## About "How to factor polynomials with 4 terms without grouping"

How to factor polynomials with 4 terms without grouping :

Here we are going to see how to factor polynomials with 4 terms without grouping.

To factor polynomials with 4 terms without grouping, we use trial and error.

Trial and error means, we should apply the values like 1, -1, 2, -2, 3, -3,..........etc.

For example, if we get 0 as remainder by applying the value x = 1, we may decide that x - 1 is a factor.

Let us look into some example problems to understand the above concept.

## How to factor polynomials with 4 terms without grouping - Examples

Example 1 :

Factorize the following polynomial

x3 - 2x2 - x + 2

Solution :

Let p(x) =  x3 - 2x2 - x + 2

Since p(x) is the cubic polynomial, it will have three linear factors.

To find those factors, we use trial and error.

If we get 0 as remainder by applying the value of x = 1, we can decide that x - 1 is one of the factors.

Similarly by applying the values -1,-2,2..... we can get the other factors

x = 1

p(1)  =  13 - 2(1)2 - 1 + 2

=  1 - 2 - 1 + 2

=  0

So (x - 1) is a factor.

x = -1

p(-1) = (-1)3 - 2(-1)2 - (-1) + 2

=  -1 - 2(1) + 1 + 2

=  -1 - 2 + 1 + 2

=  0

So (x + 1) is a factor.

x = 2

p(2)  =  (2)3 - 2(2)2 - (2) + 2

=  8 - 2(4) - 2 + 2

=  8 - 8 - 2 + 2

=  0

So (x - 2) is a factor.

Hence the factors are (x - 1) (x + 1) (x - 2).

Example 2 :

Factorize the following polynomial

x3 + 3x2 - x - 3

Solution :

Let p(x) =  x3 + 3x2 - x - 3

Since p(x) is the cubic polynomial, it will have three linear factors.

To find those factors, we use trial and error.

If we get 0 as remainder by applying the value of x = 1, we can decide that x - 1 is one of the factors.

Similarly by applying the values -1,-2,2..... we can get the other factors

x = 1

p(1)  =  13 + 3(1)2 - 1 - 3

=  1 + 3 - 1 - 3

=  0

So (x - 1) is a factor.

x = -1

p(-1)  =  (-1)3 + 3(-1)2 - (-1) - 3

=  -1 + 3(1) + 1 - 3

=  -1 + 3 + 1 - 3

=  0

So (x + 1) is a factor.

x = 2

p(2)  =  (2)3 + 3(2)2 - 2 - 3

=  8 + 3(4) - 2 - 3

=  8 + 12 - 5

0

So (x - 2) is not a factor.

x = -2

p(-2)  =  (-2)3 + 3(-2)2 - (-2) - 3

=  -8 + 3(4) + 2 - 3

=  -8 + 12 + 2 - 3

0

So (x + 2) is not a factor.

x = 3

p(3)  =  (3)3 + 3(3)2 - (3) - 3

=  27 + 27 + 3 - 3

0

So (x - 3) is not a factor.

x = -3

p(-3)  =  (-3)3 + 3(-3)2 - (-3) - 3

=  -27 + 27 + 3 - 3

=  0

So (x + 3) is a factor.

Hence the factors are (x - 1) (x + 1) (x + 3).

Example 3 :

Factorize the following polynomial

x3 + x2 - 4x - 4

Solution :

Let p(x) =  x3 + x2 - 4x - 4

Since p(x) is the cubic polynomial, it will have three linear factors.

To find those factors, we use trial and error.

If we get 0 as remainder by applying the value of x = 1, we can decide that x - 1 is one of the factors.

Similarly by applying the values -1,-2,2..... we can get the other factors

x = 1

p(1)  =  13 + 12 - 4(1) - 4

=  1 + 1 - 4 - 4

0

So (x - 1) is not a factor.

x = -1

p(-1)  =  (-1)3 + (-1)2 - 4(-1) - 4

=  -1 + 1 + 4 - 4

=  0

So (x + 1) is a factor.

x = 2

p(2)  =  23 + 22 - 4(2) - 4

=  8 + 4 - 8 - 4

=  0

So (x - 2) is a factor.

x = -2

p(-2)  =  (-2)3 + (-2)2 - 4(-2) - 4

=  -8 + 4 + 8 - 4

= 0

So (x + 2) is a factor.

Hence the factors are (x + 1) (x - 2) (x + 2).

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