In the previous section we saw the factors of polynomials. In this section we will see polynomial remainder.
Consider the following situation:
• We have a polynomial p(x)
• We have a first degree polynomial (x-a)
■ We want to know whether (x-a) is a factor of p(x)
• From what we have learned from the previous two sections in this chapter, it is easy to find the answer:
• Just find p(a).
♦ If p(a) = 0, then (x-a) is a factor of the given p(x)
Now another question arises:
■ If we find that p(a) ≠ 0, can we confirm that (x-a) is not a factor of p(x)?
• Of course we can.
• If (x-a) is a factor, then p(a) will be equal to zero
• p(x) not being equal to zero confirms that (x-a) is not a factor
An example:
Let p(x) = x2 - 3x + 3
We want to know whether (x-1) is a factor of p(x)
Solution:
• We have: p(1) = (12 - 3×1 + 3) = (1-3+3) = 1
• Here p(1) ≠ 0. So (x-1) is not a factor of (x2 - 3x + 3)
• What will happen if we divide a natural number with 'another number which is not a factor'?
For example:
• 6 is not a factor of 27. If we divide 27 by 6, we will get 'quotient 4' and 'remainder 3'
• We can write it as: 27 = 4 × 6 + 3
• In the same way, if we divide (x2 - 3x + 3) by (x-1), we will get a quotient and a remainder
■ We want to find that quotient and remainder
The following steps will help us:
1. We saw that (x-1) is not a factor of (x2 - 3x + 3)
2. But this (x-1) is a factor of (x2 - 3x + 2)
•The other factor is (x-2)
• The reader may check it using the method that we saw in the first section of this chapter
OR by simply solving it as a quadratic equation
• So we can write:
(x2 - 3x + 2) = (x-1)(x-2)
3. Now add '1' on both sides. We get:
(x2 - 3x + 2) + 1 = (x-1)(x-2) + 1
⟹ (x2 - 3x + 3) = (x-1)(x-2) + 1
4. The result in (3) is similar to:
27 = 4 × 6 + 3
• If we divide 27 by 6, we will get:
♦ 4 as the quotient
♦ 3 as the remainder
5. So it is clear:
• If we divide (x2 - 3x + 3) by (x-1), we will get:
♦ (x-2) as the quotient
♦ 1 as the remainder
Another possibility based on the above steps:
• Consider the result in step (2):
(x2 - 3x + 2) = (x-1)(x-2)
• Subtract '1' from both sides. We get:
(x2 - 3x + 2) - 1 = (x-1)(x-2) - 1
⟹ (x2 - 3x + 1) = (x-1)(x-2) - 1
• So it is clear. If we divide (x2 - 3x + 1) by (x-1), we will get:
♦ (x-2) as the quotient
♦ -1 as the remainder
Now we want a 'general method to find the quotient and remainder' which is applicable to any case.
Let us try:
1. Based on the above discussion, we know this:
• If we divide a second degree polynomial with 'a first degree polynomial (which is not a factor)', we get:
♦ A first degree polynomial as the quotient
♦ A number as remainder
2. We will use the usual representations:
• (x+b1) is the first degree polynomial which is not a factor
♦ We will be dividing p(x) with this polynomial
• (x+b2) is the quotient
• Let 'r' be the remainder
3. Then we can write:
p(x) = (x+b1)(x+b2) + r
Expanding this, we get:
p(x) = [x2 + (b1+b2)x + b1b2] + r
⟹ p(x) = x2 + (b1+b2)x + (b1b2 + r)
4. p(x) in the left side will have:
• A term in 'x2'
♦ We must equate it's coefficient to the 'coefficient of x2 ' on the right side
♦ Both will be 1. So this will not give us any benefit.
♦ We can ignore the 'coefficients of x2 ' on both sides.
