# Factorization of Polynomials

## Factorization of Polynomials

For the factorization of polynomial, factor theorem and synthetic division are very useful to find the factors of the polynomial.

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Let us see the following example:

Example 1: Factorize: x3 – 4x2 + x + 6

Solution: Constant term of this polynomial is 6 and the possible factors of 6 are: ±1, ±2, ±3, ±6.

Let, f(x) = x3 – 4x2 + x + 6

Since the degree of f(x) is 3, so there will be at most three factors.

When x = 1, f(1) = 1 – 4 + 1 + 6 = 4

(x – 1) is not a factor.

When x = -1, f(-1) = -1 - 4 - 1 + 6 = 0

(x + 1) is a factor.

When x = 2, f(2) = 8 – 4.4 + 2 + 6 = 0

(x – 2) is a factor.

When x = -2, f(-2) = -8 – 4.4 - 2 + 6 = -20

(x + 2) is not a factor.

When x = 3, f(3) = 27 – 4.9 + 3 + 6 = 0

(x – 3) is a factor.

(x + 1), (x – 2) and (x – 3) are three factors.

x3 – 4x2 + x + 6 = (x + 1)(x – 2)(x – 3).

But, instead of finding all the factors by using factor theorem, the synthetic division can be used after getting one factor with the help of factor theorem.

Example 2: Factorize: x3 – 6x2 + 11x – 6

Solution: Let, f(x) = x3–6x2+11x–6

Here constant term of the polynomial f(x) is 6 and the possible factors of 6 are: ±1, ±2, ±3, ±6.

When, x = 1, f(1) = 1 – 6 + 11 – 6 = 0. So, (x–1) is a factor of f(x).

Now, we use synthetic division to find other factor.

Quotient = x2 – 5x + 6

So the remaining factors of x3 – 6x2 + 11x – 6 are the factors of x2 – 5x + 6.

Now,

x2 – 5x + 6

= x2 – 3x – 2x + 6

= x(x – 3) – 2(x – 3)

= (x – 3)(x – 2)

Hence, x3 – 6x2 + 11x – 6 = (x – 1)(x – 2)(x – 3).

Alternative Method,

Example 3: Factorize x3 – 4x2 + x + 6.

Solution: Here, constant term is 6. Its factors are: ±1, ±2, ±3, ±6.

Let, f(x) = x3 – 4x2 + x + 6

When, x = 1, f(1) = 1 – 4 + 1 + 6 = 4

(x – 1) is not a factor.

When, x = -1, f(-1) = -1 – 4 - 1 + 6 = 0

(x + 1) is a factor.

Now, splitting the second and third terms of the expression in such a way that each of the pairs of terms so formed has a factor (x + 1), we have

x3–4x2+x+6 = x3+x2–5x2–5x+6x+6

= x2(x+1)– 5x(x+1)+ 6(x+1)

= (x + 1)(x2–5x+6)

= (x + 1) [x2–3x–2x+6]

= (x + 1)[x(x – 3)-2(x – 3)]

= (x + 1)(x – 3)(x – 2)

The general rule of factor theorem leads to the following cases which are helpful to the factorization:

1.    If there is any polynomial containing integral powers of x, the sum of whose coefficients is zero, then (x – 1) is a factor of that polynomial.

2.    If there is any polynomial containing integral powers of x, the sum of the coefficients of odd powers of x is equal to the sum of the remaining coefficients, then (x + 1) is a factor of that polynomial.

Example 4: Factorize: x3 – 11x2 + 31x – 21

Solution: Here, sum of coefficients = 1 – 11 + 31 – 21 = 0. So, (x – 1) is a factor of the polynomial x3 – 11x2 + 31x – 21. Now, splitting the second and third terms of the expression in such a way that each of the pair of terms so formed has a factor (x – 1), we have

x3 – 11x2 + 31x – 21 = x3 – x2 – 10x2 + 10x + 21x – 21

= x2(x – 1) – 10x(x – 1) + 21(x – 1)

= (x – 1)(x2 – 10x + 21)

= (x – 1)(x2 – 7x – 3x + 21)

= (x – 1)[x(x – 7) – 3(x – 7)]

= (x – 1)(x – 7)(x – 3).

Example 5: Factorize: x3 – 4x2 + x + 6

Solution: Here,

Sum of the coefficients of odd powers of x = 1 + 1 = 2.

Sum of the coefficients of even powers of x = -4 + 6 = 2

So, (x+1) is a factor of x3–4x2+x+6.

Now,

x3–4x2+x+6 = x3+x2–5x2-5x+6x+6

= x2(x+1)– 5x(x+1)+6(x+1)

= (x + 1)(x2–5x+6)

= (x + 1)(x2–3x–2x+6)

= (x + 1)[x(x – 3)–2(x – 3)]

= (x + 1)(x – 3)(x – 2).

You can comment your questions or problems regarding the factorization of polynomials here.