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) (x3 - 1), (x-1)
(ii) (x3 - 1), (x+1)
(iii) (x3 + 1), (x-1)
(iv) (x3 + 1), (x+1)
Solution:
Part (i): (x3 - 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) = (13 - 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} + 0
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor = (x+b1) = (x-1)
• So b1 = -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i) (b2+b3) = 1
(ii) (-1×b2 + b2b3 + -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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 1] = [x2 + x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 1
Part (ii): (x3 - 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)3 - 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} - 2
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]} -2
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor = (x+b1) = (x+1)
• So b1 = 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i) (b2+b3) = -1
(ii) (1×b2 + b2b3 + 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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (-1)x + 1] = [x2 - x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = -1
♦ from 3(ii) we have: (b2b3) = 1
Part (iii): (x3 + 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) = (13 + 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:
(x3 + 1) = {(x+b1) × (x+b2) × (x+b3)} + 2
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]} + 2
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor = (x+b1) = (x-1)
• So b1 = -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i) (b2+b3) = 1
(ii) (-1×b2 + b2b3 + -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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 1] = [x2 + x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 1
Part (iv): (x3 + 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)3 + 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} + 0
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor = (x+b1) = (x+1)
• So b1 = 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i) (b2+b3) = -1
(ii) (1×b2 + b2b3 + 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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (-1)x + 1] = [x2 - x + 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) = (x3 + 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) = (13 + 12 + 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} = (x3 + 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)3 + (-1)2 - 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} = (x3 + 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), (xn -1) (ii) (x-1), (xn +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), (xn - 1)
■ If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x1 - 1) = (x-1)
♦ Then p(1) = (1-1) = 0
♦ So (x-1) is a factor
• If n = 2, p(x) = (x2 - 1)
♦ Then p(1) = (12-1) = 0
♦ So (x-1) is a factor
• If n = 3, p(x) = (x3 - 1)
♦ Then p(1) = (13-1) = 0
♦ So (x-1) is a factor
• If n = 4, p(x) = (x4 - 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 (xn - 1)
Part (ii): (x-1), (xn +1)
■ If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x1 + 1) = (x+1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 2, p(x) = (x2 + 1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 3 p(x) = (x3 + 1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 4, p(x) = (x4 + 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 (xn - 1)
Part (iii): (x+1), (xn-1)
■ If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x1 - 1) = (x-1)
♦ Then p(-1) = (-1-1) = -2
♦ So (x+1) is not a factor
• If n = 2, p(x) = (x2 - 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) = (x4 - 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 (xn - 1)
Part (iv): (x+1), (xn+1)
