In the previous section we got the following information:
• We know how to divide a polynomial p(x) by a first degree polynomial of the form (x+b1)
♦ We will get the 'quotient polynomial q(x)' and the remainder 'r'
• But the method that we learned were applied only to divide second and third degree polynomials
• For higher degree polynomials, more lengthy calculations are involved. We will see them in higher classes
• But we learned a method to 'find the remainder alone'.
♦ This method will not give us the quotient
♦ This method is applicable to polynomials of any degree
• Let us apply it to a fourth degree polynomial:
Find the remainder on dividing (x4 + 2x3 - 6x2 + x + 5) by (x-2)
Solution:
1. We have: {p(x) - r} = {q(x) × (x+b1)}
Substituting for p(x) and (x+b1), we get:
{(x4 + 2x3 - 6x2 + x + 5) - r} = {q(x) × (x-2)}
2. To make the right side zero, put x = 2 on both sides. We get:
{(24 + 2×23 - 6×22 + 2 + 5) - r} = {q(x) × (2-2)}
⟹ {(16 + 16 - 24 + 2 + 5) - r} = {q(x) × (0)}
⟹ {(15) - r} = 0 ⟹ r = 15
So we have an easy method to find the remainder 'r' for polynomial of any degree. If we analyse further, we will get a still easier method to obtain 'r'. Let us try:
1. We have: {p(x) - r} = {q(x) × (x+b1)}
2. Let us isolate 'r'. We will get:
r = p(x) - {q(x) × (x+b1)}
3. Put x = -b1 in the above equation. We get:
r = p(-b1) - {q(x) × (-b1+b1)}
⟹ r = p(-b1) - {q(x) × (0)} ⟹ r = p(-b1) - {0} ⟹ r = p(-b1)
■ It cannot get any simpler than this. We will write it as a theorem:
Theorem 35.2
• We want the remainder 'r' when a polynomial p(x) is divided by (x+b1)
• The remainder 'r' is nothing but p(-b1)
An example:
Find the remainder when (x3 - 2x2 - 4x + 5) is divided by (x+2)
Solution:
• In this problem, p(x) = (x3 - 2x2 - 4x + 5) and b1 = 2
• So r = p(-b1) = p(-2) = [(-2)3 - 2(-2)2 - 4(-2) + 5] = [- 8 - 8 +8 + 5] = -3
• Note that if p(-b1) = 0, then:
♦ It means that there is no remainder
♦ It also means that (x+b1) is a factor, as we have seen earlier
♦ If it is a factor, then naturally, there will not be any remainder
In the above example, what will be the remainder if we divide p(x) by (2x-1) instead of (x+2)
Solution:
1. In (2x-1), the coefficient of x is not 1. So we have to modify it
• We can write (2x-1) as [2(x - 1⁄2)]
• The value will not change. That is:
(2x-1) = [2(x - 1⁄2)]
• So 'dividing by (2x-1)' is same as 'dividing by [2(x - 1⁄2)]'
2. So we have to find the remainder when (x3 - 2x2 - 4x + 5) is divided by [2(x - 1⁄2)]
• We will first consider (x - 1⁄2).
• That is., we will first find the remainder:
♦ when (x3 - 2x2 - 4x + 5) is divided by (x - 1⁄2)
3. We have the easy method to find it:
r = p(-b1) = p(-( - 1⁄2)) = p(1⁄2)
= [(1⁄2)3 - 2(1⁄2)2 - 4(1⁄2) + 5]
= [1⁄8 - (2 × 1⁄4) - (4 × 1⁄2) + 5]
= [1⁄8 - 1⁄2 - 2 + 5] = [- 3⁄8 + 3] = 25⁄8
4. Let q(x) be the quotient when we divide (x3 - 2x2 - 4x + 5) by (x - 1⁄2)
Then we can write:
(x3 - 2x2 - 4x + 5) = {[q(x) × (x - 1⁄2)] + 25⁄8}
5. In step (1), we modified (2x-1) into [2(x - 1⁄2)]
Now we are going to modify (x - 1⁄2):
• We can write (x - 1⁄2) as [1⁄2 × (2x - 1)]
• The value will not change. That is:
(x - 1⁄2) = [1⁄2 × (2x - 1)]
• So 'multiplying by (x - 1⁄2)' is same as 'multiplying by [1⁄2 × (2x - 1)]'
