**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: x ^{3} – 4x^{2} + x + 6*

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

*Let, f(x) =
x ^{3} – 4x^{2} + 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.*

*∴** x ^{3} – 4x^{2} + 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: x ^{3}
– 6x^{2} + 11x – 6 *

*Solution:** Let, f(x) = x ^{3}–6x^{2}+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 = x ^{2} – 5x + 6*

* So the remaining factors of x ^{3}
– 6x^{2} + 11x – 6 are the factors of x^{2} – 5x + 6.*

* Now, *

*x ^{2} – 5x +
6*

* = x ^{2} – 3x – 2x + 6*

* = x(x – 3) – 2(x – 3)*

* = (x – 3)(x – 2)*

*Hence, x ^{3}
– 6x^{2} + 11x – 6 = (x – 1)(x – 2)(x – 3).*

Alternative Method,

*Example 3: Factorize x ^{3}
– 4x^{2} + x + 6.*

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

* Let, f(x) = x ^{3} – 4x^{2}
+ 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*

* x ^{3}–4x^{2}+x+6
= x^{3}+x^{2}–5x^{2}–5x+6x+6*

* = x ^{2}(x+1)– 5x(x+1)+ 6(x+1)*

* = (x + 1)(x ^{2}–5x+6)*

* = (x + 1) [x ^{2}–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: x ^{3}
– 11x^{2} + 31x – 21*

*Solution:** Here, sum of
coefficients = 1 – 11 + 31 – 21 = 0. So, (x – 1) is a factor of the polynomial
x ^{3} – 11x^{2} + 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*

* x ^{3} – 11x^{2} + 31x
– 21 = x^{3} – x^{2} – 10x^{2} + 10x + 21x – 21*

* = x ^{2}(x – 1) – 10x(x – 1) + 21(x –
1)*

* = (x – 1)(x ^{2} – 10x + 21)*

* = (x – 1)(x ^{2} – 7x – 3x + 21)*

* = (x – 1)[x(x – 7) – 3(x – 7)]*

* = (x – 1)(x – 7)(x – 3).*

*Example 5: Factorize: x ^{3}
– 4x^{2} + 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 x ^{3}–4x^{2}+x+6.*

*Now,*

* x ^{3}–4x^{2}+x+6
= x^{3}+x^{2}–5x^{2}-5x+6x+6*

* = x ^{2}(x+1)– 5x(x+1)+6(x+1)*

* = (x + 1)(x ^{2}–5x+6)*

* = (x + 1)(**x ^{2}*

*–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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