In the previous section we saw how a second degree polynomial is split into it's factors. In this section we will see some solved examples.
Solved example 35.1
Write each polynomial below as a product of first degree polynomials. Write also the solutions of the equation p(x) = 0 in each case
(i) p(x) = x2 - 7x + 12
(ii) p(x) = x2 + 7x + 12
(iii) p(x) = x2 - 8x + 12
(iv) p(x) = x2 + 13x + 12
(v) p(x) = x2 - 2x + 1
(vi) p(x) = x2 + x - 1
(vii) p(x) = 2x2 - 5x + 2
(viii) p(x) = 6x2 - 7x + 2
Solution:
Part (i): p(x) = x2 - 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -7
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = -7
• Simplifying we get: [(b1)2 + 7b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -4 and b2 = -3
2. So we get:
• p(x) = [x2 - 7x + 12] = [(x-4)(x-3)]
• We can write: (x-4) and (x-3) are the factors of p(x) = [x2 - 7x + 12]
3. When p(x) = 0, the solutions are 4 and 3. That is:
p(4) will be zero
p(3) will be zero
• In other words, we must put 4 or 3 in the place of x, to make p(x) equal to zero
The geometric meaning of this situation can be seen in the fig.35.1 below:
• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero at x = 3 and x = 4
Part (ii): p(x) = x2 + 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 7
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = 7
• Simplifying we get: [(b1)2 - 7b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = 4 and b2 = 3
2. So we get:
p(x) = [x2 + 7x + 12] = [(x+4)(x+3)]
• We can write: (x+4) and (x+3) are the factors of p(x) = [x2 + 7x + 12]
⟹ (x-(-4)) and (x-(-3)) are the factors of p(x) = [x2 + 7x + 12]
3. When p(x) = 0, the solutions are -4 and -3. That is:
p(-4) will be zero
p(-3) will be zero
• In other words, we must put -4 or -3 in the place of x, to make p(x) equal to zero
Part (iii): p(x) = x2 - 8x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -8
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = -8
• Simplifying we get: [(b1)2 + 8b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2 and b2 = -6
2. So we get:
• p(x) = [x2 - 8x + 12] = [(x-2)(x-6)]
• We can write: (x-2) and (x-6) are the factors of p(x) = [x2 - 8x + 12]
3. When p(x) = 0, the solutions are 2 and 6. That is:
p(2) will be zero
p(6) will be zero
• In other words, we must put 2 or 6 in the place of x, to make p(x) equal to zero
Part (iv): p(x) = x2 + 13x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 13
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = 13
• Simplifying we get: [(b1)2 - 13b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = 12 and b2 = 1
2. So we get:
• p(x) = [x2 + 13x + 12] = [(x+12)(x+1)]
• We can write: (x+12) and (x+1) are the factors of p(x) = [x2 + 13x + 12]
⟹ (x-(-12)) and (x-(-1)) are the factors of p(x) = [x2 + 13x + 12]
3. When p(x) = 0, the solutions are -12 and -1. That is:
p(-12) will be zero
p(-1) will be zero
• In other words, we must put -12 or -1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.2 below:
• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero at x = -12 and x = -1
■ Note that the fig.35.2 is a little distorted because, the width of PC monitor is more than it's height. If we draw manually on a graph paper, we will get a perfect curve. However the fig. shows the geometric meaning of f(x) = 0
Part (v): p(x) = x2 - 2x + 1
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -2
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = -2
• Simplifying we get: [(b1)2 +2b1 + 1] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -1 and b2 = -1
2. So we get:
p(x) = [x2 - 2x + 1] = [(x-1)(x-1)]
• We can write: (x-1) and (x-1) are the factors of p(x) = [x2 - 2x + 1]
3. When p(x) = 0, the solutions are 1 and 1. That is:
p(1) will be zero
• In other words, we must put 1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.3 below:
• The magenta curve is a part of the plot of p(x). It's 'y value' becomes zero only at x = 1
Part (vi): p(x) = x2 + x - 1
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 1
(ii) b1b2 = -1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = -1⁄b1
(iv) Substituting this in (i) we get: (b1 + -1⁄b1) = 1
• Simplifying we get: [(b1)2 - b1 - 1] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = [1-(√5)]⁄2 and b2 = [1+(√5)]⁄2
2. So we get:
• p(x) = [x2 + x - 1] = [(x+{[1-(√5)]⁄2})(x+{[1+(√5)]⁄2})]
= [(x-{-[(√5)-1]⁄2})(x-{[(√5)+1]⁄2})]
