FACTORING POLYNOMIALS - USING FACTOR THEOREM AND SYNTHETIC DIVISION, EXAMPLES

Please study
Algebra Factoring before Factoring Polynomials
if you have not already done so.

There, we explained the terms Factors or Divisors,
Factorization of Polynomials, Prime Polynomial etc.

That knowledge is a prerequisite here.

Here, we factorize higher degree ( > 2 ) polynomials using
Factor Theorem which is based on Remainder Theorem.

Remainder Theorem

If a Polynomial in x, say f(x) is divided by (x - a), then the remainder is f(a).

Proof :
We know Dividend = Divisor x Quotient + Remainder
Let q(x) be the Quotient and r be the Remainder when f(x) is divided by (x - a).
Then we have f(x) = (x - a) x q(x) + r where in the degree of q(x) is one less than that of f(x) and r is a constant.
Since the equation is true for all real values of x, putting x = a in the equation, we get
f(a) = (a - a) x q(a) + r = 0 + r = r ⇒ r = f(a).
Thus, when f(x) is divided by (x - a), the remainder is f(a). (Proved.)
Note:

If f(x) is divided by (ax + b), then the remainder is f(-b ⁄a)

Example on Remainder Theorem : Factoring Polynomials

Find the remainder when 3x³ - 2x² + x + 2 is divided by (i) x - 1 (ii) x + 2 (iii) 2x -1 (iv) 3x + 2

Solution:
Let f(x) = 3x³ - 2x² + x + 2
(i) when f(x) is divided by (x - 1),
The remainder = f(1) = 3(1)³ - 2(1)² + (1) + 2 = 3 - 2 +1 + 2 = 4. Ans.
(ii) when f(x) is divided by (x + 2),
The remainder = f(-2) = 3(-2)³ - 2(-2)² + (-2) + 2 = 3(-8) - 2(4) - 2 + 2 = -32. Ans.
(iii) when f(x) is divided by (2x -1),
The remainder = f(1⁄2) = 3(1⁄2)³ - 2(1⁄2)² + (1⁄2) + 2 = 3(1⁄8) - 2(1⁄4) + 1⁄2 + 2 = (3 - 2 x 2 + 1 x 4 + 2 x 8)⁄8 = 19⁄8 Ans.
(iv) when f(x) is divided by (3x + 2),
The remainder = f(-2⁄3) = 3(-2⁄3)³ - 2(-2⁄3)² + (-2⁄3) + 2 = 3(-8⁄27) - 2(4⁄9) - 2⁄3 + 2 = (3 x -8 - 2 x 4 x 3 - 2 x 9 + 2 x 27)⁄27 = -12⁄27 = -4⁄9 Ans.

An important application of Remainder Theorem is the
Factor Theorem which is applied in Factoring polynomials.

Factor Theorem : Factoring Polynomials

We know that, if the remainder is zero, when a Polynomial f(x) is divided by (x - a), then (x - a) is a factor of f(x).
But, as per remainder theorem, remainder = f(a).
∴ If f(a) = 0, then (x - a) is a factor of f(x).
Also If (x - a) is a factor of f(x), then f(a) = 0.
This is Factor theorem.

(x - a) is a factor of f(x) ⇔ f(a) = 0

Factor Theorem is very useful in Factoring
Polynomials of higher (than 2) degree .

Method of Factoring Polynomials using Factor Theorem :

The method of Factoring Polynomials
involves the following steps.

STEP 1: Let the Polynomial be f(x). Check whether (x- a) is a factor of f(x) or not. To check whether (x - a) is a factor of f(x) or not, we have to check whether f(a) is zero or not. Choosing values of a is done by trial and error. To facilitate, reducing the number of trials, we take the constant term of the Polynomial, f(x), find its factors, put + and - to the factors and take those values for a and check with them, if f(a) is zero or not. This is done till we find an a such that f(a) is zero.
STEP 2: After deciding (x - a) is a factor, we divide the Polynomial f(x) by (x - a) to get the Quotient. This division is better done by Synthetic Division which is explained in Solved Example 1 of Set 3 below.
STEP 3: we, then Factorize the Quotient for further factors. This Factorizing is done either by Factor theorem (Repeat from Step 1) or by other methods depending upon the degree of the Quotient.

The method of Factoring Polynomials
will be clear by the following set of Solved Examples.

