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Notable products

Note: It is advisable to apply the parentheses technique for any notable product that has many unknowns, exponents and/or some negative sign in a single term. At the end of the post is this technique.

What is a notable product?

A notable product is the multiplication between 2 or more binomials resulting in a binomial, a trinomial, or a polynomial. Next we will look at the different expressions of notable products.

We will also explain the development that leads to these formulas, so you can see that all these notable product results are only a resolution of multiplication, addition and/or subtraction.

Binomial squared or square of a binomial

This notable product is the sum of two terms squared and is equal to:

  • The square of the first term, plus
  • twice the product of the first term by the second term, plus
  • the square of the second term.

In more visual terms, it goes like this:

\left( x + y \right)^{2} = x^{2} + 2xy + y^{2}

\left( x - y \right)^{2} = x^{2} - 2xy + y^{2}

Let’s break down the behavior of the binomial squared

We will multiply term by term so that we can observe how the behavior is, remember the parenthesis technique if you think it is necessary:

(x + y)^{2}= (x + y)(x + y)

Now we take each term and we will multiply it step by step, we start with the first term x and we will multiply it by the x of the second parenthesis, then we will multiply it by the y of the second parenthesis. Then we will take the y from the first parenthesis and multiply it first by the x of the second parenthesis and then by the y of the second parenthesis:

x \cdot x + x \cdot y + y\cdot x + y \cdot y

We carry out the corresponding multiplications:

x^{2} +xy + yx + y^{2}

Finally we do a sum to obtain our final result, as we can see, the xy and the yx are added:

x^{2} + 2xy + y^{2}

Examples of the binomial squared

Example 1. \left( 3 + y \right)^{2}

We perform the behavior of the binomial squared

\left( 3 + y \right)^{2} = 3^{2} + 2 \cdot 3 \cdot y + y^{2}

Finally we carry out the corresponding operations to arrive at our final result:

= 3^{2} + 6y + y^{2}

Example 2. \left( 3x^{2} + 17\right)^{2}

For this case we will use the parenthesis technique:

\left( \left(3x^{2} \right) + 17\right)^{2}

Now we easily perform the behavior of the binomial squared:

= \left( 3x^{2} \right)^{2} + 2 \cdot \left( 3x^{2} \right) \cdot \left( 17 \right) + 17^{2}

We carry out the corresponding operations to obtain our final result:

9x^{4} + 102x^{2} + 289

Conjugate binomial

The conjugate binomial is the product of the sum of two terms multiplied by the subtraction of these terms, which is equal to:

  1. square of the first term minus
  2. the square of the second term.

Visualmente es así:

(x + y) (x - y) = x^{2} - y^{2}

Let’s break down the behavior of the conjugate binomial

We will multiply term by term so that we can observe how the behavior is, remember the parenthesis technique if you think it is necessary, in this case we will use the parenthesis technique:

(x + y)(x - y) = (x + y)(x + (-y))

Now we express each multiplication of the binomials:

x \cdot x + x \cdot(-y) + x\cdot y + y\cdot(-y)

We carry out the corresponding multiplications:

x^{2} -xy + xy - y^{2}

Finally we perform a subtraction to obtain our final result, as we can see, the -xy and the xy are canceled:

x^{2} - y^{2}

Examples of a conjugated binomial

Example 1. \left( x + 4 \right) \left(x - 4 \right)

\left( x + 4 \right) \left( x - 4 \right) = x^{2} - \left( 4 \right)^{2}

= x^{2} - 16

Example 2. \left( y^{3} + 13\right) \left(y^{3} - 13 \right)

\left( y^{3} + 13\right) \left(y^{3} - 13 \right) = \left( y^{3}\right)^{2} - \left( 13 \right)^{2}

= y^{6} - 169

Binomial with a common term

The product of two binomials where there is a similar term is equal to:

  1. Square of the first term plus
  2. the sum of the different terms multiplied by the similar term plus
  3. the product of the different terms.

Visually it is like this:

\left( x + a \right) \left( x + b \right) = x^{2} + \left( a + b \right) x + a \cdot b

Let’s break down the binomial with a common term

Let’s use the previous expression to demonstrate the breakdown of the binomial with a common term:

(x + a) (x +b)

We will express each multiplication of the expression:

(x + a)(x + b) = x\cdot x + x \cdot b + a \cdot x + a \cdot b

Now we do each multiplication:

x^{2} + xb + ax + ab

As we can see, we can express the xb + ax as x (a + b) to obtain our final result:

x^{2}  + x(a + b) + ab

And we didn’t say it but what we did in the end was factoring.