6. In step (4), we have a multiplication using (x - 1⁄2) on the right side.
• We will use [1⁄2× (2x - 1)] instead. Then step (4) becomes:
(x3 - 2x2 - 4x + 5) = {[q(x) × 1⁄2 × (2x - 1)] + 25⁄8}
7. Let 'q(x) × 1⁄2 ' = r(x)
• Then the result in step (6) becomes:
(x3 - 2x2 - 4x + 5) = {[r(x) × (2x - 1)] + 25⁄8}
8. The result in step (7) indicates that:
• When (x3 - 2x2 - 4x + 5) is divided by (2x - 1), we get:
♦ The quotient r(x)
♦ The remainder 25⁄8.
■ But this is just what we want:
The remainder when (x3 - 2x2 - 4x + 5) is divided by (2x - 1)
9. So we can write:
• The remainder when (x3 - 2x2 - 4x + 5) is divided by (x - 1⁄2) is 25⁄8
• The remainder when (x3 - 2x2 - 4x + 5) is divided by (2x - 1) is also 25⁄8.
The problem is solved
10. Note that, the quotient q(x) in step (4) has become r(x) in step (7)
• This is because q(x) was multiplied by a factor
• But we need not worry about the quotients. It is the remainder that we want
In the problems we have seen so far,
• The polynomial (of any degree) p(x) which is divided is called the dividend
• The first degree polynomial with which p(x) is divided is called the divisor
See images here.
• We will often encounter problems in which the 'coefficient of x' in the divisor is not 1.
♦ The divisor (2x-1) that we saw above is an example
• So we need to develop a general method which can be used in such problems. We will write it as a theorem.
Theorem 35.3:
1. The general form of a divisor (first degree polynomial) can be written as (ax+b1)
2. This (ax+b1) can be written as [a(x+b1⁄a)]
• The value will not change
3. Now, the remainder when p(x) is divided by (x+b1⁄a) is p(-b1⁄a)
■ The remainder when p(x) is divided by (ax+b1) will also be p(-b1⁄a)
Another useful result is derived below. We will write that also as a theorem.
Theorem 35.4:
• In some cases, we will also want to know whether (ax+b1) is a factor
• We have seen that (ax+b1) is same as [a(x+b1⁄a)]
• So, if (ax+b1) is a factor, [a(x+b1⁄a)] will also be a factor
• If [a(x+b1⁄a)] is a factor, both 'a' and '(x+b1⁄a)' will be factors individually
• Example: 12 is a factor of 96
♦ 12 can be written as (4×3)
♦ Both 4 and 3 are factors of 96 individually
• So we can write:
To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
■ Now, how do we check whether (x+b1⁄a) is a factor?
• We know the method:
If p(-b1⁄a) = 0, then (x+b1⁄a) is a factor
We will see some solved examples
Solved example 35.6
For each pair of polynomials below, check whether the first is a factor of the second. If not a factor, find the remainder on dividing the second by the first.
(i) (x-1), (x3 + 4x2 - 3x - 6)
(ii) (x+1), (x3 + 4x2 - 3x - 6)
(iii) (x-2), (x3 + 3x2 - 4x - 12)
(iv) (x+2), (x3 + 3x2 - 4x - 12)
(v) (2x-1), (2x3 - x2 - 8x + 6)
(vi) (3x-1), (3x3 - 10x2 + 9x - 2)
Solution:
Part (i):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x-1) = (x+(-1))
