Factorization of Polynomials

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.

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.     

Synthetic division

          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.


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