♦ But ensure that coefficients are '1' on both sides
• A term in 'x'
♦ Equate it's coefficient to the 'coefficient of x' on the right side, which is (b1+b2)
• A constant term
♦ Equate it to the constant term on the right side, which is (b1b2 + r)
5. Thus we will get two equations.
• The two unknowns are 'b2' and 'r'. We can easily find them by solving the equations
An example:
Divide (x2 - 3x - 10) by (x-2). Find the quotient and remainder
Solution:
1. First we must make sure that (x-2) is not a factor.
• We will get a remainder only if (x-2) is not a factor
• p(2) = (22 - 3×2 + 1) = (4-6+1) = -1
• p(2) ≠ 0. So (x-2) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2) = -3
(ii) (b1b2 + r) = -10
3. Substituting b1 = -2, we get:
(i) (-2+b2) = -3
(ii) (-2×b2 + r) = -10
4. From 3(i) we get: b2 = -3+2 = -1
Substituting this in 3(ii) we get:
(-2×-1 + r) = -10 ⟹ (2+r) = -10 ⟹ r = -12
5. So the quotient is: (x+b2) = (x-1)
The remainder is: r = -12
Another example:
Divide (x2 - 3x - 8) by (x+2). Find the quotient and remainder
Solution:
1. First we must make sure that (x+2) is not a factor.
• We will get a remainder only if (x+2) is not a factor
• p(-2) = ((-2)2 - 3×(-2) + 1) = (4+6+1) = 11
• p(-2) ≠ 0. So (x+2) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2) = -3
(ii) (b1b2 + r) = -8
3. Substituting b1 = 2, we get:
(i) (2+b2) = -3
(ii) (2×b2 + r) = -8
4. From 3(i) we get: b2 = -3-2 = -5
Substituting this in 3(ii) we get:
(2×-5 + r) = -8 ⟹ (-10+r) = -8 ⟹ r = 2
5. So the quotient is: (x+b2) = (x-5)
The remainder is: r = 2
Now we move to the next level. We are going to divide a third degree polynomial
Let us write the steps:
1. Based on the discussions so far in this chapter, we know this:
• If we divide a third degree polynomial with 'a first degree polynomial (which is not a factor)', we get:
♦ A second degree polynomial as the quotient
♦ A number as remainder
2. We will use the usual representations:
• (x+b1) is the first degree polynomial which is not a factor
♦ We will be dividing p(x) with this polynomial
• [(x+b2)(x+b3)] is the quotient
♦ Note that, [(x+b2)(x+b3)] is the general form of a second degree polynomial
♦ It is just written as 'product of two first degree polynomials' for convenience in calculations
• Let 'r' be the remainder
3. Then we can write:
p(x) = {(x+b1)[(x+b2)(x+b3)]} + r
Example: 94 = {(5)×[(6)×(3)]} + 4
• 94 is divided by '5', which is not a factor
• The quotient is 18 and remainder is 4
• The quotient is further factorized as '6×3'
Expanding the expression, we get:
p(x) = {(x+b1)[x2 + (b2+b3)x + b2b3]} + r
= {x3 + (b1)x2 +(b2+b3)x2 + (b2+b3)(b1)x + (b2b3)x + b1b2b3]} + r
⟹ p(x)= {x3 + (b1+b2+b3)x2 + (b1b2+b2b3+b1b3)x + b1b2b3} + r
⟹ p(x)= {x3 + (b1+b2+b3)x2 + (b1b2+b2b3+b1b3)x + (b1b2b3 + r)}
4. p(x) in the left side will have:
• A term in 'x3'
♦ We must equate it's coefficient to the 'coefficient of x3 ' on the right side
♦ Both will be 1. So this will not give us any benefit.
♦ We can ignore the 'coefficients of x3 ' on both sides.