• So b1 = -1 ⟹ (-b1) = 1
2. So p(-b1) = p(1) = (13 + 4×12 - 3×1 - 6) = (1+4-3-6) = -4
• p(1) ≠ 0. So (x-1) is not a factor
3. If (x+b1) is not a factor, there will be a remainder.
• From theorem 35.2 that we saw at the beginning of this section,
♦ This remainder 'r' is nothing but p(-b1)
• But p(-b1) is calculated in step (2) above
4. So r = p(-b1) = p(1) = -4
Part (ii):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x+1)
• So b1 = 1 ⟹ (-b1) = -1
2. So p(-b1) = p(-1) = ((-1)3 + 4×(-1)2 - 3×(-1) - 6) = (-1+4+3-6) = 0
• p(-b1) = p(-1) = 0. So (x+1) is a factor
3. If (x+1) is a factor, there will not be a remainder
Part (iii):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x-2) = (x+(-2))
• So b1 = -2 ⟹ (-b1) = 2
2. So p(-b1) = p(2) = (23 + 3×22 - 4×2 - 12) = (8+12-8-12) = 0
• p(-b1) = p(2) = 0. So (x-2) is a factor
3. If (x-2) is a factor, there will not be a remainder
Part (iv):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x+2)
• So b1 = 2 ⟹ (-b1) = -2
2. So p(-b1) = p(-2) = ((-2)3 + 3×(-2)2 - 4×(-2) - 12) = (-8+12+8-12) = 0
• p(-b1) = p(-2) = 0. So (x+2) is a factor
3. If (x+2) is a factor, there will not be a remainder
Part (v): (2x-1), (2x3 - x2 - 8x + 6)
1. To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
• This is based on theorem 35.4 written above in this section
• In our case, (ax+b1) = (2x-1) = (2x+(-1))
♦ So we get: a = 2 and b1 = -1
• Then (x+b1⁄a) = (x+(-1⁄2))
• So (b1⁄a) = (-1⁄2) ⟹ (-b1⁄a) = (1⁄2)
2. If (x+b1) is a factor, then p(-b1) = 0
• Similarly, if (x+b1⁄a) is a factor, then p(-b1⁄a) = 0
3. (2x3 - x2 - 8x + 6) can be written as: [2(x3 - 1⁄2×x2 - 4x + 3)]
• Consider (x3 - 1⁄2×x2 - 4x + 3):
• p(-b1⁄a) = p(1⁄2) = (1⁄2)3 - 1⁄2×(1⁄2)2 - 4×(1⁄2) + 3
= 1⁄8 - 1⁄8 - 2 + 3 = 1
• p(1⁄2) ≠ 0. So (x-1⁄2) is not a factor
4. (x-1⁄2) is not a factor of (x3 - 1⁄2×x2 - 4x + 3)
• So (x-1⁄2) is not a factor of [2(x3 - 1⁄2×x2 - 4x + 3)]
Example:
• 4 is not a factor of 21
• So 4 is not a factor of (n×21)
♦ What ever value 'n' takes
• That is., (x-1⁄2) is not a factor of (2x3 - x2 - 8x + 6)
5. If (x-1⁄2) is not a factor of (2x3 - x2 - 8x + 6), then
• (2x-1) is also not a factor of (2x3 - x2 - 8x + 6)
♦ This is based on theorem 35.4 written above in this section
• That means, if we divide (2x3 - x2 - 8x + 6) by (2x-1) , there will be a remainder
• Our next task is to find this remainder
6. As before, we write (2x3 - x2 - 8x + 6) as: [2(x3 - 1⁄2×x2 - 4x + 3)]
• We will first find the remainder 'r' when (x3 - 1⁄2×x2 - 4x + 3) is divided by (2x-1)
♦ We can apply theorem 35.3 written above in this section
• We have: r = p(-b1⁄a) = p(1⁄2) = 1
♦ This result is obtained from step (3) above
7. Consider the general result:
p(x) = {[quotient × divisor] + r}
• Applying this, we get:
(x3 - 1⁄2×x2 - 4x + 3) = {[quotient × (2x-1)] + 1}
• Multiplying both sides by '2', we get:
2×(x3 - 1⁄2×x2 - 4x + 3) = 2×{[quotient × (2x-1)] + 1}
⟹ (2x3 - x2 - 8x + 6) = {2×[quotient × (2x-1)] + 2×1}
⟹ (2x3 - x2 - 8x + 6) = {[(2×quotient) × (2x-1)] + 2}
8. This means that, when (2x3 - x2 - 8x + 6) is divided by (2x-1), the remainder is 2
• But this is the same task mentioned in step (5)
• So the remainder that we are trying to find is 2
