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Binomial theorem

Binomial calculator raised to some exponent

The structure of the calculator is as follows:

$$\left( \text{Monomial 1} + \text{Monomial 2} \right)^{\text{Exponent}} = \text{Result}$$

  • You can enter decimal values, but it is best to only enter whole values
  • Do not enter letters or other characters. Except for the character “-” that indicates that an integer is negative
Binomial theorem program Monomial 1:

Monomial 2:

Exponent:

Answer:

Let’s go with the theory of the binomial theorem

The same binomial theorem is known as the binomial formula because, that is, a formula. The binomial theorem is only valid in terms of an integer and positive power of a binomial. Let’s see the first five values of the power:

$$
\begin{array}{c c l}
(x + y)^{1} & = & x + y \\
(x + y)^{2} & = & x^{2} + 2xy + y^{2} \\
(x + y)^{3} & = & x^{3} + 3x^{2}y + 3xy^{2} + y^{3} \\
(x + y)^{4} & = & x^{4} + 4x^{3}y + 6x^{2}y^{2} + 4xy^{3} + y^{4}\\
(x + y)^{5} & = & x^{5} + 5x^{4}y + 10x^{3}y^{2} + 10x^{2}y^{3} + 5xy^{4} + y^{5}
\end{array}
$$

There is a lot of theory that proves the binomial theorem, but we are going to go directly with the way in which you can write binomials with larger exponents, we need the pascal triangle and understand something very simple. The first thing we need to understand is that when we raise a binomial to a certain power, we will observe the following behavior:

$$(x+y)^{n} = x^{n} + a\ x^{n-1}y + b \ x^{n-2} y^{2}  + \dots  + b \ x^{2}y^{n-2} + a \ x y^{n-1} + y^{n}$$

Assuming that we do not know the coefficients that each monomial will have, what I want you to observe is that the exponent of the first term, $x$, starts with the value of the binomial power and is subtracted one by one until that its exponent reaches zero. The opposite occurs with the second term of the binomial, $y$, its exponent starts at zero (because a number raised to zero is equal to one) and the exponent is increased one by one until it reaches the same value as the exponent of the binomial.

Well, since we have very clear the aforementioned, let’s go to the pascal triangle, we are going to place the values that the coefficients of the development of a certain binomial would take depending on its exponent, let’s see:

$$
\begin{array}{c c c}
n = 0 & \quad & 1 \\
n = 1 & & 1 \quad 1\\
n = 2 & & 1 \quad 2 \quad 1 \\
n = 3 & & 1 \quad 3 \quad 3 \quad 1 \\
n = 4 & & 1 \quad 4 \quad 6 \quad 4 \quad 1 \\
n = 5 & & 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\
n = 6 & & 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1
\end{array}
$$

Great, once we have the pascal triangle, let’s go with our example, develop the following binomial:

$$(x+y)^{6}$$

To do this the first thing we are going to do is to place the ascending and descending order of the exponents and then we are going to add the coefficients that are obtained from the pascal triangle when $n = 6$:

$$
\begin{array}{c c c c c c c c c c c c c}
x^{6} & \ & x^{5} y & \ & x^{4} y^{2} & \ & x^{3} y^{3} & \ & x^{2} y^{4} & \ & x y^{5} & \ & y^{6} \\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\
1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\
x^{6} & \ & 6 x^{5} y & \ & 15 x^{4} y^{2} & \ & 20 x^{3} y^{3} & \ & 15 x^{2} y^{4} & \ & 6 x y^{5} & \ & y^{6}
\end{array}
$$
$$
\begin{array}{c}
\Downarrow \\
Finally \\
\Downarrow \\
x^{6} + 6 x^{5} y + 15 x^{4} y^{2} + 20 x^{3} y^{3} + 15 x^{2} y^{4} + 6 x y^{5} + y^{6}
\end{array}
$$

With that you can develop binomials with larger exponents!

What happens if a term in my binomial is negative?

Quiet, it is much easier than it sounds, what is going to be done is to encapsulate the negative in a parenthesis and we will place a positive sign. Let’s solve the following binomial:

$$(x-y)^{6}$$

It is easier to group the negative with a parenthesis and place a positive sign because the brain is easier. Let’s see how it looks:

$$(x+ (-y))^{6}$$

Great, our second term is now $(-y)$. It only remains to write how our developed binomial will be, so let’s write almost the same as we wrote above:

$$
\begin{array}{c c c c c c c c c c c c c}
x^{6} & \ & x^{5} (-y) & \ & x^{4} (-y)^{2} & \ & x^{3} (-y)^{3} & \ & x^{2} (-y)^{4} & \ & x (-y)^{5} & \ & (-y)^{6} \\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\
1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1 \\
\downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow & & \downarrow \\
x^{6} & \ & 6 x^{5} (-y) & \ & 15 x^{4} (-y)^{2} & \ & 20 x^{3} (-y)^{3} & \ & 15 x^{2} (-y)^{4} & \ & 6 x (-y)^{5} & \ & (-y)^{6}
\end{array}
$$
$$
\begin{array}{c}
\Downarrow \\
Almost \  finally \\
\Downarrow \\
x^{6} + 6 x^{5} (-y) + 15 x^{4} (-y)^{2} + 20 x^{3} (-y)^{3} + 15 x^{2} (-y)^{4} + 6 x (-y)^{5} + (-y)^{6}
\end{array}
$$

The only thing missing is to develop the parentheses of each $(-y)$. Remember that a negative raised to an even power will always be positive, and if it is raised to an odd power it will always be negative.

$$x^{6} – 6 x^{5} y + 15 x^{4} y^{2} – 20 x^{3} y^{3} + 15 x^{2} y^{4} – 6 x y^{5} + y^{6}$$

Now, you know how to develop binomials with very large exponents.

Before you leave for the time being

I leave you a PDF where you can find the binomials from $n = 0$ to $n = 20$ and you will also find the formula of the binomial theorem, just click here $\Rightarrow$ Binomial theorem and formulas

Thank you for being at this moment with us : )

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