• We can write: (x-{-[(√5)-1]⁄2}) and (x-{[(√5)+1]⁄2}) are the factors of p(x) = [x2 + x - 1]
3. When p(x) = 0, the solutions are {-[(√5)-1]⁄2} and {[(√5)+1]⁄2}. That is:
p({-[(√5)-1]⁄2}) will be zero
p({[(√5)+1]⁄2}) will be zero
• In other words, we must put {-[(√5)-1]⁄2} or {[(√5)+1]⁄2} in the place of x, to make p(x) equal to zero
Part (vii): p(x) = 2x2 - 5x + 2
The question can be modified as: 2[x2 - (5⁄2)x + 1]
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -5⁄2
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = -5⁄2
• Simplifying we get: [2(b1)2 + 5b1 + 2] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2 and b2 = -1⁄2
2. So we get:
p(x) = [x2 - (5⁄2)x + 1] = [(x-2)(x-(1⁄2))]
• We can write: (x-2) and (x-(1⁄2)) are the factors of p(x) = [x2 - (5⁄2)x + 1]
3. When p(x) = 0, the solutions are 2 and (1⁄2). That is:
p(2) will be zero
p(1⁄2) will be zero
• In other words, we must put 2 or (1⁄2) in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.4 below:
• The outer magenta curve is a part of the plot of p(x) = [x2 - (5⁄2)x + 1]
• The inner magenta curve is a part of the plot of p(x) = [2x2 - 5x + 2]
• In both cases, 'y value' becomes zero at x = 2 and x = 1⁄2
Part (viii): p(x) = 6x2 - 7x + 2
The question can be modified as: 6[x2 - (7⁄6)x + 1⁄3]
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -7⁄6
(ii) b1b2 = 1⁄3
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄3b1
(iv) Substituting this in (i) we get: (b1 + 1⁄3b1) = -7⁄6
• Simplifying we get: [18(b1)2 + 21b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2⁄3 and b2 = -1⁄2
2. So we get:
• p(x) = [x2 - (7⁄6)x + 1⁄3] = [(x-(2⁄3))(x-(1⁄2))]
• We can write: (x-(2⁄3)) and (x-(1⁄2)) are the factors of p(x) = [x2 - (7⁄6)x + 1⁄3]
3. When p(x) = 0, the solutions are (2⁄3) and (1⁄2). That is:
p(2⁄3) will be zero
p(1⁄2) will be zero
• In other words, we must put (2⁄3) or (1⁄2) in the place of x, to make p(x) equal to zero
Solved example 35.2
Find a second degree polynomial p(x) such that p(1) = 0 and p(-2) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put 1 in place of x, that polynomial will become zero
♦ So 1 is a solution of p(x)
♦ So (x-1) is a factor of p(x)
• If we put -2 in place of x, that polynomial will become zero
♦ So -2 is a solution of p(x)
♦ So (x-(-2)) is a factor of p(x)
⟹ (x+2) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = (x-1)(x+2)
= x2 + x - 2
Solved example 35.3
Find a second degree polynomial p(x) such that p(1+√3) = 0 and p(1-√3) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put (1+√3) in place of x, that polynomial will become zero
♦ So (1+√3) is a solution of p(x)
♦ So (x-(1+√3)) is a factor of p(x)
• If we put (1-√3) in place of x, that polynomial will become zero
♦ So (1-√3) is a solution of p(x)
♦ So (x-(1-√3)) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-(1+√3)][(x-(1-√3)]
5. Let A = (1+√3) and B = (1-√3)
Then we get: p(x) = [x-A][x-B]
= x2 - (A+B)x + AB
6. A+B = (1+√3) + (1-√3) = 2
AB = (1+√3)(1-√3) = [12-(√3)2] = [1-3] = -2
7. Then the result in (5) becomes:
p(x) = x2 - (A+B)x + AB
= x2 -2x - 2
Solved example 35.4
Find a third degree polynomial p(x) such that p(1) = 0, p(√2) = 0 and p(-√2) = 0
Solution:
1. Suppose we have a third degree polynomial p(x)
• If we put (1) in place of x, that polynomial will become zero
♦ So 1 is a solution of p(x)
♦ So (x-1) is a factor of p(x)
• If we put (√2) in place of x, that polynomial will become zero
♦ So (√2) is a solution of p(x)
♦ So (x-(√2)) is a factor of p(x)
• If we put (-√2) in place of x, that polynomial will become zero
♦ So (-√2) is a solution of p(x)
♦ So (x-(-√2)) is a factor of p(x) same as (x+(√2)) is a factor of p(x)
2. We got three factors of the third degree polynomial p(x)
• All the three factors are first degree polynomials
3. Any third degree polynomial will have only three factors
• And each of those factors will be a first degree polynomial
• In our case we obtained all the three factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-1][x-(√2)][(x+(√2)]
5. Let us first multiply the last two factors:
[x-(√2)][(x+(√2)] = [x2-(√2)2] = [x2-2]
Now we get:
• p(x) = [x-1][x2-2] = x3 - x2 -2x + 2
• The geometric representation of p(x) can be seen in the fig.35.5 below:
■ Note that the magenta curve meets the x axis at 3 points
• One point is on the negative side of the x axis.
♦ Between -1.5 and -1. It is closer to -1.5. This point is (x = -√2).