Part (vi): (3x-1), (3x3 - 10x2 + 9x - 2)
1. To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
• This is based on theorem 35.4 written above in this section
• In our case, (ax+b1) = (3x-1) = (3x+(-1))
♦ So we get: a = 3 and b1 = -1
• Then (x+b1⁄a) = (x+(-1⁄3))
• So (b1⁄a) = (-1⁄3) ⟹ (-b1⁄a) = (1⁄3)
2. If (x+b1) is a factor, then p(-b1) = 0
• Similarly, if (x+b1⁄a) is a factor, then p(-b1⁄a) = 0
3. (3x3 - 10x2 + 9x - 2) can be written as: [3(x3 - 10⁄3×x2 + 3x - 2⁄3)]
• Consider (x3 - 10⁄3×x2 + 3x - 2⁄3):
• p(-b1⁄a) = p(1⁄3) = (1⁄3)3 - 10⁄3×(1⁄3)2 + 3×(1⁄3) - 2⁄3
= 1⁄27 - 10⁄27 + 1 - 2⁄3
= -9⁄27 + 1⁄3
= -1⁄3 + 1⁄3 = 0
• p(1⁄3) = 0. So (x - 1⁄3) is a factor
• So (3x-1) is a factor
In the next section, we will see a few more solved examples.
• We know how to divide a polynomial p(x) by a first degree polynomial of the form (x+b1)
♦ We will get the 'quotient polynomial q(x)' and the remainder 'r'
• But the method that we learned were applied only to divide second and third degree polynomials
• For higher degree polynomials, more lengthy calculations are involved. We will see them in higher classes
• But we learned a method to 'find the remainder alone'.
♦ This method will not give us the quotient
♦ This method is applicable to polynomials of any degree
• Let us apply it to a fourth degree polynomial:
Find the remainder on dividing (x4 + 2x3 - 6x2 + x + 5) by (x-2)
Solution:
1. We have: {p(x) - r} = {q(x) × (x+b1)}
Substituting for p(x) and (x+b1), we get:
{(x4 + 2x3 - 6x2 + x + 5) - r} = {q(x) × (x-2)}
2. To make the right side zero, put x = 2 on both sides. We get:
{(24 + 2×23 - 6×22 + 2 + 5) - r} = {q(x) × (2-2)}
⟹ {(16 + 16 - 24 + 2 + 5) - r} = {q(x) × (0)}
⟹ {(15) - r} = 0 ⟹ r = 15
So we have an easy method to find the remainder 'r' for polynomial of any degree. If we analyse further, we will get a still easier method to obtain 'r'. Let us try:
1. We have: {p(x) - r} = {q(x) × (x+b1)}
2. Let us isolate 'r'. We will get:
r = p(x) - {q(x) × (x+b1)}
3. Put x = -b1 in the above equation. We get:
r = p(-b1) - {q(x) × (-b1+b1)}
⟹ r = p(-b1) - {q(x) × (0)} ⟹ r = p(-b1) - {0} ⟹ r = p(-b1)
■ It cannot get any simpler than this. We will write it as a theorem:
Theorem 35.2
• We want the remainder 'r' when a polynomial p(x) is divided by (x+b1)
• The remainder 'r' is nothing but p(-b1)
An example:
Find the remainder when (x3 - 2x2 - 4x + 5) is divided by (x+2)
Solution:
• In this problem, p(x) = (x3 - 2x2 - 4x + 5) and b1 = 2
• So r = p(-b1) = p(-2) = [(-2)3 - 2(-2)2 - 4(-2) + 5] = [- 8 - 8 +8 + 5] = -3
• Note that if p(-b1) = 0, then:
♦ It means that there is no remainder
♦ It also means that (x+b1) is a factor, as we have seen earlier
♦ If it is a factor, then naturally, there will not be any remainder
In the above example, what will be the remainder if we divide p(x) by (2x-1) instead of (x+2)
Solution:
1. In (2x-1), the coefficient of x is not 1. So we have to modify it
• We can write (2x-1) as [2(x - 1⁄2)]
• The value will not change. That is:
(2x-1) = [2(x - 1⁄2)]
• So 'dividing by (2x-1)' is same as 'dividing by [2(x - 1⁄2)]'
2. So we have to find the remainder when (x3 - 2x2 - 4x + 5) is divided by [2(x - 1⁄2)]
• We will first consider (x - 1⁄2).
