Multiplying Two Binomials

Class work: Grade 10 FMP Unit 3 Skill 4

Class work solutions: Skill 4 Worksheet Solutions

When multiplying a binomial (an expression with two terms) by another binomial, we end up with four terms.

Oftentimes, some of the terms can be gathered and so your final expression might have three or even less terms.

The terms of a binomial can be anything – integers, fractions, x’s, y’s, abc’s, sines, cosines – any thing that has mathematical meaning or that you create to give mathematical meaning.

Example 1: Integers – a simpler way to multiply

Here’s an example where both terms in both binomials are just integers.

27 \times 34=(20+7)(30+4)

Let’s suppose the numbers are actually the length and width of a rectangle.

By letting 27=20+7 and 34=30+4 we create four simple calculations.

7 \times 30 = 210

7 \times 4 = 28

20\times 30=600

20\times 4 = 80

Therefore the total area of this rectangle, or the answer to 27 \times 34 is 210+28+600+80 which is 890+28=918.

In other words, (20+7)(30+4)=600+80+210+28. In the examples, we see how to go from the brackets to the four terms without drawing the rectangle.

Example 2: the terms are a, b, c and d

In general, to multiply a binomial by binomial we are multiplying

(a+b)(c+d)

Where the a, b, c and d can be any number or variable or mixture of numbers and variables.

Using a rectangle diagram, let the width be a+b units; the length be c+d units.

The total area of the rectangle is ac+ad+bc+bd.

Now, of course, since this diagram has been made on square paper, we can quantify a, b, c and d. Let’s count the squares: a=13; b=4; c=5; d=7.

In other words, the length is a+b=13+4=17; the width is c+d=5+7=12.

The area is 17\times 12 which is equal to 13\times5+13\times7+4\times 5+ 4\times 7 = 65+91+20+28=156+48=204.

The point of this example, is that whatever the values of a, b, c and d, the value of

(a+b)(c+d)

can be calculated as the sum of

ac+ad+bc+bd.

Example 3: The classic

Although the a, b, c and d can be any values, this course is geared towards studying quadratic equations which involve an x^2 term.

Most of the expressions from here on look like (x+4)(x+5). Here is a rectangle model of this expression:

Notice that (x+4)(x+5) can be calculated with four smaller calculations:

x^2+4x+5x+20.

We notice that 4x+5x=9x

Therefore,

(x+4)(x+5)=x^2+9x+20

Example 4: Without drawing the rectangle

Suppose we wish to expand and simplify:

(x+2)(x+6)

without drawing the rectangle.

From the rectangle diagrams, we notice that everything in the first bracket multiplies everything in the second bracket. We can write this as follows:

    \[(x+2)(x+6)\\=x(x+6)+2(x+6)\\=x^2+6x+2x+12\\=x^2+8x+12\]

Example 5: The algorithm FOIL: First Outside Inside Last

We multiply the first terms in each bracket to get x^2

We multiply the outside terms in each bracket to get 6x

We multiply the inside terms in each bracket to get 2x

We multiply the last terms in each bracket to get 12.

Then we add those terms together and simplify to x^2+8x+12.

Example 6: Watch out!

Remember that a square number means multiply the number by the number? So…

    \[(x-5)^2\]

means

    \[(x-5)(x-5)\]

The correct expansion of this example is x^2-5x-5x+25=x^2-10x+25.

Example 7: Extending the principle

In example 4 we said that ‘everything in the first bracket multiplies everything in the second.’

With a little rectangle drawing, we can see how this principle extends to any number of terms in any number of brackets. Let’s expand

(x+y+3)(x+2)

    \[(x+y+3)(x+2)\\=x(x+2)+y(x+2)+3(x+2)\\=x^2+2x+yx+2y+3x+6\\=x^2+5x+yx+2y+6\]

For three brackets, we simply multiply out the first two then we have two brackets. Etc.

    \[(x+2)(x+3)(x+6)\\=(x^2+5x+6)(x+6))\\=x^2(x+6)+5x(x+6)+6(x+6)\\=x^3+6x^2+5x^2+30x+6x+36\\=x^3+11x^2+36x+36\]

More here.

Practice (Skill #4)

You may need to use paper and pencil before entering your answer, particularly for the second exercise (green).

Example 8: Use more brackets if you need to

    \[-(x-10)(2x-1)\]

    \[=-[2x^2-21x+10]\]

    \[=-2x^2+21x-10\]


3. factors and products

Next: Factor Trinomial