Polynomial Expansion

Polynomial expansion refers to expressing a polynomial in the expanded form by multiplying its factors. It is commonly used in algebra to simplify algebraic expressions and also in the context of the Binomial Theorem.

For example, multiplying the polynomials (x + 2) and (x + 3), we get

(x + 2) and (x + 3) = x² + 5x + 6, which is in the expanded form

Thus, for expanding, we multiply its factors and combine the like terms, resulting in a new polynomial with more terms.

The process often involves using the distributive property of multiplication over addition, combining like terms, or applying algebraic identities.

The general expanded form of a polynomial is:

P(x) = a_nxⁿ + a_{n – 1}x^{n – 1} + … + a₁x + a₀ 

Here, 

• p(x) is the polynomial
• a_n, a_n-1, …, a₁ are coefficients of xⁿ, x^n-1, …, x 
• a_n (≠ 0) is the leading coefficient
• a₀ is constant and ∀ a_n ∈ ℝ

Polynomials can be expanded in the following ways:

Using Distributive Property

As we know, we multiply two or more factors for expanding the polynomials. When multiplying a polynomial by a monomial or another polynomial, we use the distributive property of multiplication over addition.

According to the distributive property: A(B + C) = AB + AC

For Monomials Multiplied by Polynomials

Let us expand 2xy(x⁵ + 4x) 

On multiplying the monomial with each term of the binomial using the distributive property, we get

= (2xy)(x⁵) + (2xy)(4x)

= 2x⁶y + 8x²y

For Polynomials Multiplied by Polynomials

Let us expand (x – 3)(x² – 2x + 1)

On multiplying each term of the first polynomial by the second polynomial, we get

= x(x² – 2x + 1) – 3(x² – 2x + 1)

Now, on multiplying the monomial with each term of the trinomial, we get

= (x)(x²) + (x)(-2x) + (x)(1) + (-3)(x²) + (-3)(-2x) + (-3)(1)

= x³ – 2x² + x – 3x² + 6x – 3

Combining the like terms,

= x³ + (-2x² – 3x²) + (x + 6x) – 3

= x³ – 5x² + 7x – 3

For Binomials Multiplied by Binomials (By the FOIL Method)

When expanding two binomials, we use a special form of the distributive property, commonly referred to as the FOIL method. This method helps remember the multiplication process easily.

The FOIL method stands for:

F ⇒ Firsts

O ⇒ Outers

I ⇒ Inners

L ⇒ Lasts

Let us expand the polynomials (2x + 5)(x – 1)

Using the FOIL method,

F ⇒ (2x)(x) = 2x² 

O ⇒ (2x)(-1) = -2x 

I ⇒ (5)(x) = 5x

L ⇒ (5)(-1) = -5

Thus, (2x + 5)(x – 1) = 2x² – 2x + 5x – 5

Now, combining like terms,

(2x + 5)(x – 1) = 2x² + (-2x + 5x) – 5 = 2x² + 3x – 5

Using the Binomial Theorem

For higher power of binomials, we use the Binomial Theorem to expand the expression.

For a binomial raised to a power n, the binomial theorem is mathematically written as: 

${\left( x+y\right) ^{n}=\sum ^{n}_{k=0}\begin{pmatrix} n \\ k \end{pmatrix}x^{n-k}y^{k}}$

Here,

• x and y are the terms in the binomial
• n is the exponent
• ${\begin{pmatrix} n \\ k \end{pmatrix}}$ is the binomial coefficient, representing the number of ways to choose k terms from n 

Let us expand (x + 2)³ using the binomial theorem.

(x + 2)³ = ${\begin{pmatrix} 3 \\ 0 \end{pmatrix}x^{3}2^{0}+\begin{pmatrix} 3 \\ 1 \end{pmatrix}x^{3}2^{1}+\begin{pmatrix} 3 \\ 2 \end{pmatrix}x^{1}2^{2}+\begin{pmatrix} 3 \\ 1 \end{pmatrix}x^{2}2^{3}}$

⇒ (x + 2)³ = 1 ⋅ x³ + 3 ⋅ x² ⋅ 2 + 3 ⋅ x ⋅ 4 + 1 ⋅ 8

⇒ (x + 2)³ = x³ + 6x² + 12x + 8

Thus, (x + 2)³ = x³ + 6x² + 12x + 8

Using the Pascal’s Triangle

We can also use Pascal’s triangle to expand expressions of the form (x + y)ⁿ, where each row of the triangle corresponds to the coefficients required for the expansion. 

Let us expand (x + 2)³.

First, we find the coefficients of the expansion using Pascal’s triangle.

For n = 3, the 3^rd-row coefficients of the triangle are 1, 3, 3, 1

Now, combining the coefficients with powers of x and 2, we get

(x + 2)³ = 1 ⋅ (x)³ + 3 ⋅ (x)² ⋅ (2) + 3 ⋅ (x) ⋅ (2)² + 1 ⋅ (2)³ 

⇒ (x + 2)³ = x³ + 6x² + 12x + 8

Thus, (x + 2)³ = x³ + 6x² + 12x + 8

Solved Examples