Saturday, February 24, 2018

Chapter 35.4 - Solved examples on Polynomial Remainder

In the previous section we saw the easy method to find the remainder. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 35.7
For each pair of polynomials below, find the quotient and remainder on dividing the first by the second
(i) (x- 1)(x-1)
(ii) (x- 1)(x+1)
(iii) (x+ 1)(x-1)
(iv) (x+ 1)(x+1)
Solution:
Part (i)(x- 1)(x-1)
1. First we must make sure that (x-1) is not a factor. 
• We will get a remainder only if (x-1) is not a factor
• p(1) = (1- 1) = (1-1) = 0
• p(1) = 0. So (x-1) is a factor
• We will not get a remainder
• But there will be a quotient, and we can find it
2. The general form is:
p(x) = {(x+b1) × (x+b2× (x+b3)} + r
• Substituting the known values, we get:
(x- 1) = {(x+b1) × (x+b2× (x+b3)} + 0
⟹ x+ 0x2 + 0x - 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1bb2bb1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor =  (x+b1) = (x-1)
• So b= -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i)  (b2+b3) = 1
(ii) (-1×bb2b+ -1×b3) = 0
 [-1×(b2+b3) b2b3] = 0
 [-1×(1) b2b3] = 0 ( (b2+b3) =1)
 [b2b3] = 1
4. The quotient is: (x+b2× (x+b3[x+ (b2+b3)x + b2b3]
[x+ (1)x + 1] = [xx + 1
    ♦ ∵ from 3(i) we have: (b2+b3) = 1   
    ♦ from 3(ii) we have: (b2b3) = 1

Part (ii)(x- 1)(x+1)
1. First we must make sure that (x+1) is not a factor. 
• We will get a remainder only if (x+1) is not a factor
• p(-1) = ((-1)- 1) = (-1-1) = -2
• p(-1)  0. So (x+1) is not a factor
• We will get a quotient and remainder
• The remainder r = p(-1) = -2
2. The general form is:
p(x) = {(x+b1) × (x+b2× (x+b3)} + r
• Substituting the known values, we get:
(x- 1) = {(x+b1) × (x+b2× (x+b3)} - 2
⟹ x+ 0x2 + 0x - 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]} -2
⟹ x+ 0x2 + 0x + 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1bb2bb1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor =  (x+b1) = (x+1)
• So b= 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i)  (b2+b3) = -1
(ii) (1×bb2b+ 1×b3) = 0
 [1×(b2+b3) b2b3] = 0
 [1×(-1) b2b3] = 0 ( (b2+b3) = -1)
 [b2b3] = 1
4. The quotient is: (x+b2× (x+b3[x+ (b2+b3)x + b2b3]
[x+ (-1)x + 1] = [xx + 1
    ♦ ∵ from 3(i) we have: (b2+b3) = -1   
    ♦ from 3(ii) we have: (b2b3) = 1

Part (iii)(x+ 1)(x-1)
1. First we must make sure that (x-1) is not a factor. 
• We will get a remainder only if (x-1) is not a factor
• p(1) = (1+ 1) = (1+1) = 2
• p(1)  0. So (x-1) is not a factor
• We will get a quotient and remainder
• The remainder r = p(1) = 2
2. The general form is:
p(x) = {(x+b1) × (x+b2× (x+b3)} + r
• Substituting the known values, we get:
(x+ 1) = {(x+b1) × (x+b2× (x+b3)} + 2
⟹ x+ 0x2 + 0x + 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]} + 2
⟹ x+ 0x2 + 0x - 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1bb2bb1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor =  (x+b1) = (x-1)
• So b= -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i)  (b2+b3) = 1
(ii) (-1×bb2b+ -1×b3) = 0
 [-1×(b2+b3) b2b3] = 0
 [-1×(1) b2b3] = 0 ( (b2+b3) = 1)
 [b2b3] = 1
4. The quotient is: (x+b2× (x+b3[x+ (b2+b3)x + b2b3]
[x+ (1)x + 1] = [xx + 1
    ♦ ∵ from 3(i) we have: (b2+b3) = 1
    ♦ from 3(ii) we have: (b2b3) = 1