■ If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x1 + 1) = (x+1)
♦ Then p(-1) = (-1+1) = 0
♦ So (x+1) is a factor
• If n = 2, p(x) = (x2 + 1)
♦ Then p(-1) = ((-1)2+1) = (1+1) = 2
♦ So (x+1) is not a factor
• If n = 3, p(x) = (x3 + 1)
♦ Then p(-1) = ((-1)3+1) = (-1+1) = 0
♦ So (x+1) is a factor
• If n = 4, p(x) = (x2 + 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 (xn + 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) = (x1 - 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) = (x2 - 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) = (x3 - 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) = (x4 - 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 (xn - 1)
Solved example 35.10
Prove that if (x2 - 1) is a factor of (ax3 + bx2 + cx + d), then a = -c and b = -d
Solution:
1. (ax3 + bx2 + cx + d) is a third degree polynomial. So there will be three first degree factors. We can write it as:
(ax3 + bx2 + cx + d) = a[x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = a[(x+b1) × (x+b2) × (x+b3)]
⟹ [x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = [(x+b1) × (x+b2) × (x+b3)]
2. Given that (x2 - 1) is 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:
[x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = [(x+1) × (x-1) × (x+b3)]
4. Let p(x) = [x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)]
• (x+1) is a factor of p(x). So we get: p(-1) = 0
• We can write:
[(-1)3 + (b⁄a)(-1)2 + (c⁄a)(-1) + (d⁄a)] = 0
⟹ [-1 + (b⁄a) - (c⁄a) + (d⁄a)] = 0
⟹ [(b⁄a) + (d⁄a)] = [1 + (c⁄a)]
5. Similarly, (x-1) is a factor of p(x). So we get: p(1) = 0
• We can write:
[(1)3 + (b⁄a)(1)2 + (c⁄a)(1) + (d⁄a)] = 0
⟹[1 + (b⁄a) + (c⁄a) + (d⁄a)] = 0
⟹[{1 + (c⁄a)} + {(b⁄a) + (d⁄a)}] = 0
6. But from (4), we have: (b⁄a) + (d⁄a) = 1 + (c⁄a)
substituting this in (5), we get:
[{1 + (c⁄a)} + {1 + (c⁄a)}] = 0
⟹ [{2 + 2(c⁄a)}] = 0 ⟹ [1 + (c⁄a)] = 0
⟹ [c⁄a] = -1 ⟹ c = -a ⟹ a = -c
7. Substituting a = -c in (4), we get:
[(b⁄a) + (d⁄a)] = [1 + (c⁄a)]
⟹ [-(b⁄c) - (d⁄c)] = [1 - (c⁄c)] ⟹ [-(b+d)⁄c] = [1-1] = 0
⟹ -(b+d) = 0 ⟹ (b+d) = 0 ⟹ b = -d
Solved example 35.11
What first degree polynomial added to (2x3 - 3x2 + 5x + 1) gives a multiple of (x2-1)?
Solution:
1. A first degree polynomial is to be added to (2x3 - 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:
(2x3 - 3x2 + 5x + 1) = {q(x) × (x2-1)} + r
• Bringing r to the left side, we get:
{(2x3 - 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 {(2x3 - 3x2 + 5x + 1) - r}
4. So we have to add '-r' to (2x3 - 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 {(2x3 - 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 {(2x3 - 3x2 + 5x + 1) + (ax+b)}, Then:
(i) (x+1) is a factor of {(2x3 - 3x2 + 5x + 1) + (ax+b)}
(ii) (x-1) is a factor of {(2x3 - 3x2 + 5x + 1) + (ax+b)}
8. (i) If (x+1) is a factor, then p(-1) = 0
So we get: {(2×(-1)3 - 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: {(2×13 - 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:
{(2x3 - 3x2 + 5x + 1) + (ax+b)} = {(2x3 - 3x2 + 5x + 1) + (-7x+2)} = (2x3 - 3x2 - 2x + 3)
Solved example 35.12
Check whether (4x3 + 4x2 - x - 1) is a multiple of (2x+1)
Solution:
• We have to check whether (4x3 + 4x2 - x - 1) is a multiple of (2x+1)
• That is., we have to check whether (2x+1) is a multiple of (4x3 + 4x2 - x - 1)
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 in a previous section
• In our case, (ax+b1) = (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. (4x3 + 4x2 - x - 1) can be written as: [4(x3 + x2 - (1⁄4)x - 1⁄4)]
• Consider (x3 + x2 - (1⁄4)x - 1⁄4):
• p(-b1⁄a) = p(-1⁄2) = (-1⁄2)3 + (-1⁄2)2 - {(1⁄4)×(-1⁄2)} - 1⁄4
= -1⁄8 + 1⁄4 + 1⁄8 - 1⁄4 = 0
• p(-1⁄2) = 0. So (x+(1⁄2)) is a factor
• So (2x+1) is a factor
In the next section, we will see a discussion on Probability.
