# POLYNOMIAL FACTORING - USING FACTOR THEOREM AND SYNTHETIC DIVISION, EXAMPLES

Method of Polynomial Factoring
if you have not already done so.

Remainder Theorem, Factor Theorem
and explained the Method of factoring
with Example.

Here, we will apply the method to solve problems.
The knowledge of Synthetic Division is made use of here.
That knowledge is a prerequisite here.
So, please learn the method before proceeding further.

Now, Problems on Polynomial Factoring follow.

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## Example 1 of Polynomial Factoring

Solve the following problem on Polynomial Factoring

Factorize 3x4 - 10x3 + 5x2 + 10x - 8

Solution to Example 1 of Polynomial Factoring

Let f(x) = 3x4 - 10x3 + 5x2 + 10x - 8
Here constant term = -8. Its factors are +1,-1,+2, -2, +4, -4, +8, -8 So we check witha = +1,-1,+2, -2, +4, -4, +8, -8, whether f(a) is zero or not.
f(1) = 3 - 10 + 5 + 10 - 8 = 0 < BR> ⇒ x - (1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (3x4 - 10x3 + 5x2 + 10x - 8) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

```1  |   +3    -10    +5    +10    -8
|   +0    +3     -7    -2     +8
|------------------------------------
|   +3    -7     -2    +8     +0
```

∴ Quotient = 3x3 - 7x2 - 2x + 8
Thus, (3x4 - 10x3 + 5x2 + 10x - 8) ÷ (x - 1) = 3x3 - 7x2 - 2x + 8
∴ f(x) = (x - 1)(3x3 - 7x2 - 2x + 8)

Let p(x) = 3x3 - 7x2 - 2x + 8This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = -8. Its factors are +1,-1,+2, -2, +4, -4, +8, -8 So we check witha = +1,-1,+2, -2, +4, -4, +8, -8, whether f(a) is zero or not.
p(1) = 3 - 7 -2 + 8 = 2 ≠ 0
p(-1) = 3(-1)3 - 7(-1)2 -2(-1) + 8 = -3 - 7 + 2 + 8 = 0.
x -(-1) = (x + 1) is a factor of p(x).
Now let us divide p(x) = (3x3 - 7x2 - 2x + 8) by (x + 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

```-1  |   +3    -7    -2    +8
|   +0    -3    +10   -8
|-----------------------------
|   +3    -10   +8    +0
```

∴ Quotient = (+3)x2 + (-10)x + (8) = 3x2 - 10x + 8
Thus, p(x) ÷ (x + 1) = 3x2 - 10x + 8
∴ p(x) = (x + 1)(3x2 - 10x + 8)
We can factorize 3x2 - 10x + 8 using the knowledge of factorization of Quadratic Polynomials.
3x2 - 10x + 8 = 3x2 - 6x - 4x + 8= 3x(x - 2) - 4(x - 2) = (x - 2)(3x - 4)
∴ p(x) = (x + 1)(x - 2)(3x - 4)
∴ f(x) = (x - 1)p(x) = (x - 1)(x + 1)(x - 2)(3x - 4).
Thus, 3x4 - 10x3 + 5x2 + 10x - 8
= (x - 1)(x + 1)(x - 2)(3x - 4).
Thus Polynomial Factoring of
3x4 - 10x3 + 5x2 + 10x - 8
by using Factor Theorem gave the Factors as
(x - 1)(x + 1)(x - 2)(3x - 4). Ans.

Thus, Example 1 of Polynomial Factoring is solved.

## Example 2 of Polynomial Factoring

Solve the following problem on Polynomial Factoring

Factorize x4 + 2x3 - 8x2 + 30x - 25 Ans. (x - 1)(x + 5)(x2 - 2x + 5)

Solution to Example 2 of Polynomial Factoring

Let f(x) = x4 + 2x3 - 8x2 + 30x - 25
Here constant term = -25. Its factors are +1,-1,+5, -5, +25, -25, So we check witha = +1,-1,+5, -5, +25, -25 whether f(a) is zero or not.
f(1) = 1 + 2 - 8 + 30 - 25 = 0 < BR> ⇒ x - (1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (x4 + 2x3 - 8x2 + 30x - 25) by (x - 1) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

```1  |   +1    +2    -8    +30    -25
|   +0    +1    +3    -5     +25
|------------------------------------
|   +1    +3    -5    +25    +0
```

∴ Quotient = x3 + 3x2 - 5x + 25
Thus, (x4 + 2x3 - 8x2 + 30x - 25) ÷ (x - 1) = x3 + 3x2 - 5x + 25
∴ f(x) = (x - 1)(x3 + 3x2 - 5x + 25)

Let p(x) = x3 + 3x2 - 5x + 25This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = +25. Its factors are +1,-1,+5, -5. So we check witha = +1,-1,+5, -5, whether f(a) is zero or not.
p(1) = 1 + 3 -5 + 25 = 24 ≠ 0
p(-1) = (-1)3 + 3(-1)2 -5(-1) + 25 = 32 ≠ 0.
p(5) = 53 + 3(5)2 - 5(5) + 25 ≠ 0 p(-5) = (-5)3 + 3(-5)2 - 5(-5) + 25 = -125 + 75 + 25 +25 = 0
x -(-5) = (x + 5) is a factor of p(x).
Now let us divide p(x) = (x3 + 3x2 - 5x + 25) by (x + 5) usingSynthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

```-5  |   +1    +3    -5    +25
|   +0    -5    +10   -25
|-----------------------------
|   +1    -2    +5    +0
```

∴ Quotient = (+1)x2 + (-2)x + (5) = x2 - 2x + 5
Thus, p(x) ÷ (x + 5) = x2 - 2x + 5
∴ p(x) = (x + 5)(x2 - 2x + 5)
x2 - 2x + 5 has no real factors, since 5 has no factors whose sum is -2
∴ p(x) = (x + 5)(x2 - 2x + 5)
∴ f(x) = (x - 1)p(x) = (x - 1)((x + 5)(x2 - 2x + 5).
Thus, x4 + 2x3 - 8x2 + 30x - 25
= (x - 1)((x + 5)(x2 - 2x + 5).
Thus Factoring Polynomial x4 + 2x3 - 8x2 + 30x - 25 by using Factor Theorem gave the Factors as(x - 1)((x + 5)(x2 - 2x + 5). Ans.

Thus, Example 2 of Polynomial Factoring is solved.

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## Exercise on Polynomial Factoring

Solve the Following Problems on Polynomial Factoring :

1. Factorize 2x4 - 5x2 + 5x - 2
2. Factorize x4 + 4x3 + 3x2 - 4x - 4

For Answers of these problems on Polynomial Factoring
See at the bottom of the Page.

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## Answers to Exercise on Polynomial Factoring

Answers to the problems in Exercise
on Polynomial Factoring are given below.

1. (x - 1)(x + 2)(2x2 - 2x + 1)
2. (x - 1)(x + 1)(x + 2)2