Part (iv)(x+ 1)(x+1)
1. First we must make sure that (x+1) is not a factor. 
• We will get a remainder only if (x+1) is not a factor
• p(-1) = ((-1)+ 1) = (-1+1) = 0
• p(-1) = 0. So (x+1) is a factor
• We will not get a remainder
• But there will be a quotient, and we can find it
2. The general form is:
p(x) = {(x+b1) × (x+b2× (x+b3)} + r
• Substituting the known values, we get:
(x- 1) = {(x+b1) × (x+b2× (x+b3)} + 0
⟹ x+ 0x2 + 0x + 1 = {(x+b1) × [x+ (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1bb2bb1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor =  (x+b1) = (x+1)
• So b= 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i)  (b2+b3) = -1
(ii) (1×bb2b+ 1×b3) = 0
 [1×(b2+b3) b2b3] = 0
 [1×(-1) b2b3] = 0 ( (b2+b3) = -1)
 [b2b3] = 1
4. The quotient is: (x+b2× (x+b3[x+ (b2+b3)x + b2b3]
[x+ (-1)x + 1] = [xx + 1
    ♦ ∵ from 3(i) we have: (b2+b3) = -1   
    ♦ from 3(ii) we have: (b2b3) = 1

Solved example 35.8
By adding a number to p(x) = (x+ x2 + x), a new polynomial q(x) is to be formed.
(i) What number should be added, so that (x-1) is a factor of q(x)?
(ii) What number should be added, so that (x+1) is a factor of q(x)?
Solution:
Part (i):
1. Let us first check whether (x-1) is a factor of p(x)
• For that, we must find p(1)
• p(1) = (11+ 1) = (1+1+1) = 3
• p(1)  0. So (x-1) is not a factor of p(x)
2. So, if we divide p(x) by (x-1), we will get a remainder 'r'
• This situation can be written as:
p(x) = {(x-1) × quotient} + r
3. But we know that the remainder 'r' will be equal to p(1)
• We have already calculated p(1) as 3
• So we have r = 3
4. Substituting the value of r in (2), we get:
p(x) = {(x-1) × quotient} + 3
 {p(x) - 3} = {(x-1) × quotient}
5. On the left side, we have a new polynomial obtained by adding '-3' to p(x)
• Let us call this new polynomial as 'q(x)'
• The right side indicates that, if we divide 'q(x)' by (x-1), there will not be a remainder
    ♦ That means, (x-1) is a factor of q(x)
• So the number to be added is -3
• The new polynomial q(x) = {p(x) - 3} = (x+ x2 + x - 3)

Part (ii):
1. Let us first check whether (x+1) is a factor of p(x)
• For that, we must find p(-1)
• p(-1) = ((-1)+ (-1)- 1) = (-1+1-1) = -1
• p(-1)  0. So (x+1) is not a factor of p(x)
2. So, if we divide p(x) by (x+1), we will get a remainder 'r'
• This situation can be written as:
p(x) = {(x+1) × quotient} + r
3. But we know that the remainder 'r' will be equal to p(-1)
• We have already calculated p(1) as -1
• So we have r = -1
4. Substituting the value of r in (2), we get:
p(x) = {(x-1) × quotient} - 1
 {p(x) + 1} = {(x-1) × quotient}
5. On the left side, we have a new polynomial obtained by adding '1' to p(x)
• Let us call this new polynomial as 'q(x)'
• The right side indicates that, if we divide 'q(x)' by (x+1), there will not be a remainder
    ♦ That means, (x+1) is a factor of q(x)
• So the number to be added is 1
• The new polynomial q(x) = {p(x) + 1} = (x+ x2 + x + 1)

