

${\color{Purple} L=1*-2=-2}$

Finally,

$(x+1) (x-2)={\color{Red} x^2}-{\color{Blue} 2x}+{\color{Orange} x}-{\color{Purple} 2}=\underline{x^2-x-2}$

On Example 4,

Multiply (x-1) and (x+2)

${\color{Red} F=x*x=x^2}$

${\color{Blue} O=x*2=2*x=2x}$

${\color{Orange} I=-1*x=-x}$

${\color{Purple} L=-1*2=-2}$

For the final result

$(x-1)(x+2)={\color{Red} x^2}+{\color{Blue} 2x}-{\color{Orange} x}-{\color{Purple} 2}=\underline{x^2+x-2}$

Take note that the FOIL sequence didn't change but only the binomials.

Now let's have short exercises.

$1.)\ (x+2)(x+3)=$

Supplying the FOIL:

$F=x^2$

$O=3x$

$I=2x$

$L=6$

Therefore:

$=x^2+3x+2x+6$ , just gather and combine the results.

$=\underline{x^2+5x+6}$

$2.)\ (x-3)(x-3)=$

Supplying the FOIL:

$F=x^2$

$O=-3x$

$I=-3x$

$L=9$

Therefore:

$=x^2-3x-3x+9$ , just gather and combine the results.

$=\underline{x^2-6x+9}$

$3.)\ (a+5)(a-1)=$ we can use any variable so we use "a" this time. The FOIL method won't change.

Supplying the FOIL:

$F=a^{2}$

$O=-1*a=-a$

$I=5*a=5a$

$l=-1*5=-5$

Therefore:

$=a^2-1a+5a-5$

$=\underline{a^2+4a-5}$

$4.)\ (m-3)(m+4)=$ we can use any variable so we use "m" this time. The FOIL method won't change.

Supplying the FOIL:

$F=m^2$

$O=m*4=4m$

$I=-3*m=-3m$

$L=-3*4=-12$

Therefore:

$(m-3)(m+4)=m^2+4m-3m-12$ , just gather and combine the results.

$(m-3)(m+4)=\underline{m^2+m-12}$

$5.)\ (b+3)(b-7)=$

Supplying the FOIL:

$F=b*b=b^2$

$O=-7*b=-7b$

$I=3*b=3b$

$L=3*-7=-21$

Therefore:

$(b+3)(b-7)=b^2-7b+3b-21$ , just gather and combine the results.

$(b+3)(b-7)=\underline{b^2-4b-21}$

$6.)\ (2+c)(4+c)=$

Supplying the FOIL:

$F=2*4=8$

$O=2*c=2c$

$I=c*4=4c$

$L=c*c=c^2$

Therefore:

$(2+c)(4+c)=8+2c+4c+c^2$ , just gather and combine the results.

$(2+c)(4+c)=8+6c+c^2$ , rearrange this in descending order

$(2+c)(4+c)=\underline{c^2+6c+8}$

$7.)\ (x-7)(x-3)=$

Supplying the FOIL:

$F=x*x=x^2$

$O=x*-3=-3x$

$I=-7*x=-7x$

$L=-7*-3=21$

Therefore:

$(x-7)(x-3)=x^2-3x-7x+21$ , just gather and combine the results.

$(x-7)(x-3)=\underline{x^2-10x+21}$

$8.)\ (t-6)(t+4)=$

Supplying the FOIL:

$F=t*t=t^2$

$O=t*4=4t$

$I=-6*t=-6t$

$L=-6*4=-24$

Therefore:

$(t-6)(t+4)=t^2+4t-6t-24$ , just gather and combine the results.

$(t-6)(t+4)=\underline{t^2-2t-24}$

$9.)\ (u+6)(u-5)=$

Supplying the FOIL:

$F=u*u=u^2$

$O=u*-5=-5u$

$I=-6*u=-6u$

$L=6*-5=-30$

Therefore:

$(u+6)(u-5)=u^2-5u+6u-30$ , just gather and combine the results.

$(u+6)(u-5)=\underline{u^2+u-30}$

$10.)\ (ax+2)(x+3)=$

Supplying the FOIL:

$F=ax*x=ax^2$

$O=ax*3=3ax$

$I=2*x=2x$

$L=2*3=6$

Therefore:

$(ax+2)(x+3)=\underline{ax^2+3ax+2x+6}$

$11.)\ (abc-4)(cba+3)=$

Supplying the FOIL:

$F=a^2b^2c^2$

$O=3*abc=3abc$

$I=-4*abc=-4abc$

$L=-4*3=-12$

Therefore:

$(abc-4)(cba+3)=a^2b^2c^2+3abc-4abc-12$

$(abc-4)(cba+3)=\underline{a^2b^2c^2-abc-12}$

Now you are ready to take your own. We prepared an equation with some parts missing for you to fill. Just follow the FOIL method's sequence you know.

Practice Exercises:

Supply all the question marks in the given equations.

$1.)\ (y+3)(y+2)=y^2+2y+3y+{\color{DarkRed} ?}={\color{DarkRed} ?}$

$2.)\ (m+6)(m-3)={\color{DarkRed} ?}-3m+6m-18={\color{DarkRed} ?}$

$3.)\ (z-1)(z+9)=z^2+{\color{DarkRed} ?}-z-9={\color{DarkRed} ?}$

$4.)\ (mn-7)(m-2)={\color{DarkRed} ?}-2mn-7m+14={\color{DarkRed} ?}$

Short Quiz:

Multiply the given binomials by the FOIL method.

$1.)\ (y-2)(y+7)=$

$2.)\ (7-s)(s-7)=$

$3.)\ (t-6)(7+t)=$

$4.)\ (5y-2)(3y+2)=$

$5.)\ (2-5y)(2+3y)=$

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