Example 1 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize x³ + 5x² + 7x + 3

Solution to Example 1 of Factoring Polynomials

Let f(x) = x³ + 5x² + 7x + 3
Here constant term = 3. Its factors are +1, -1, +3, -3. So we check with a = +1, -1, +3, -3, whether f(a) is zero or not.
f(1) is obviously not zero, as all the coefficients are positive.
f(-1) = (-1)³ + 5(-1)² + 7(-1) + 3 = -1 + 5(1) -7 + 3 = 0
∴ x -(-1) = (x + 1) is a factor of f(x).
After one factor is arrived at, we can divide the given Polynomial with this factor, using the method explained in Polynomials.

However, we have a more convenient method called SYNTHETIC DIVISION which is explained below.
To divide x³ + 5x² + 7x + 3 by (x + 1)
Using Synthetic Division:

-1  |   +1    +5    +7    +3
    |   +0    -1    -4    -3  
    |-----------------------------
    |   +1    +4    +3    +0 

Note: You might have noted, steps 8,9,10 are repetition of steps 5, 6, 7.

Now, the third row numbers ( +1, +4, +3; last '0' is ignored.) form the coefficients of the Quotient in descending order starting with the power of the variable one less than the highest power of the Dividend.

∴ Here, the Quotient = x² + 4x + 3.
Thus, (x³ + 5x² + 7x + 3) ÷ (x + 1) = x² + 4x + 3
∴ x³ + 5x² + 7x + 3 = (x + 1)(x² + 4x + 3)
We can factorize x² + 4x + 3 using the knowledge of factoring Polynomials of second degree.
x² + 4x + 3 = x² + x + 3x + 3 = x(x + 1) + 3(x + 1) = (x + 1)(x + 3)
∴ x³ + 5x² + 7x + 3 = (x + 1)(x + 1)(x + 3)
= (x + 1)²(x + 3).
Thus Factoring Polynomial x³ + 5x² + 7x + 3 by using Factor Theorem gave the Factors as (x + 1)²(x + 3). Ans.

Example 2 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize 4x³ + 8x² - 6x - 12

Solution to Example 2 of Factoring Polynomials

Let f(x) = 4x³ + 8x² - 6x - 12
Here constant term = -12. Its factors are +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +12, -12. So we check with a = +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, whether f(a) is zero or not.
f(1) = 4 + 8 -6 -12 = -6 ≠ 0; f(-1) = 4(-1)³ + 8(-1)² -6(-1) -12 = -4 + 8 +6 -12 = -2 ≠ 0;
f(2) = 4(2)³ + 8(2)² - 6(2) - 12 ≠ 0;
f(-2) = 4(-2)³ + 8(-2)² - 6(-2) - 12 = -32 +32 +12 -12 = 0;
⇒ x -(-2) = (x + 2) is a factor of f(x).
Now let us divide f(x) = (4x³ + 8x² - 6x - 12) by (x + 2) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

-2  |   +4    +8    -6    -12
    |   +0    -8    +0    +12  
    |-----------------------------
    |   +4    +0    -6    +0 

∴ (4x³ + 8x² - 6x - 12) = (x + 2)(2x + √6)(2x - √6).
Thus Factoring Polynomial 4x³ + 8x² - 6x - 12 by using Factor Theorem gave the Factors as (x + 2)(2x + √6)(2x - √6). Ans.

Example 3 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize 2x³ - 7x² - 10x + 24

Solution to Example 3 of Factoring Polynomials

Let f(x) = 2x³ - 7x² - 10x + 24
Here constant term = +24. Its factors are +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +8, -8, +12, -12, +24, -24. So we check with a = +1,-1,+2, -2, +3, -3, +4, -4, +6, -6, +8, -8, +12, -12, +24, -24 whether f(a) is zero or not.
f(1) = 2 - 7 -10 +24 = +9 ≠ 0; f(-1) = 2(-1)³ - 7(-1)² -10(-1) -12 = -2 - 7 + 10 -12 = -11 ≠ 0;
f(2) = 2(2)³ - 7(2)² - 10(2) + 24 ≠ 0;
f(-2) = 2(-2)³ - 7(-2)² - 10(-2) + 24 = -16 -28 +20 + 24 = 0;
⇒ x -(-2) = (x + 2) is a factor of f(x).
Now let us divide f(x) = (2x³ - 7x² - 10x + 24) by (x + 2) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

-2  |   +2    -7    -10    +24
    |   +0    -4    +22    -24  
    |-----------------------------
    |   +2    -11   +12    +0 

∴ (2x³ - 7x² - 10x + 24) = (x + 2)(x - 4)(2x - 3).
Thus Factoring Polynomial 2x³ - 7x² - 10x + 24 by using Factor Theorem gave the Factors as (x + 2)(x - 4)(2x - 3). Ans.