♦ But ensure that coefficients are '1' on both sides
• A term in 'x2'
♦ Equate it's coefficient to the 'coefficient of x2' on the right side, which is (b1+b2+b3)
• A term in 'x'
♦ Equate it's coefficient to the 'coefficient of x' on the right side, which is (b1b2+b2b3+b1b3)
• A constant term
♦ Equate it to the constant term on the right side, which is (b1b2b3 + r)
5. Thus we will get three equations.
• We have b1, b2, b3 and r
♦ But b1 is already known
• So we have 3 equations and 3 unknowns.
• We can easily find the unknowns by solving the equations
An example:
Divide (x3 - 2x2 - x + 4) by (x-3). Find the quotient and remainder
Solution:
1. First we must make sure that (x-3) is not a factor.
• We will get a remainder only if (x-3) is not a factor
• p(3) = (33 - 2×(3)2 - 3 + 4) = (27-18-3+4) = 10
• p(3) ≠ 0. So (x-3) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2+b3) = -2
(ii) (b1b2 + b2b3 + b1b3) = -1
(iii) b1b2b3+ r = 4
3. Substituting b1 = -3, we get:
(i) (b2+b3) = 1
(ii) (-3×b2 + b2b3 + -3×b3) = -1
⟹ [-3×(b2+b3) + b2b3] = -1
⟹ [-3×(1) + b2b3] = -1 (∵ (b2+b3) =1)
⟹ [b2b3] = 2
(iii) -3 × 2 + r = 4
⟹ -6 + r = 4
⟹ r = 10
• So the remainder 'r' = 10
4. The quotient is: [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 2] = [x2 + x + 2]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 2
Another example:
Divide (x3 - 2x2 - x + 9) by (x-3). Find the quotient and remainder
Solution:
1. First we must make sure that (x-3) is not a factor.
• We will get a remainder only if (x-3) is not a factor
• p(3) = (33 - 2×(3)2 - 3 + 9) = (27-18-3+9) = 15
• p(3) ≠ 0. So (x-3) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2+b3) = -2
(ii) (b1b2 + b2b3 + b1b3) = -1
(iii) b1b2b3+ r = 9
3. Substituting b1 = -3, we get:
(i) (b2+b3) = 1
(ii) (-3×b2 + b2b3 + -3×b3) = -1
⟹ [-3×(b2+b3) + b2b3] = -1
⟹ [-3×(1) + b2b3] = -1 (∵ (b2+b3) =1)
⟹ [b2b3] = 2
(iii) -3 × 2 + r = 9
⟹ -6 + r = 9
⟹ r = 15
• So the remainder 'r' = 15
4. The quotient is: [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 2] = [x2 + x + 2]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 2
In this section so far,
• We have divided a second degree polynomial by a 'first degree polynomial (x+b1) which is not a factor'
♦ We obtained a polynomial as quotient
♦ We obtained a remainder also
• We have divided a third degree polynomial by a 'first degree polynomial (x+b1) which is not a factor'
♦ We obtained a polynomial as quotient
♦ We obtained a remainder also
• In general, we can divide any polynomial p(x) by (x+b1).
♦ We will get another polynomial q(x) as the quotient
♦ If the degree of p(x) is 'n', then degree of q(x) will be '(n-1)'
♦ We will get 'r' as the remainder
♦ If 'r' is zero, it simply means that (x+b1) is a factor of p(x)
• So we can write:
p(x) = {q(x) × (x+b1)} + r
• Let us bring 'r' to the left side. We get:
{p(x) - r} = {q(x) × (x+b1)}
■ This is an interesting situation.
• On the left side, we have a new polynomial, obtained just by subtracting 'r' from the original polynomial p(x)
• From what is available on the right side, it is clear that, (x+b1) is a factor of this new polynomial
♦ If we divide the new polynomial by (x+b1), there will not be any remainder
Example: 94 = {(5)×(18)} + 4
• 94 is divided by '5', which is not a factor
• The quotient is 18 and remainder is 4
■ 94 - 4 = 90
• 90 = {(5)×(18)}
• 5 is a factor of 90. There is no remainder
Let us apply this new technique in a problem:
• Is (x-3) a factor of (x3 - 2x2 + x + 2)?