• That is., we will first find the remainder:
♦ when (x3 - 2x2 - 4x + 5) is divided by (x - 1⁄2)
3. We have the easy method to find it:
r = p(-b1) = p(-( - 1⁄2)) = p(1⁄2)
= [(1⁄2)3 - 2(1⁄2)2 - 4(1⁄2) + 5]
= [1⁄8 - (2 × 1⁄4) - (4 × 1⁄2) + 5]
= [1⁄8 - 1⁄2 - 2 + 5] = [- 3⁄8 + 3] = 25⁄8
4. Let q(x) be the quotient when we divide (x3 - 2x2 - 4x + 5) by (x - 1⁄2)
Then we can write:
(x3 - 2x2 - 4x + 5) = {[q(x) × (x - 1⁄2)] + 25⁄8}
5. In step (1), we modified (2x-1) into [2(x - 1⁄2)]
Now we are going to modify (x - 1⁄2):
• We can write (x - 1⁄2) as [1⁄2 × (2x - 1)]
• The value will not change. That is:
(x - 1⁄2) = [1⁄2 × (2x - 1)]
• So 'multiplying by (x - 1⁄2)' is same as 'multiplying by [1⁄2 × (2x - 1)]'
6. In step (4), we have a multiplication using (x - 1⁄2) on the right side.
• We will use [1⁄2× (2x - 1)] instead. Then step (4) becomes:
(x3 - 2x2 - 4x + 5) = {[q(x) × 1⁄2 × (2x - 1)] + 25⁄8}
7. Let 'q(x) × 1⁄2 ' = r(x)
• Then the result in step (6) becomes:
(x3 - 2x2 - 4x + 5) = {[r(x) × (2x - 1)] + 25⁄8}
8. The result in step (7) indicates that:
• When (x3 - 2x2 - 4x + 5) is divided by (2x - 1), we get:
♦ The quotient r(x)
♦ The remainder 25⁄8.
■ But this is just what we want:
The remainder when (x3 - 2x2 - 4x + 5) is divided by (2x - 1)
9. So we can write:
• The remainder when (x3 - 2x2 - 4x + 5) is divided by (x - 1⁄2) is 25⁄8
• The remainder when (x3 - 2x2 - 4x + 5) is divided by (2x - 1) is also 25⁄8.
The problem is solved
10. Note that, the quotient q(x) in step (4) has become r(x) in step (7)
• This is because q(x) was multiplied by a factor
• But we need not worry about the quotients. It is the remainder that we want
In the problems we have seen so far,
• The polynomial (of any degree) p(x) which is divided is called the dividend
• The first degree polynomial with which p(x) is divided is called the divisor
See images here.
• We will often encounter problems in which the 'coefficient of x' in the divisor is not 1.
♦ The divisor (2x-1) that we saw above is an example
• So we need to develop a general method which can be used in such problems. We will write it as a theorem.