Example 4 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize 2x³ - 11x² + 17x - 6

Solution to Example 4 of Factoring Polynomials

Let f(x) = 2x³ - 11x² + 17x - 6
Here constant term = -6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check with a = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not.
f(1) = 2 - 11 +17 -6 = +2 ≠ 0; f(-1) = 2(-1)³ - 11(-1)² +17(-1) -6 = -2 + 11 - 17 -6 = -14 ≠ 0;
f(2) = 2(2)³ - 11(2)² +17(2) - 6 = 16 - 44 + 34 - 6 = 0;
⇒ x -(2) = (x - 2) is a factor of f(x).
Now let us divide f(x) = (2x³ - 11x² + 17x - 6) by (x - 2) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

2  |   +2    -11    +17    -6
   |   +0    +4     -14    +6  
   |-----------------------------
   |   +2    -7     +3     +0 

∴ (2x³ - 11x² + 17x - 6) = (x - 2)(x - 3)(2x - 1).
Thus Factoring Polynomial 2x³ - 11x² + 17x - 6 by using Factor Theorem gave the Factors as (x - 2)(x - 3)(2x - 1). Ans.

Example 5 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize x⁴ + 5x³ + 5x² - 5x - 6

Solution to Example 5 of Factoring Polynomials

Let f(x) = x⁴ + 5x³ + 5x² - 5x - 6
Here constant term = -6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check with a = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not.
f(1) = 1 + 5 + 5 - 5 - 6 = 0
⇒ x -(1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (x⁴ + 5x³ + 5x² - 5x - 6 ) by (x - 1) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

1  |   +1    +5    +5    -5    -6
   |   +0    +1    +6    +11   +6  
   |------------------------------------
   |   +1    +6    +11   +6    +0 

Let p(x) = x³ + 6x² + 11x + 6 This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = 6. Its factors are +1,-1,+2, -2, +3, -3, +6, -6. So we check with a = +1,-1,+2, -2, +3, -3, +6, -6 whether f(a) is zero or not.
p(1) is obviously not zero, as all the coefficients are positive.
p(-1) = (-1)³ + 6(-1)² + 11(-1) + 6 = -1 + 6 -11 +6 = 0
⇒ x -(-1) = (x + 1) is a factor of p(x).
Now let us divide p(x) = (x³ + 6x² + 11x + 6) by (x + 1) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

-1  |   +1    +6    +11    +6
    |   +0    -1    -5     -6  
    |-----------------------------
    |   +1    +5    +6     +0 

Example 6 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize 4x⁴ - 12x³ + 7x² + 3x - 2

Solution to Example 6 of Factoring Polynomials

Let f(x) = 4x⁴ - 12x³ + 7x² + 3x - 2
Here constant term = -2. Its factors are +1,-1,+2, -2. So we check with a = +1,-1,+2, -2, whether f(a) is zero or not.
f(1) = 4 - 12 + 7 + 3 - 2 = 0 < BR> ⇒ x - (1) = (x - 1) is a factor of f(x).
Now let us divide f(x) = (4x⁴ - 12x³ + 7x² + 3x - 2) by (x - 1) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

1  |   +4    -12    +7    +3    -2
   |   +0    +4     -8    -1    +2  
   |------------------------------------
   |   +4    -8     -1    +2    +0 

Let p(x) = 4x³ - 8x² - x + 2 This is a third Degree Polynomial, which is to be factorized using Facor Theorem.
Here constant term = 2. Its factors are +1,-1,+2, -2. So we check with a = +1,-1,+2, -2, whether f(a) is zero or not.
p(1) = 4 - 8 -1 + 2 = -3 ≠ 0
p(-1) = 4(-1)³ - 8(-1)² -(-1) + 2 = -4 - 8 + 1 + 2 ≠ 0
p(2) = 4(2)³ - 8(2)² -(2) + 2 = 32 - 32 - 2 + 2 = 0.
⇒ x -(2) = (x - 2) is a factor of p(x).
Now let us divide p(x) = (4x³ - 8x² - x + 2) by (x - 2) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

2  |   +4    -8    -1    +2
   |   +0    +8    +0    -2  
   |-----------------------------
   |   +4    +0    -1    +0 

Example 7 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Factorize 3x⁴ - 10x³ + 5x² + 10x - 8

Solution to Example 7 of Factoring Polynomials

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 with a = +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) using Synthetic 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 

Let p(x) = 3x³ - 7x² - 2x + 8 This 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 with a = +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) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

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

Example 8 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

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

Solution to Example 8 of Factoring Polynomials

Let f(x) = x⁴ + 2x³ - 8x² + 30x - 25
Here constant term = -25. Its factors are +1,-1,+5, -5, +25, -25, So we check with a = +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) using Synthetic 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 

Let p(x) = x³ + 3x² - 5x + 25 This 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 with a = +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) using Synthetic Division (For detailed explanation of Synthetic Division see Example 1 above.)

