# Standard Identities of Binomials and Trinomials

We already know the basics of algebra, we know why we use algebra and what are the general terms one needs to know, in order to solve an algebraic equation. Next, we move to algebraic identities. Let us know the standard identities of a binomials and trinomials equations.

## What are Algebraic Identities?

Those equations of algebra which are true for every value of the variables present in the equation are as algebraic identities. The algebraic identities are also helpful in the factorization of polynomials. The utility factor in the computation of algebraic expression is found this way. You may have read some of them in previous grades, however, we will try and brush up the previous few lessons and then proceed with some examples of algebraic identities.

For Example:Â The identityÂ (x+y)2 = x2+2xy+y2Â will be the same for all values of x and y.

### Browse more Topics under Algebraic Expressions And Identities

Because an identity stays the same for every value of its variables, one can substitute the terms of one side of the equation, with the terms of the other side, as shown in the example above, where we replaced an instance of (x+y)2 with the instance of x2 + 2xy + y2 on the other side, and vice versa.

Using these binomials and trinomials identities cleverly can help find shortcuts to several algebraic problems, making it easier to manipulate algebra.

### Standard Algebraic Identities

The source of standard algebraic identities is the Binomial Theorem. The binomial theorem also known as binomial expansion is derived by expanding the powers of binomials or sums total of two terms. The coefficients used along with the terms for expanding are called binomial coefficients. The theorem and its generalizations are useful in proving theories, results and solving combinatorics problems, calculus, algebra, and several other mathematical problems.

The standard algebraic identities are:Â

• (a + b)2 = a2 + 2ab + b2
• (a â€“ b)2 = a2 â€“ 2ab + b2
• Â a2 â€“ b2 = (a + b)(a â€“ b)
• (x + a)(x + b) = x2 + (a + b) x + ab
• (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
• (a + b)3 = a3 + b3 + 3ab (a + b)
• (a â€“ b)3 = a3 â€“ b3 â€“ 3ab (a â€“ b)
• a3 + b3 + c3â€“ 3abc = (a + b + c)(a2 + b2 + c2 â€“ ab â€“ bc â€“ ca)

### Trinomial Identities

Apart from the algebraic identities mentioned above one more useful identity frequently used is the one given by the equation:

(x –Â  y) (x â€“ z) Â = x2 – (y + z) x + yz

Notice that, in the above given equation, on the right side, the sum total, say S = y+z and the product, sayÂ  P = yz of y and z are present. When the equation is expressed in a form x2 â€“ Sx + Â P, and Â S and P are representing a sum and product of two numericals or expressions, this expression is known as second degree trinomial which is to say x2 â€“ Sx + P Â Therefore, when we write the above given example in reverse it can be seen as factorization for second degree trinomials, that is to say:Â  x2 – Sx + P= (x â€“ y) (x – z),

Here, S= y + z and P = yz. The above noted factorization method is also called ViÃ¨teâ€™s formula, as a commemoration to FranÃ§ois Viete, the French mathematician. He was a sixteenth century mathematician who pioneered in using alphabets to represent variables. He is often called the father of algebra.

Example of Trinomials Identity :Â (a + b + c)2Â = a2Â + b2Â + c2Â + 2 Â· a Â· b + 2 Â· a Â· c + 2 Â· b Â· c

(x2Â âˆ’ x + 1)2Â = (x2)2Â + (âˆ’x)2Â + 12Â + 2Â Â·Â x2Â Â·Â (âˆ’x) + 2 x2Â Â·Â 1 + 2Â Â·Â (âˆ’x)Â Â·Â 1

= x4Â + x2Â + 1 âˆ’ 2x3Â + 2x2Â âˆ’ 2x

= x4âˆ’ 2x3Â + 3x2Â âˆ’ 2x + 1

## Solved Example For You

Q. Factorize: x2 + 9y2 – 25m2 – 16n2 + 16xy + 40mn

a.Â (2x+3y+5m+4m)(xâˆ’3yâˆ’2m+4n)Â  Â  Â  Â  Â  b.Â (2x+3y+4mâˆ’4m)(x+3yâˆ’5mâˆ’4n)

c.Â (x+3y+5mâˆ’4n)(x+3yâˆ’5m+4n)Â  Â  Â  Â  Â  Â  Â  d.Â (x+y+3mâˆ’4m)(x+3yâˆ’+5m+4n)

Ans:Â c.Â (x+3y+5mâˆ’4n)(x+3yâˆ’5m+4n)

x2 + 9y2 – 25m2 – 16n2 + 6xy + 40mn = x2 + 6xy + 9y2 -25m2 + 40mn – 16 n2Â

= ( x2 + 6xy + 9y2) – (25m2 – 40 mn + 16n2) = ( x + 3y)2 – (5m-4n)2Â

= [(x + 3y) + (5m-4n)] [(x + 3y) – (5m – 4n)] = (x + 3y + 5m – 4n) (x + 3y – 5m + 4n

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