♦ If it is not a factor, what number must be subtracted from (x3 - 2x2 + x + 2) to make (x-3) a factor?
Solution:
1. We have: p(x) = (x3 - 2x2 + x + 2)
• p(3) = (33 - 2×(3)2 + 3 + 2) = (27-18+3+2) = 14
• p(3) ≠ 0. So (x-3) is not a factor
2. So if we divide p(x) by (x-3), there will be a remainder 'r'.
• We can write: p(x) = {q(x) × (x-3)} + r
♦ Where q(x) is the quotient
♦ r is the remainder
3. Bring r to the left side:
{p(x) - r} = {q(x) × (x-3)}
• We know how to find q(x) and r.
♦ We have seen the methods when the given p(x) is of second degree or third degree
• But in this problem, our task is to find 'r' only. We need not find the quotient q(x)
♦ This is because, if we subtract 'r', (x-3) will become a factor
4. Is there a method to find 'r' alone? Let us try:
(i) Expand the expression in (3). We get:
{(x3 - 2x2 + x + 2) - r} = {q(x) × (x-3)}
(ii) If we put x = 3 in this expression, the right side of '=' sign will become zero.
This is shown below:
{(33 - 2×32 + 3 + 2) - r} = {q(x) × (3-3)}
⟹ {(27 - 18 + 3 + 2) - r} = {q(x) × (0)}
⟹ {14-r} = 0
⟹ r = 14
5. The original polynomial is p(x) = (x3 - 2x2 + x + 2)
• Subtract 14 to get the new polynomial:
(x3 - 2x2 + x + 2) - 14 = (x3 - 2x2 + x - 12)
• (x-3) will be a factor of the new polynomial p'(x) = (x3 - 2x2 + x - 12)
Check:
p'(3) = (33 - 2×32 + 3 - 12) = (27-18+3-12) = 0
In the next section, we will see an easier method to find the remainder 'r'.
Consider the following situation:
• We have a polynomial p(x)
• We have a first degree polynomial (x-a)
■ We want to know whether (x-a) is a factor of p(x)
• From what we have learned from the previous two sections in this chapter, it is easy to find the answer:
• Just find p(a).
♦ If p(a) = 0, then (x-a) is a factor of the given p(x)
Now another question arises:
■ If we find that p(a) ≠ 0, can we confirm that (x-a) is not a factor of p(x)?
• Of course we can.
• If (x-a) is a factor, then p(a) will be equal to zero
• p(x) not being equal to zero confirms that (x-a) is not a factor
An example:
Let p(x) = x2 - 3x + 3
We want to know whether (x-1) is a factor of p(x)
Solution:
• We have: p(1) = (12 - 3×1 + 3) = (1-3+3) = 1
• Here p(1) ≠ 0. So (x-1) is not a factor of (x2 - 3x + 3)
• What will happen if we divide a natural number with 'another number which is not a factor'?
For example:
• 6 is not a factor of 27. If we divide 27 by 6, we will get 'quotient 4' and 'remainder 3'
• We can write it as: 27 = 4 × 6 + 3
• In the same way, if we divide (x2 - 3x + 3) by (x-1), we will get a quotient and a remainder
■ We want to find that quotient and remainder
The following steps will help us:
1. We saw that (x-1) is not a factor of (x2 - 3x + 3)
2. But this (x-1) is a factor of (x2 - 3x + 2)
•The other factor is (x-2)
• The reader may check it using the method that we saw in the first section of this chapter
OR by simply solving it as a quadratic equation
• So we can write:
(x2 - 3x + 2) = (x-1)(x-2)
3. Now add '1' on both sides. We get:
(x2 - 3x + 2) + 1 = (x-1)(x-2) + 1
⟹ (x2 - 3x + 3) = (x-1)(x-2) + 1
4. The result in (3) is similar to:
27 = 4 × 6 + 3
• If we divide 27 by 6, we will get:
♦ 4 as the quotient
♦ 3 as the remainder
5. So it is clear:
• If we divide (x2 - 3x + 3) by (x-1), we will get:
♦ (x-2) as the quotient
♦ 1 as the remainder
Another possibility based on the above steps:
• Consider the result in step (2):
(x2 - 3x + 2) = (x-1)(x-2)
• Subtract '1' from both sides. We get:
(x2 - 3x + 2) - 1 = (x-1)(x-2) - 1
⟹ (x2 - 3x + 1) = (x-1)(x-2) - 1
• So it is clear. If we divide (x2 - 3x + 1) by (x-1), we will get:
♦ (x-2) as the quotient
♦ -1 as the remainder
Now we want a 'general method to find the quotient and remainder' which is applicable to any case.