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

Example 9 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

Find a and b in order that x³ - 6x² + ax + b is exactly divisible by x² - 3x + 2

Solution to Example 9 of Factoring Polynomials

Let f(x) = x³ - 6x² + ax + b We can factorize x² - 3x + 2 using the knowledge of factorization of Quadratic Polynomials.
x² - 3x + 2 = x² - x - 2x + 2 = x(x - 1) - 2(x - 1) = (x - 1)(x - 2)
f(x)[ = x³ - 6x² + ax + b] is exactly divisible by x² - 3x + 2
⇒ f(x) is exactly divisible by (x - 1)(x - 2)
⇒ (x - 1) and (x - 2) are factors of f(x)
⇒ f(1) = 0 and f(2) = 0
f(1) = 1 - 6 + a + b = a + b - 5 = 0 ⇒ a + b = 5...........(i)
f(2) = (2)³ - 6(2)² + a(2) + b = 8 - 24 + 2a + b = 2a + b - 16
f(2) = 0 ⇒ 2a + b - 16 = 0 ⇒ 2a + b = 16..........(ii)
(ii) - (i) gives (2a + b) - (a + b) = 16 - 5 ⇒ 2a + b - a - b = 11⇒ a = 11
Using this value of a in (i), we get
11 + b = 5 ⇒ b = 5 - 11 = -6
Thus a = 11, b = -6. Thus Factor Theorem in Factoring Polynomials helped us to find a and b, for x³ - 6x² + ax + b to be exactly divisible by x² - 3x + 2 as a = 11 and b = -6. Ans.

Example 10 of Factoring Polynomials

Solve the following problem on Factoring Polynomials

If a and b are unequal and x² + ax + b and x² + bx + a have a common factor, show that a + b + 1 = 0

Solution to Example 10 of Factoring Polynomials

Let f(x) = x² + ax + b; and p(x) = x² + bx + a
Let (x - k) be the common factor of f(x) and p(x). ⇒ f(k) = 0 and p(k) = 0
f(k) = 0 ⇒ k² + ak + b = 0............(i)
p(k) = 0 ⇒ k² + bk + a = 0............(ii)
(i) - (ii) gives (k² + ak + b) - (k² + bk + a) = 0 - 0 = 0
⇒ ak + b - bk - a = 0 ⇒ k(a - b) -1(a - b) = 0 ⇒ (a - b)(k- 1) = 0
⇒ ( (k- 1) = 0 [ Since by data a and b are unequal ⇒ (a - b) ≠ 0 ]
⇒ k = 1.
Substituting this value of k in (i), we get
(1)² + a(1) + b = 0 ⇒ a + b + 1 = 0 Thus, If a and b are unequal and x² + ax + b and x² + bx + a have a common factor, we could show that a + b + 1 = 0 using Factor Theorem in Factoring polynomials.

Exercise on Factoring Polynomials

Solve the Following Problems on Factoring polynomials :

1. Factorize x³ - 4x² + 5x - 2
2. Factorize x³ - x² - 5x - 3
3. Factorize x³ + 6x² + 11x + 6
4. Factorize 2x³ - x² - 15x + 18
5. Factorize x⁴ + 2x³ - 7x² - 8x + 12
6. Factorize x⁴ + 3x³ - 7x² - 27x - 18
7. Factorize 2x⁴ - 5x² + 5x - 2
8. Factorize x⁴ + 4x³ + 3x² - 4x - 4
9. Find l and m in order that x⁴ - x³ + lx² + mx + 4 is exactly divisible by x² - x - 2
10. If f(x) = x² + 5x + a and p(x) = x² + 3x + b have a common factor then (i) find the common factor (ii) show that (a - b)² = 2(3a - 5b)

For Answers See at the bottom of the Page.

Answers to Exercise on Factoring Polynomials

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

1. (x - 1)²(x - 2)
2. (x + 1)²(x - 3)
3. (x + 1)(x + 2)(x + 3}
   
4. (x - 2)(2x - 3)(x + 3}
5. (x - 1)(x - 2)(x + 2)(x + 3)
6. (x - 3)(x + 1)(x + 2)(x + 3)
   
7. (x - 1)(x + 2)(2x² - 2x + 1)
8. (x - 1)(x + 1)(x + 2)²
   
9. l = -4, m = 2
10. (i) x + (a - b)⁄2