Question

Note: This is NOT a question. This is a tutorial.

How to square a binomial and a polynomial? See comments below to see how!

Answer 1

how many of this have you done?

Answer 2

SQUARE OF A BINOMIAL

A square of a binomial is characterized by the form \((a+b)^2\). This can be expanded by

multiplying \((a+b)\) by itself. So, \((a+b)^2 = (a+b)(a+b)\). If we use the distributive

property here, we'll have \((a)(a) + (a)(b) + (b)(a) + (b)(b)\). If we

simplify it further, we'll have \((a)(a)+(a)(b)+(b)(a)+(b)(b)]=a^2+ab+ab+b^2\). If we combine the like terms, we'll have \(a^2+ab+ab+b^2=a^2+2ab+b^2\). Notice anything? The first term of the expanded form is also the square of the first term of the binomial we expanded. The last term of the expanded form is the square of the last term of the binomial we expanded. But what about the middle term of the expanded form? 2ab. Does it look familiar? 2ab is the product of the two terms in the binomial we expanded multiplied by 2. This is the shortcut to the square of a binomial. For example, we have \((2x+1)^2\). From what we learned in the above paragraph, the expanded form of this is...the square of the first term plus the square of the last term plus the product of the terms in the binomial multiplied by 2. What is the square of the first term? The first term is 2x so the square is \(4x^2\). What about the square of the last term? The last term is 1 so the square is simply 1 again. Now, the product of the two terms is \(2x\times 1 = 2x\). Now multiply it by 2 so we have \(2x \times 2 = 4x\). And so, the expanded form of \((2x+1)^2 = 4x^2 + 4x + 1\).



SQUARE OF A POLYNOMIAL

Much like the square of the binomial, there is also a shortcut for finding the square of

any polynomial. For example, we have \((a+b+c)^2\). If we expand this, we have \((a

+ b + c)^2 = (a + b + c)(a + b + c)\). By distributive property, we'll have \((a)(a)+(a)(b)+(a)(c)+(b)(a)+(b)(b)+(b)(c)+(c)(a)+(c)(b)+(c)(c)\). Simplifying further, we'll have \(a^2 + ab + ac + ab + b^2 + bc + ac + bc + c^2\). If we combine like terms, we'll have \(a^2 + 2ab + 2ac + b^2 + 2bc + c^2\). Notice anything? The terms of the expanded form consist of the square of the first term of the polynomial we expanded, the square of the second term of the polynomial we expanded, the third term of the polynomial we expanded, the product of the first term and the second term of the polynomial we expanded multiplied by 2, the product of the first term and the last term of the polynomial we expanded multiplied by 2, and the product of the second term and the last term of the polynomial we expanded multiplied by 2. This is true for all squares of polynomials.

For example, we have \((x^2 - 2x + 1)^2\). First we add the squares of each term. The square of the first term is \((x^2)^2 = x^4\). The square of the second term is \((-2x)^2 = 4x^2\). Note: be careful in the squaring of negatives and of variables with exponents. The square of the last term is \(1^2 = 1\). Now, we get the product of the first term and the second term. \((x^2)(-2x) = -2x^3\). Multiply it by 2 so we have \((-2x^3)(2) = -4x^3\). Now, we get the product of the first term and the last term. \((x^2)(1) = x^2\). Multiplly it by 2. \((x^2)(2) = 2x^2\). Lastly, we get the product of the second term and the last term. \((-2x)(1) = -2x\). Multiply it by 2. \((-2x)(2) = -4x\). Now, we combine all of these. So, \((x^2 - 2x + 1)^2 = x^4 + 4x^2 + 1 - 4x^3 + 2x^2 - 4x\). Notice how we have common terms? We add them so we'll have \(x^4 + 4x^2 + 1 - 4x^3 + 2x^2 - 4x = x^4 - 4x^3 + 6x^2 - 4x + 1\). Note that we must arrange them by descending order of exponents. Want to check? We'll use the distributive property to check.

\((x^2 - 2x + 1)^2 = (x^2 - 2x + 1)(x^2 - 2x + 1)\)

\((x^2 - 2x + 1)(x^2 - 2x + 1)\)

\((x^2)(x^2)+(x^2)(-2x)+(x^2)(1)+(-2x)(x^2)+(-2x)(-2x)+(-2x)(1)+\)
\((1)(x^2)+(1)(-2x)+(1)(1)\)

\(x^4 - 2x^3 + x^2 -2x^3 + 4x^2 - 2x + x^2 - 2x + 1\)

\(x^4 - 4x^3 + 6x^2 - 4x +1\)

is this the answer we got? \(\checkmark\)

Answer 3

@imranmeah91 i dont have stocks of them...i do them impromptu :P

Answer 4

@Mimi_x3 your request

Answer 5

An alternate method which is funner. :P
\[T_k = \left(\begin{matrix}n \\ k\end{matrix}\right)a ^{n-k}b ^{k}\]
\[(a+b)^2 => \sum_{k=0}^{2}\left(\begin{matrix}2 \\ k\end{matrix}\right)*(a)^{2-k}*(b)^{k}\]
Assuming you are able to read sigma notations..so its \[a^2+2ab+b^2\]

Answer 6

right..fun :P

Answer 7

Isn't it?
want a funner example? :P

Answer 8

This is awesome!!!