Solved example 35.7
For each pair of polynomials below, find the quotient and remainder on dividing the first by the second
(i) (x3 - 1), (x-1)
(ii) (x3 - 1), (x+1)
(iii) (x3 + 1), (x-1)
(iv) (x3 + 1), (x+1)
Solution:
Part (i): (x3 - 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) = (13 - 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} + 0
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor = (x+b1) = (x-1)
• So b1 = -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i) (b2+b3) = 1
(ii) (-1×b2 + b2b3 + -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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 1] = [x2 + x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 1
Part (ii): (x3 - 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)3 - 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} - 2
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]} -2
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor = (x+b1) = (x+1)
• So b1 = 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i) (b2+b3) = -1
(ii) (1×b2 + b2b3 + 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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (-1)x + 1] = [x2 - x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = -1
♦ from 3(ii) we have: (b2b3) = 1
Part (iii): (x3 + 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) = (13 + 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:
(x3 + 1) = {(x+b1) × (x+b2) × (x+b3)} + 2
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]} + 2
⟹ x3 + 0x2 + 0x - 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = -1
3. We have: Divisor = (x+b1) = (x-1)
• So b1 = -1
• Substituting b1 = -1 in the equations obtained in (2), we get:
(i) (b2+b3) = 1
(ii) (-1×b2 + b2b3 + -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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (1)x + 1] = [x2 + x + 1]
♦ ∵ from 3(i) we have: (b2+b3) = 1
♦ from 3(ii) we have: (b2b3) = 1
Part (iv): (x3 + 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)3 + 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:
(x3 - 1) = {(x+b1) × (x+b2) × (x+b3)} + 0
⟹ x3 + 0x2 + 0x + 1 = {(x+b1) × [x2 + (b2+b3)x + b2b3]}
• Comparing the coefficients, we get:
(i) (b1+b2+b3) = 0
(ii) (b1b2 + b2b3 + b1b3) = 0
(iii) b1b2b3 = 1
3. We have: Divisor = (x+b1) = (x+1)
• So b1 = 1
• Substituting b1 = 1 in the equations obtained in (2), we get:
(i) (b2+b3) = -1
(ii) (1×b2 + b2b3 + 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) = [x2 + (b2+b3)x + b2b3]
= [x2 + (-1)x + 1] = [x2 - x + 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) = (x3 + 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) = (13 + 12 + 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} = (x3 + 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)3 + (-1)2 - 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} = (x3 + 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), (xn -1) (ii) (x-1), (xn +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), (xn - 1)
■ If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x1 - 1) = (x-1)
♦ Then p(1) = (1-1) = 0
♦ So (x-1) is a factor
• If n = 2, p(x) = (x2 - 1)
♦ Then p(1) = (12-1) = 0
♦ So (x-1) is a factor
• If n = 3, p(x) = (x3 - 1)
♦ Then p(1) = (13-1) = 0
♦ So (x-1) is a factor
• If n = 4, p(x) = (x4 - 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 (xn - 1)
Part (ii): (x-1), (xn +1)
■ If (x-1) is a factor, p(1) = 0
• If n = 1, p(x) = (x1 + 1) = (x+1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 2, p(x) = (x2 + 1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 3 p(x) = (x3 + 1)
♦ Then p(1) = (1+1) = 2
♦ So (x-1) is not a factor
• If n = 4, p(x) = (x4 + 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 (xn - 1)
Part (iii): (x+1), (xn-1)
■ If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x1 - 1) = (x-1)
♦ Then p(-1) = (-1-1) = -2
♦ So (x+1) is not a factor