Theorem 35.3:
1. The general form of a divisor (first degree polynomial) can be written as (ax+b1)
2. This (ax+b1) can be written as [a(x+b1⁄a)]
• The value will not change
3. Now, the remainder when p(x) is divided by (x+b1⁄a) is p(-b1⁄a)
■ The remainder when p(x) is divided by (ax+b1) will also be p(-b1⁄a)
Another useful result is derived below. We will write that also as a theorem.
Theorem 35.4:
• In some cases, we will also want to know whether (ax+b1) is a factor
• We have seen that (ax+b1) is same as [a(x+b1⁄a)]
• So, if (ax+b1) is a factor, [a(x+b1⁄a)] will also be a factor
• If [a(x+b1⁄a)] is a factor, both 'a' and '(x+b1⁄a)' will be factors individually
• Example: 12 is a factor of 96
♦ 12 can be written as (4×3)
♦ Both 4 and 3 are factors of 96 individually
• So we can write:
To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
■ Now, how do we check whether (x+b1⁄a) is a factor?
• We know the method:
If p(-b1⁄a) = 0, then (x+b1⁄a) is a factor
We will see some solved examples
Solved example 35.6
For each pair of polynomials below, check whether the first is a factor of the second. If not a factor, find the remainder on dividing the second by the first.
(i) (x-1), (x3 + 4x2 - 3x - 6)
(ii) (x+1), (x3 + 4x2 - 3x - 6)
(iii) (x-2), (x3 + 3x2 - 4x - 12)
(iv) (x+2), (x3 + 3x2 - 4x - 12)
(v) (2x-1), (2x3 - x2 - 8x + 6)
(vi) (3x-1), (3x3 - 10x2 + 9x - 2)
Solution:
Part (i):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x-1) = (x+(-1))
• So b1 = -1 ⟹ (-b1) = 1
2. So p(-b1) = p(1) = (13 + 4×12 - 3×1 - 6) = (1+4-3-6) = -4
• p(1) ≠ 0. So (x-1) is not a factor
3. If (x+b1) is not a factor, there will be a remainder.
• From theorem 35.2 that we saw at the beginning of this section,
♦ This remainder 'r' is nothing but p(-b1)
• But p(-b1) is calculated in step (2) above
4. So r = p(-b1) = p(1) = -4
Part (ii):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x+1)
• So b1 = 1 ⟹ (-b1) = -1
2. So p(-b1) = p(-1) = ((-1)3 + 4×(-1)2 - 3×(-1) - 6) = (-1+4+3-6) = 0
• p(-b1) = p(-1) = 0. So (x+1) is a factor
3. If (x+1) is a factor, there will not be a remainder
Part (iii):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x-2) = (x+(-2))
• So b1 = -2 ⟹ (-b1) = 2
2. So p(-b1) = p(2) = (23 + 3×22 - 4×2 - 12) = (8+12-8-12) = 0
• p(-b1) = p(2) = 0. So (x-2) is a factor
3. If (x-2) is a factor, there will not be a remainder
Part (iv):
1. If (x+b1) is a factor, then p(-b1) = 0
• In our case, (x+b1) = (x+2)
• So b1 = 2 ⟹ (-b1) = -2
2. So p(-b1) = p(-2) = ((-2)3 + 3×(-2)2 - 4×(-2) - 12) = (-8+12+8-12) = 0
• p(-b1) = p(-2) = 0. So (x+2) is a factor
3. If (x+2) is a factor, there will not be a remainder
Part (v): (2x-1), (2x3 - x2 - 8x + 6)