Let us try:
1. Based on the above discussion, we know this:
• If we divide a second degree polynomial with 'a first degree polynomial (which is not a factor)', we get:
♦ A first degree polynomial as the quotient
♦ A number as remainder
2. We will use the usual representations:
• (x+b1) is the first degree polynomial which is not a factor
♦ We will be dividing p(x) with this polynomial
• (x+b2) is the quotient
• Let 'r' be the remainder
3. Then we can write:
p(x) = (x+b1)(x+b2) + r
Expanding this, we get:
p(x) = [x2 + (b1+b2)x + b1b2] + r
⟹ p(x) = x2 + (b1+b2)x + (b1b2 + r)
4. p(x) in the left side will have:
• A term in 'x2'
♦ We must equate it's coefficient to the 'coefficient of x2 ' on the right side
♦ Both will be 1. So this will not give us any benefit.
♦ We can ignore the 'coefficients of x2 ' on both sides.
♦ But ensure that coefficients are '1' on both sides
• A term in 'x'
♦ Equate it's coefficient to the 'coefficient of x' on the right side, which is (b1+b2)
• A constant term
♦ Equate it to the constant term on the right side, which is (b1b2 + r)
5. Thus we will get two equations.
• The two unknowns are 'b2' and 'r'. We can easily find them by solving the equations
An example:
Divide (x2 - 3x - 10) by (x-2). Find the quotient and remainder
Solution:
1. First we must make sure that (x-2) is not a factor.
• We will get a remainder only if (x-2) is not a factor
• p(2) = (22 - 3×2 + 1) = (4-6+1) = -1
• p(2) ≠ 0. So (x-2) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2) = -3
(ii) (b1b2 + r) = -10
3. Substituting b1 = -2, we get:
(i) (-2+b2) = -3
(ii) (-2×b2 + r) = -10
4. From 3(i) we get: b2 = -3+2 = -1
Substituting this in 3(ii) we get:
(-2×-1 + r) = -10 ⟹ (2+r) = -10 ⟹ r = -12
5. So the quotient is: (x+b2) = (x-1)
The remainder is: r = -12
Another example:
Divide (x2 - 3x - 8) by (x+2). Find the quotient and remainder
Solution:
1. First we must make sure that (x+2) is not a factor.
• We will get a remainder only if (x+2) is not a factor
• p(-2) = ((-2)2 - 3×(-2) + 1) = (4+6+1) = 11
• p(-2) ≠ 0. So (x+2) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2) = -3
(ii) (b1b2 + r) = -8
3. Substituting b1 = 2, we get:
(i) (2+b2) = -3
(ii) (2×b2 + r) = -8
4. From 3(i) we get: b2 = -3-2 = -5
Substituting this in 3(ii) we get:
(2×-5 + r) = -8 ⟹ (-10+r) = -8 ⟹ r = 2
5. So the quotient is: (x+b2) = (x-5)
The remainder is: r = 2
Now we move to the next level. We are going to divide a third degree polynomial
Let us write the steps:
1. Based on the discussions so far in this chapter, we know this:
• If we divide a third degree polynomial with 'a first degree polynomial (which is not a factor)', we get:
♦ A second degree polynomial as the quotient
♦ A number as remainder
2. We will use the usual representations:
• (x+b1) is the first degree polynomial which is not a factor
♦ We will be dividing p(x) with this polynomial
• [(x+b2)(x+b3)] is the quotient
♦ Note that, [(x+b2)(x+b3)] is the general form of a second degree polynomial
♦ It is just written as 'product of two first degree polynomials' for convenience in calculations
• Let 'r' be the remainder
3. Then we can write:
p(x) = {(x+b1)[(x+b2)(x+b3)]} + r
Example: 94 = {(5)×[(6)×(3)]} + 4
• 94 is divided by '5', which is not a factor
• The quotient is 18 and remainder is 4
• The quotient is further factorized as '6×3'
Expanding the expression, we get:
p(x) = {(x+b1)[x2 + (b2+b3)x + b2b3]} + r
= {x3 + (b1)x2 +(b2+b3)x2 + (b2+b3)(b1)x + (b2b3)x + b1b2b3]} + r
⟹ p(x)= {x3 + (b1+b2+b3)x2 + (b1b2+b2b3+b1b3)x + b1b2b3} + r
⟹ p(x)= {x3 + (b1+b2+b3)x2 + (b1b2+b2b3+b1b3)x + (b1b2b3 + r)}
4. p(x) in the left side will have:
• A term in 'x3'
♦ We must equate it's coefficient to the 'coefficient of x3 ' on the right side
♦ Both will be 1. So this will not give us any benefit.
♦ We can ignore the 'coefficients of x3 ' on both sides.
♦ But ensure that coefficients are '1' on both sides
• A term in 'x2'
♦ Equate it's coefficient to the 'coefficient of x2' on the right side, which is (b1+b2+b3)
• A term in 'x'
♦ Equate it's coefficient to the 'coefficient of x' on the right side, which is (b1b2+b2b3+b1b3)
• A constant term
♦ Equate it to the constant term on the right side, which is (b1b2b3 + r)
5. Thus we will get three equations.
• We have b1, b2, b3 and r
♦ But b1 is already known
• So we have 3 equations and 3 unknowns.
• We can easily find the unknowns by solving the equations
An example:
Divide (x3 - 2x2 - x + 4) by (x-3). Find the quotient and remainder
Solution:
1. First we must make sure that (x-3) is not a factor.
• We will get a remainder only if (x-3) is not a factor
• p(3) = (33 - 2×(3)2 - 3 + 4) = (27-18-3+4) = 10
• p(3) ≠ 0. So (x-3) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2+b3) = -2
(ii) (b1b2 + b2b3 + b1b3) = -1
(iii) b1b2b3+ r = 4
3. Substituting b1 = -3, we get:
(i) (b2+b3) = 1
(ii) (-3×b2 + b2b3 + -3×b3) = -1
⟹ [-3×(b2+b3) + b2b3] = -1
⟹ [-3×(1) + b2b3] = -1 (∵ (b2+b3) =1)
⟹ [b2b3] = 2
(iii) -3 × 2 + r = 4
⟹ -6 + r = 4
⟹ r = 10
• So the remainder 'r' = 10
4. The quotient is: [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 2] = [x2 + x + 2]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 2
Another example:
Divide (x3 - 2x2 - x + 9) by (x-3). Find the quotient and remainder
Solution:
1. First we must make sure that (x-3) is not a factor.
• We will get a remainder only if (x-3) is not a factor
• p(3) = (33 - 2×(3)2 - 3 + 9) = (27-18-3+9) = 15
• p(3) ≠ 0. So (x-3) is not a factor
2. Comparing the coefficients, we get:
(i) (b1+b2+b3) = -2
(ii) (b1b2 + b2b3 + b1b3) = -1
(iii) b1b2b3+ r = 9
3. Substituting b1 = -3, we get:
(i) (b2+b3) = 1
(ii) (-3×b2 + b2b3 + -3×b3) = -1
⟹ [-3×(b2+b3) + b2b3] = -1
⟹ [-3×(1) + b2b3] = -1 (∵ (b2+b3) =1)
⟹ [b2b3] = 2
(iii) -3 × 2 + r = 9
⟹ -6 + r = 9
⟹ r = 15
• So the remainder 'r' = 15
4. The quotient is: [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 2] = [x2 + x + 2]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 2
In this section so far,
• We have divided a second degree polynomial by a 'first degree polynomial (x+b1) which is not a factor'
♦ We obtained a polynomial as quotient
♦ We obtained a remainder also
• We have divided a third degree polynomial by a 'first degree polynomial (x+b1) which is not a factor'
♦ We obtained a polynomial as quotient
♦ We obtained a remainder also
• In general, we can divide any polynomial p(x) by (x+b1).
♦ We will get another polynomial q(x) as the quotient
♦ If the degree of p(x) is 'n', then degree of q(x) will be '(n-1)'
♦ We will get 'r' as the remainder
♦ If 'r' is zero, it simply means that (x+b1) is a factor of p(x)
• So we can write:
p(x) = {q(x) × (x+b1)} + r
• Let us bring 'r' to the left side. We get:
{p(x) - r} = {q(x) × (x+b1)}
■ This is an interesting situation.
• On the left side, we have a new polynomial, obtained just by subtracting 'r' from the original polynomial p(x)
• From what is available on the right side, it is clear that, (x+b1) is a factor of this new polynomial
♦ If we divide the new polynomial by (x+b1), there will not be any remainder
Example: 94 = {(5)×(18)} + 4
• 94 is divided by '5', which is not a factor
• The quotient is 18 and remainder is 4
■ 94 - 4 = 90
• 90 = {(5)×(18)}
• 5 is a factor of 90. There is no remainder
Let us apply this new technique in a problem:
• Is (x-3) a factor of (x3 - 2x2 + x + 2)?
♦ If it is not a factor, what number must be subtracted from (x3 - 2x2 + x + 2) to make (x-3) a factor?
Solution:
1. We have: p(x) = (x3 - 2x2 + x + 2)
• p(3) = (33 - 2×(3)2 + 3 + 2) = (27-18+3+2) = 14
• p(3) ≠ 0. So (x-3) is not a factor
2. So if we divide p(x) by (x-3), there will be a remainder 'r'.
• We can write: p(x) = {q(x) × (x-3)} + r
♦ Where q(x) is the quotient
♦ r is the remainder
3. Bring r to the left side:
{p(x) - r} = {q(x) × (x-3)}
• We know how to find q(x) and r.
♦ We have seen the methods when the given p(x) is of second degree or third degree
• But in this problem, our task is to find 'r' only. We need not find the quotient q(x)
♦ This is because, if we subtract 'r', (x-3) will become a factor
4. Is there a method to find 'r' alone? Let us try:
(i) Expand the expression in (3). We get:
{(x3 - 2x2 + x + 2) - r} = {q(x) × (x-3)}
(ii) If we put x = 3 in this expression, the right side of '=' sign will become zero.
This is shown below:
{(33 - 2×32 + 3 + 2) - r} = {q(x) × (3-3)}
⟹ {(27 - 18 + 3 + 2) - r} = {q(x) × (0)}
⟹ {14-r} = 0
⟹ r = 14
5. The original polynomial is p(x) = (x3 - 2x2 + x + 2)
• Subtract 14 to get the new polynomial:
(x3 - 2x2 + x + 2) - 14 = (x3 - 2x2 + x - 12)
• (x-3) will be a factor of the new polynomial p'(x) = (x3 - 2x2 + x - 12)
Check:
p'(3) = (33 - 2×32 + 3 - 12) = (27-18+3-12) = 0
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