Solved example 35.9
In each pair of polynomials below, find what kind of natural number n must be, so that the first is a factor of the second
(i) (x-1), (x-1) (ii) (x-1), (x+1) (iii) (x+1), (xn-1) (iv) (x+1), (xn+1) (v) (x2-1), (xn-1) 
Solution:
Natural numbers are 1, 2, 3, . . . (See fig.16.1)
In the given problem, n can be any natural number. We will see each case:
Part (i)(x-1), (x- 1)
 If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x- 1) = (x-1)
    ♦ Then p(1) = (1-1) = 0
    ♦ So (x-1) is a factor
• If n = 2, p(x) = (x- 1)
    ♦ Then p(1) = (12-1) = 0
    ♦ So (x-1) is a factor
• If n = 3, p(x) = (x- 1)
    ♦ Then p(1) = (13-1) = 0
    ♦ So (x-1) is a factor
• If n = 4, p(x) = (x- 1)
    ♦ Then p(1) = (14-1) = 0
    ♦ So (x-1) is a factor
_    _    _    _
_    _    _    _
■ So for all natural numbers 1, 2, 3, 4, . . . , 
(x-1) will be a factor of (x- 1)

Part (ii)(x-1), (x+1)
 If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x+ 1) = (x+1)
    ♦ Then p(1) = (1+1) = 2
    ♦ So (x-1) is not a factor
• If n = 2, p(x) = (x+ 1)
    ♦ Then p(1) = (1+1) = 2
    ♦ So (x-1) is not a factor
• If n = 3 p(x) = (x+ 1)
    ♦ Then p(1) = (1+1) = 2
    ♦ So (x-1) is not a factor
• If n = 4, p(x) = (x+ 1)
    ♦ Then p(1) = (1+1) = 2
    ♦ So (x-1) is not a factor
_    _    _    _
_    _    _    _
■ So for all natural numbers 1, 2, 3, 4, . . . , 
(x-1) will be not a factor of (x- 1)

Part (iii)(x+1), (xn-1)
 If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x- 1) = (x-1)
    ♦ Then p(-1) = (-1-1) = -2
    ♦ So (x+1) is not a factor
• If n = 2, p(x) = (x- 1)
    ♦ Then p(-1) = ((-1)2-1) = (1-1) = 0
    ♦ So (x+1) is a factor
• If n = 3, p(x) = (x3- 1)
    ♦ Then p(-1) = ((-1)3-1) = (-1-1) = -2
    ♦ So (x+1) is not a factor
• If n = 4, p(x) = (x- 1)
    ♦ Then p(-1) = ((-1)4-1) = (1-1) = 0
    ♦ So (x+1) is a factor
_    _    _    _
_    _    _    _
■ So for all even natural numbers 2, 4, 6, 8,  . . . , 
(x+1) will be a factor of (x- 1)

Part (iv)(x+1), (xn+1)
 If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x+ 1) = (x+1)
    ♦ Then p(-1) = (-1+1) = 0
    ♦ So (x+1) is a factor
• If n = 2, p(x) = (x+ 1)
    ♦ Then p(-1) = ((-1)2+1) = (1+1) = 2
    ♦ So (x+1) is not a factor
• If n = 3, p(x) = (x+ 1)
    ♦ Then p(-1) = ((-1)3+1) = (-1+1) = 0
    ♦ So (x+1) is a factor
• If n = 4, p(x) = (x+ 1)
    ♦ Then p(-1) = ((-1)4+1) = (1+1) = 2
    ♦ So (x+1) is not a factor
_    _    _    _
_    _    _    _
■ So for all odd natural numbers 1, 3, 5, 7,  . . . , 
(x+1) will be a factor of (x+ 1)