1. To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
• This is based on theorem 35.4 written above in this section
• In our case, (ax+b1) = (2x-1) = (2x+(-1))
♦ So we get: a = 2 and b1 = -1
• Then (x+b1⁄a) = (x+(-1⁄2))
• So (b1⁄a) = (-1⁄2) ⟹ (-b1⁄a) = (1⁄2)
2. If (x+b1) is a factor, then p(-b1) = 0
• Similarly, if (x+b1⁄a) is a factor, then p(-b1⁄a) = 0
3. (2x3 - x2 - 8x + 6) can be written as: [2(x3 - 1⁄2×x2 - 4x + 3)]
• Consider (x3 - 1⁄2×x2 - 4x + 3):
• p(-b1⁄a) = p(1⁄2) = (1⁄2)3 - 1⁄2×(1⁄2)2 - 4×(1⁄2) + 3
= 1⁄8 - 1⁄8 - 2 + 3 = 1
• p(1⁄2) ≠ 0. So (x-1⁄2) is not a factor
4. (x-1⁄2) is not a factor of (x3 - 1⁄2×x2 - 4x + 3)
• So (x-1⁄2) is not a factor of [2(x3 - 1⁄2×x2 - 4x + 3)]
Example:
• 4 is not a factor of 21
• So 4 is not a factor of (n×21)
♦ What ever value 'n' takes
• That is., (x-1⁄2) is not a factor of (2x3 - x2 - 8x + 6)
5. If (x-1⁄2) is not a factor of (2x3 - x2 - 8x + 6), then
• (2x-1) is also not a factor of (2x3 - x2 - 8x + 6)
♦ This is based on theorem 35.4 written above in this section
• That means, if we divide (2x3 - x2 - 8x + 6) by (2x-1) , there will be a remainder
• Our next task is to find this remainder
6. As before, we write (2x3 - x2 - 8x + 6) as: [2(x3 - 1⁄2×x2 - 4x + 3)]
• We will first find the remainder 'r' when (x3 - 1⁄2×x2 - 4x + 3) is divided by (2x-1)
♦ We can apply theorem 35.3 written above in this section
• We have: r = p(-b1⁄a) = p(1⁄2) = 1
♦ This result is obtained from step (3) above
7. Consider the general result:
p(x) = {[quotient × divisor] + r}
• Applying this, we get:
(x3 - 1⁄2×x2 - 4x + 3) = {[quotient × (2x-1)] + 1}
• Multiplying both sides by '2', we get:
2×(x3 - 1⁄2×x2 - 4x + 3) = 2×{[quotient × (2x-1)] + 1}
⟹ (2x3 - x2 - 8x + 6) = {2×[quotient × (2x-1)] + 2×1}
⟹ (2x3 - x2 - 8x + 6) = {[(2×quotient) × (2x-1)] + 2}
8. This means that, when (2x3 - x2 - 8x + 6) is divided by (2x-1), the remainder is 2
• But this is the same task mentioned in step (5)
• So the remainder that we are trying to find is 2
Part (vi): (3x-1), (3x3 - 10x2 + 9x - 2)
1. To check whether (ax+b1) is a factor, we need to check whether (x+b1⁄a) is a factor
• This is based on theorem 35.4 written above in this section
• In our case, (ax+b1) = (3x-1) = (3x+(-1))
♦ So we get: a = 3 and b1 = -1
• Then (x+b1⁄a) = (x+(-1⁄3))
• So (b1⁄a) = (-1⁄3) ⟹ (-b1⁄a) = (1⁄3)
2. If (x+b1) is a factor, then p(-b1) = 0
• Similarly, if (x+b1⁄a) is a factor, then p(-b1⁄a) = 0
3. (3x3 - 10x2 + 9x - 2) can be written as: [3(x3 - 10⁄3×x2 + 3x - 2⁄3)]
• Consider (x3 - 10⁄3×x2 + 3x - 2⁄3):
• p(-b1⁄a) = p(1⁄3) = (1⁄3)3 - 10⁄3×(1⁄3)2 + 3×(1⁄3) - 2⁄3
= 1⁄27 - 10⁄27 + 1 - 2⁄3
= -9⁄27 + 1⁄3
= -1⁄3 + 1⁄3 = 0
• p(1⁄3) = 0. So (x - 1⁄3) is a factor
• So (3x-1) is a factor
