Examples

Find the Factors Using the Factor Theorem
x4-2x3-10x2+7x+4x42x310x2+7x+4 , x-4x4
Step 1
Divide x4-2x3-10x2+7x+4x-4x42x310x2+7x+4x4 using synthetic division and check if the remainder is equal to 00. If the remainder is equal to 00, it means that x-4x4 is a factor for x4-2x3-10x2+7x+4x42x310x2+7x+4. If the remainder is not equal to 00, it means that x-4x4 is not a factor for x4-2x3-10x2+7x+4x42x310x2+7x+4.
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Step 1.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
4411-22-10107744
  
Step 1.2
The first number in the dividend (1)(1) is put into the first position of the result area (below the horizontal line).
4411-22-10107744
  
11
Step 1.3
Multiply the newest entry in the result (1)(1) by the divisor (4)(4) and place the result of (4)(4) under the next term in the dividend (-2)(2).
4411-22-10107744
 44 
11
Step 1.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
4411-22-10107744
 44 
1122
Step 1.5
Multiply the newest entry in the result (2)(2) by the divisor (4)(4) and place the result of (8)(8) under the next term in the dividend (-10)(10).
4411-22-10107744
 4488 
1122
Step 1.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
4411-22-10107744
 4488 
1122-22
Step 1.7
Multiply the newest entry in the result (-2)(2) by the divisor (4)(4) and place the result of (-8)(8) under the next term in the dividend (7)(7).
4411-22-10107744
 4488-88 
1122-22
Step 1.8
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
4411-22-10107744
 4488-88 
1122-22-11
Step 1.9
Multiply the newest entry in the result (-1)(1) by the divisor (4)(4) and place the result of (-4)(4) under the next term in the dividend (4)(4).
4411-22-10107744
 4488-88-44
1122-22-11
Step 1.10
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
4411-22-10107744
 4488-88-44
1122-22-1100
Step 1.11
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
1x3+2x2+(-2)x-11x3+2x2+(2)x1
Step 1.12
Simplify the quotient polynomial.
x3+2x2-2x-1x3+2x22x1
x3+2x2-2x-1x3+2x22x1
Step 2
The remainder from dividing x4-2x3-10x2+7x+4x-4x42x310x2+7x+4x4 is 00, which means that x-4x4 is a factor for x4-2x3-10x2+7x+4x42x310x2+7x+4.
x-4x4 is a factor for x4-2x3-10x2+7x+4x42x310x2+7x+4
Step 3
Find all the possible roots for x3+2x2-2x-1x3+2x22x1.
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Step 3.1
If a polynomial function has integer coefficients, then every rational zero will have the form pqpq where pp is a factor of the constant and qq is a factor of the leading coefficient.
p=±1p=±1
q=±1q=±1
Step 3.2
Find every combination of ±pq±pq. These are the possible roots of the polynomial function.
±1±1
±1±1
Step 4
Set up the next division to determine if x-1x1 is a factor of the polynomial x3+2x2-2x-1x3+2x22x1.
x3+2x2-2x-1x-1x3+2x22x1x1
Step 5
Divide the expression using synthetic division to determine if it is a factor of the polynomial. Since x-1x1 divides evenly into x3+2x2-2x-1x3+2x22x1, x-1x1 is a factor of the polynomial and there is a remaining polynomial of x2+3x+1x2+3x+1.
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Step 5.1
Place the numbers representing the divisor and the dividend into a division-like configuration.
111122-22-11
  
Step 5.2
The first number in the dividend (1)(1) is put into the first position of the result area (below the horizontal line).
111122-22-11
  
11
Step 5.3
Multiply the newest entry in the result (1)(1) by the divisor (1)(1) and place the result of (1)(1) under the next term in the dividend (2)(2).
111122-22-11
 11 
11
Step 5.4
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
111122-22-11
 11 
1133
Step 5.5
Multiply the newest entry in the result (3)(3) by the divisor (1)(1) and place the result of (3)(3) under the next term in the dividend (-2)(2).
111122-22-1
 13 
13
Step 5.6
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
112-2-1
 13 
131
Step 5.7
Multiply the newest entry in the result (1) by the divisor (1) and place the result of (1) under the next term in the dividend (-1).
112-2-1
 131
131
Step 5.8
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
112-2-1
 131
1310
Step 5.9
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
1x2+3x+1
Step 5.10
Simplify the quotient polynomial.
x2+3x+1
x2+3x+1
Step 6
Find all the possible roots for x2+3x+1.
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Step 6.1
If a polynomial function has integer coefficients, then every rational zero will have the form pq where p is a factor of the constant and q is a factor of the leading coefficient.
p=±1
q=±1
Step 6.2
Find every combination of ±pq. These are the possible roots of the polynomial function.
±1
±1
Step 7
The final factor is the only factor left over from the synthetic division.
x2+3x+1
Step 8
The factored polynomial is (x-4)(x-1)(x2+3x+1).
(x-4)(x-1)(x2+3x+1)
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