Part (v)(x2-1), (xn-1)
■ (x2-1) can be split as [(x+1)(x-1)]
• If  (x2-1) is a factor, then both (x+1) and (x-1) should be factors
    ♦ If (x+1) is a factor, p(-1) = 0
    ♦ If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (xn-1) = (x- 1) = (x-1)
    ♦ Then p(-1) = (-1-1) = -2
                p(1) = (1-1) = 0
    ♦ So (x+1) is not a factor. But (x-1) is a factor
    ♦ If  (x2-1) is a factor, then both (x+1) and (x-1) should be factors
    ♦ Thus (x2-1) is not a factor
• If n = 2, p(x) = (xn-1) = (x- 1)
    ♦ Then p(-1) = ((-1)2-1) = (1-1) = 0
                p(1) = (12-1) = (1-1) = 0
    ♦ So (x+1) is a factor. Also (x-1) is a factor
    ♦ If  (x2-1) is a factor, then both (x+1) and (x-1) should be factors
    ♦ Thus (x2-1) is a factor
• If n = 3, p(x) = (xn-1) = (x- 1)
    ♦ Then p(-1) = ((-1)3-1) = (-1-1) = -2
                p(1) = (13-1) = (1-1) = 0
    ♦ So (x+1) is not a factor. But (x-1) is a factor
    ♦ If  (x2-1) is a factor, then both (x+1) and (x-1) should be factors
    ♦ Thus (x2-1) is not a factor
• If n = 4, p(x) = (xn-1) = (x- 1)
    ♦ Then p(-1) = ((-1)4-1) = (1-1) = 0
                p(1) = (14-1) = (1-1) = 0
    ♦ So (x+1) is a factor. Also (x-1) is a factor
    ♦ If  (x2-1) is a factor, then both (x+1) and (x-1) should be factors
    ♦ Thus (x2-1) is a factor
_    _    _    _
_    _    _    _
■ So for all even natural numbers 2, 4, 6, 8,  . . . , 
(x2-1) will be a factor of (x- 1)

Solved example 35.10
Prove that if (x2 - 1) is a factor of (ax+ bx2 + cx + d), then a = -c and b = -d
Solution:
1. (ax+ bx2 + cx + d) is a third degree polynomial. So there will be three first degree factors. We can write it as:
(ax+ bx2 + cx + d) = a[x+ (ba)x2 + (ca)x + (da)] = a[(x+b1) × (x+b2× (x+b3)]
⟹ [x+ (ba)x2 + (ca)x + (da)] = [(x+b1) × (x+b2× (x+b3)]
2. Given that (x2 - 1is a factor. 
This (x2 - 1) can be split as [(x+1) × (x-1)]
3. So we have two out of the three factors. We can write:
[x+ (ba)x2 + (ca)x + (da)] = [(x+1) × (x-1) × (x+b3)]
4. Let p(x) = [x+ (ba)x2 + (ca)x + (da)]
• (x+1) is a factor of p(x). So we get: p(-1) = 0
• We can write:
[(-1)+ (ba)(-1)2 + (ca)(-1) + (da)] = 0
⟹ [-1 + (ba) - (ca) + (da)] = 0
⟹ [(ba) + (da)] = [1 +  (ca)]
5. Similarly, (x-1) is a factor of p(x). So we get: p(1) = 0
• We can write:
[(1)+ (ba)(1)2 + (ca)(1) + (da)] = 0
[1 + (ba) + (ca) + (da)] = 0
[{1 (ca)} + {(ba) + (da)}] = 0
6. But from (4), we have: (ba) + (da) =  1  (ca)
substituting this in (5), we get:
[{1 (ca)} + {1  (ca)}] = 0
⟹ [{2 + 2(ca)}] = 0  ⟹ [1 + (ca)] = 0
⟹ [ca] = -1 ⟹ c = -a ⟹ a = -c
7. Substituting a = -c in (4), we get:
[(ba) + (da)] = [1 (ca)]
⟹ [-(bc) - (dc)] = [1 (cc)⟹ [-(b+d)c] = [1-1] = 0
⟹ -(b+d) = 0 ⟹ (b+d) = 0 ⟹ b = -d