• If n = 2, p(x) = (x2 - 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) = (x4 - 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 (xn - 1)
Part (iv): (x+1), (xn+1)
■ If (x+1) is a factor, p(-1) = 0
• If n = 1, p(x) = (x1 + 1) = (x+1)
♦ Then p(-1) = (-1+1) = 0
♦ So (x+1) is a factor
• If n = 2, p(x) = (x2 + 1)
♦ Then p(-1) = ((-1)2+1) = (1+1) = 2
♦ So (x+1) is not a factor
• If n = 3, p(x) = (x3 + 1)
♦ Then p(-1) = ((-1)3+1) = (-1+1) = 0
♦ So (x+1) is a factor
• If n = 4, p(x) = (x2 + 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 (xn + 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) = (x1 - 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) = (x2 - 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) = (x3 - 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) = (x4 - 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 (xn - 1)
Solved example 35.10
Prove that if (x2 - 1) is a factor of (ax3 + bx2 + cx + d), then a = -c and b = -d
Solution:
1. (ax3 + bx2 + cx + d) is a third degree polynomial. So there will be three first degree factors. We can write it as:
(ax3 + bx2 + cx + d) = a[x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = a[(x+b1) × (x+b2) × (x+b3)]
⟹ [x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = [(x+b1) × (x+b2) × (x+b3)]
2. Given that (x2 - 1) is 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:
[x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)] = [(x+1) × (x-1) × (x+b3)]
4. Let p(x) = [x3 + (b⁄a)x2 + (c⁄a)x + (d⁄a)]
• (x+1) is a factor of p(x). So we get: p(-1) = 0
• We can write:
[(-1)3 + (b⁄a)(-1)2 + (c⁄a)(-1) + (d⁄a)] = 0
⟹ [-1 + (b⁄a) - (c⁄a) + (d⁄a)] = 0
⟹ [(b⁄a) + (d⁄a)] = [1 + (c⁄a)]
5. Similarly, (x-1) is a factor of p(x). So we get: p(1) = 0
• We can write:
[(1)3 + (b⁄a)(1)2 + (c⁄a)(1) + (d⁄a)] = 0
⟹[1 + (b⁄a) + (c⁄a) + (d⁄a)] = 0
⟹[{1 + (c⁄a)} + {(b⁄a) + (d⁄a)}] = 0
6. But from (4), we have: (b⁄a) + (d⁄a) = 1 + (c⁄a)
substituting this in (5), we get:
[{1 + (c⁄a)} + {1 + (c⁄a)}] = 0
⟹ [{2 + 2(c⁄a)}] = 0 ⟹ [1 + (c⁄a)] = 0
⟹ [c⁄a] = -1 ⟹ c = -a ⟹ a = -c
7. Substituting a = -c in (4), we get:
[(b⁄a) + (d⁄a)] = [1 + (c⁄a)]
⟹ [-(b⁄c) - (d⁄c)] = [1 - (c⁄c)] ⟹ [-(b+d)⁄c] = [1-1] = 0
⟹ -(b+d) = 0 ⟹ (b+d) = 0 ⟹ b = -d
Solved example 35.11
What first degree polynomial added to (2x3 - 3x2 + 5x + 1) gives a multiple of (x2-1)?
Solution:
1. A first degree polynomial is to be added to (2x3 - 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:
(2x3 - 3x2 + 5x + 1) = {q(x) × (x2-1)} + r
• Bringing r to the left side, we get:
{(2x3 - 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 {(2x3 - 3x2 + 5x + 1) - r}
4. So we have to add '-r' to (2x3 - 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 {(2x3 - 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 {(2x3 - 3x2 + 5x + 1) + (ax+b)}, Then:
(i) (x+1) is a factor of {(2x3 - 3x2 + 5x + 1) + (ax+b)}
(ii) (x-1) is a factor of {(2x3 - 3x2 + 5x + 1) + (ax+b)}
8. (i) If (x+1) is a factor, then p(-1) = 0
So we get: {(2×(-1)3 - 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: {(2×13 - 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:
{(2x3 - 3x2 + 5x + 1) + (ax+b)} = {(2x3 - 3x2 + 5x + 1) + (-7x+2)} = (2x3 - 3x2 - 2x + 3)
Solved example 35.12
Check whether (4x3 + 4x2 - x - 1) is a multiple of (2x+1)
Solution:
• We have to check whether (4x3 + 4x2 - x - 1) is a multiple of (2x+1)
• That is., we have to check whether (2x+1) is a multiple of (4x3 + 4x2 - x - 1)
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 in a previous section
• In our case, (ax+b1) = (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. (4x3 + 4x2 - x - 1) can be written as: [4(x3 + x2 - (1⁄4)x - 1⁄4)]
• Consider (x3 + x2 - (1⁄4)x - 1⁄4):
• p(-b1⁄a) = p(-1⁄2) = (-1⁄2)3 + (-1⁄2)2 - {(1⁄4)×(-1⁄2)} - 1⁄4
= -1⁄8 + 1⁄4 + 1⁄8 - 1⁄4 = 0
• p(-1⁄2) = 0. So (x+(1⁄2)) is a factor
• So (2x+1) is a factor
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