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.