Solved example 35.11
What first degree polynomial added to (2x - 3x2 + 5x + 1) gives a multiple of (x2-1)?
Solution:
1. A first degree polynomial is to be added to (2x - 3x2 + 5x + 1)
• The new polynomial obtained after the addition will be a multiple of (x2-1)
• So before the addition, it is not a multiple of (x2-1)
• So before the addition, if we divide by (x2-1), there will be a quotient q(x) and a remainder 'r'
2. Then we can write:
(2x - 3x2 + 5x + 1) = {q(x) × (x2-1)} + r
• Bringing r to the left side, we get:
{(2x - 3x2 + 5x + 1) - r} = {q(x) × (x2-1)} 
3. Now the left side as a whole is a multiple of (x2-1)
• That means (x2-1) is a factor of {(2x - 3x2 + 5x + 1) - r}
4. So we have to add '-r' to (2x - 3x2 + 5x + 1)
• (x2-1) will be a factor of the 'new polynomial obtained after the addition'
5. We have to find '-r'. It is given that '-r' is a first degree polynomial
• The general form of the first degree polynomial is (ax+b)
• So we can write: -r = (ax+b)  
6. Thus (3) becomes:
• (x2-1) is a factor of {(2x - 3x2 + 5x + 1) + (ax+b)}
7. (x2-1) can be split up as [(x+1)(x-1)]
So if (x2-1) is a factor of {(2x - 3x2 + 5x + 1) + (ax+b)}, Then:   
(i) (x+1) is a factor of {(2x - 3x2 + 5x + 1) + (ax+b)} 
(ii) (x-1) is a factor of {(2x - 3x2 + 5x + 1) + (ax+b)} 
8. (i) If (x+1) is a factor, then p(-1) = 0
So we get: {(2×(-1) - 3×(-1)2 + 5×(-1) + 1) + (a×-1+b)} = 0
{-2 - 3 + -5 + 1 - a+b} = 0 ⟹ {-9-a+b} = 0 ⟹ (a-b) = -9
(ii) If (x-1) is a factor, then p(1) = 0
So we get: {(2×1- 3×12 + 5×1 + 1) + (a×1+b)} = 0
{2 - 3 + 5 + 1 + a+b} = 0 ⟹ {5+a+b} = 0 ⟹ (a+b) = -5
9. Adding 8(i) and 8(ii), we get: 2a = -14  a = -7
• Subtracting 8(ii) from 8(i), we get: -b-b = -9 - (-5) ⟹ -2b = -4 ⟹ b = 2
• So the polynomial to be added = -r = (ax+b) = (-7×x+2) = (-7x+2)
10. The polynomial obtained after addition will be:
{(2x - 3x2 + 5x + 1) + (ax+b)} = {(2x - 3x2 + 5x + 1) + (-7x+2)} = (2x - 3x2 - 2x + 3)

Solved example 35.12
Check whether (4x+ 4x2 - x - 1) is a multiple of (2x+1)
Solution:
• We have to check whether (4x+ 4x2 - x - 1) is a multiple of (2x+1)
• That is., we have to check whether (2x+1) is a multiple of (4x+ 4x2 - x - 1)
1. To check whether (ax+b1is a factor, we need to check whether (x+b1ais a factor
• This is based on theorem 35.4 written in a previous section 
• In our case, (ax+b1) = (2x+1) 
    ♦ So we get: a = 2 and b= 1
• Then (x+b1a) = (x+(12)) 
• So (b1a) = (12 (-b1a) = (-12)
2. If (x+b1is a factor, then p(-b1) = 0
• Similarly, if (x+b1ais a factor, then p(-b1a) = 0
3. (4x+ 4x2 - x - 1) can be written as: [4(xx2 - (14)x - 14)]  
• Consider (xx2 - (14)x - 14):
• p(-b1a) = p(-12) = (-12)(-12)2 - {(14)×(-12)} - 14 
= -1118 - 14 = 0
• p(-12) = 0. So (x+(12)) is a factor
• So (2x+1) is a factor


In the next section, we will see a discussion on Probability.


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