Example of binomial with a common term

Example 1. (x + 3)(x + 12)

\left( x + 3 \right) \left( x + 12 \right) = x^{2} + \left(3 + 12 \right) x + 3 \cdot 12

= x^{2} + 15x + 36

Example 2. (x^{4} + 7)(x^{4} - 16)

Remember the parenthesis technique:

(x^{4} + 7)(x^{4} - 16) \Rightarrow (x^{4} + 7)(x^{4} + (-16))

Now we can proceed to carry out the behavior of the binomial with a common term:

=\left( x^{4} \right)^{2} + x^{4} \left( 7 + (-16) \right) + (7)(-16)

We carry out the respective operations:

=x^{8} + x^{4}\left( 7- 16\right) + (-112)

=x^{8} + x^{4}\left(-9\right) - 112

Now we will get our final result:

x^{8} - 9x^{4} - 112

Parenthesis technique

When there are several unknowns in a term

When we have terms that have several unknowns or even have exponents, it is best to express that term by enclosing it in parentheses and carry out the operations with said parentheses, let’s see a relatively complex example:

\left( 9x^{2}yz^{4} + 4ay^{5} \right)^{2} \Rightarrow \left( \left( 9x^{2}yz^{4}\right) + \left( 4ay^{5}\right) \right)^{2}

Simply what we have to do is solve using the parentheses:

\left( \left( 9x^{2}yz^{4}\right) + \left( 4ay^{5}\right) \right)^{2} = \left( 9x^{2}yz^{4}\right)^{2} + 2 \cdot \left( 9x^{2}yz^{4} \right) \cdot \left( 4ay^{5} \right) + \left( 4ay^{5}\right)^{2}

Now we will carry out the corresponding multiplications to obtain our final result:

\left( 9x^{2}yz^{4} + 4ay^{5} \right)^{2} = 81x^{4}y^{2}z^{8} + 72ax^{2}y^{6}z^{4} + 16a^{2}y^{10}

When you don’t know what to do with the negative

Several times it turns out that negative signs are present in some term of the binomial and sometimes it is confusing when performing the corresponding operations. So what we have to do is isolate that term with a parenthesis.

But before continuing, I want to mention that subtractions can still be represented as additions, I want you to take this into account. Now, let’s use the previous example and change the sign:

\left( 9x^{2}yz^{4} - 4ay^{5} \right)^{2} \Rightarrow \left( \left( 9x^{2}yz^{4}\right) + \left(-4ay^{5}\right) \right)^{2}

We are going to solve as we solved in the previous example, only we will keep the negative inside the parentheses of the term \left (4ay ^ {5} \right) :

\left( \left( 9x^{2}yz^{4}\right) + \left(- 4ay^{5}\right) \right)^{2} = \left( 9x^{2}yz^{4}\right)^{2} + 2 \cdot \left( 9x^{2}yz^{4} \right) \cdot \left(- 4ay^{5} \right) + \left(- 4ay^{5}\right)^{2}

Once the corresponding multiplications have been carried out, we will have the following result:

\left( 9x^{2}yz^{4} + 4ay^{5} \right)^{2} = 81x^{4}y^{2}z^{8} - 72ax^{2}y^{6}z^{4} + 16a^{2}y^{10}

Other notable products

With the techniques that were explained above you can solve any binomial that comes your way, but in any case we will attach some more notable products:

  1. \left( ax + b \right)\left( cx + d \right) = acx^{2} + \left( ad + bc \right)x + bd
  2. \left( x + y \right)^{3} = x^{3} + 3x^{2}y + 3xy^{2} + b^{3}
  3. \left( x - y \right)^{3} = x^{3} - 3x^{2}y + 3xy^{2} - y^{3}
  4. \left( x + y \right)\left( x^{2} - xy + y^{2} \right) = x^{3} + y^{3}
  5. \left( x - y \right)\left( x^{2} + xy + y^{2} \right) = x^{3} - y^{3}

Remember to use the parenthesis technique when necessary!

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