♦ We know that √2 = 1.414. That is why, the point is closer to 1.5
• Two points are on the positive side of the x axis
♦ One is at exact (x = 1)
♦ The other point is between 1 and 1.5, closer to 1.5. It is (x = √2)
Solved example 35.5
Prove that the polynomial x2 + x + 1 cannot be written as a product of first degree polynomials
Solution:
1. Let us first assume that the given polynomial can be written as a product of first degree polynomials
• Let those first degree polynomials be (x+b1) and (x+b2)
2. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 1
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = 1
• Simplifying we get: [(b1)2 + -b1 + 1] = 0
• This is a quadratic equation in (b1)
3. But there is no number that will satisfy [(b1)2 + -b1 + 1] = 0
• That means b1 does not exist
• So we cannot use 2(i) to obtain b2 either
• Both b1 and b2 does not exist.
• Consequently, there is no (x+b1) and (x+b2)
• Thus our assumption in (1) is wrong
In the next section, we will see polynomial remainder.
Solved example 35.1
Write each polynomial below as a product of first degree polynomials. Write also the solutions of the equation p(x) = 0 in each case
(i) p(x) = x2 - 7x + 12
(ii) p(x) = x2 + 7x + 12
(iii) p(x) = x2 - 8x + 12
(iv) p(x) = x2 + 13x + 12
(v) p(x) = x2 - 2x + 1
(vi) p(x) = x2 + x - 1
(vii) p(x) = 2x2 - 5x + 2
(viii) p(x) = 6x2 - 7x + 2
Solution:
Part (i): p(x) = x2 - 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -7
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = -7
• Simplifying we get: [(b1)2 + 7b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -4 and b2 = -3
2. So we get:
• p(x) = [x2 - 7x + 12] = [(x-4)(x-3)]
• We can write: (x-4) and (x-3) are the factors of p(x) = [x2 - 7x + 12]
3. When p(x) = 0, the solutions are 4 and 3. That is:
p(4) will be zero
p(3) will be zero
• In other words, we must put 4 or 3 in the place of x, to make p(x) equal to zero
The geometric meaning of this situation can be seen in the fig.35.1 below:
 |
| Fig.35.1 |
Part (ii): p(x) = x2 + 7x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 7
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = 7
• Simplifying we get: [(b1)2 - 7b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = 4 and b2 = 3
2. So we get:
p(x) = [x2 + 7x + 12] = [(x+4)(x+3)]
• We can write: (x+4) and (x+3) are the factors of p(x) = [x2 + 7x + 12]
⟹ (x-(-4)) and (x-(-3)) are the factors of p(x) = [x2 + 7x + 12]
3. When p(x) = 0, the solutions are -4 and -3. That is:
p(-4) will be zero
p(-3) will be zero
• In other words, we must put -4 or -3 in the place of x, to make p(x) equal to zero
Part (iii): p(x) = x2 - 8x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -8
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = -8
• Simplifying we get: [(b1)2 + 8b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2 and b2 = -6
2. So we get:
• p(x) = [x2 - 8x + 12] = [(x-2)(x-6)]
• We can write: (x-2) and (x-6) are the factors of p(x) = [x2 - 8x + 12]
3. When p(x) = 0, the solutions are 2 and 6. That is:
p(2) will be zero
p(6) will be zero
• In other words, we must put 2 or 6 in the place of x, to make p(x) equal to zero
Part (iv): p(x) = x2 + 13x + 12
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 13
(ii) b1b2 = 12
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 12⁄b1
(iv) Substituting this in (i) we get: (b1 + 12⁄b1) = 13
• Simplifying we get: [(b1)2 - 13b1 + 12] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = 12 and b2 = 1
2. So we get:
• p(x) = [x2 + 13x + 12] = [(x+12)(x+1)]
• We can write: (x+12) and (x+1) are the factors of p(x) = [x2 + 13x + 12]
⟹ (x-(-12)) and (x-(-1)) are the factors of p(x) = [x2 + 13x + 12]
3. When p(x) = 0, the solutions are -12 and -1. That is:
p(-12) will be zero
p(-1) will be zero
• In other words, we must put -12 or -1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.2 below:
 |
| Fig.35.2 |
■ Note that the fig.35.2 is a little distorted because, the width of PC monitor is more than it's height. If we draw manually on a graph paper, we will get a perfect curve. However the fig. shows the geometric meaning of f(x) = 0
Part (v): p(x) = x2 - 2x + 1
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -2
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = -2
• Simplifying we get: [(b1)2 +2b1 + 1] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -1 and b2 = -1
2. So we get:
p(x) = [x2 - 2x + 1] = [(x-1)(x-1)]
• We can write: (x-1) and (x-1) are the factors of p(x) = [x2 - 2x + 1]
3. When p(x) = 0, the solutions are 1 and 1. That is:
p(1) will be zero
• In other words, we must put 1 in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.3 below:
 |
| Fig.35.3 |
Part (vi): p(x) = x2 + x - 1
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 1
(ii) b1b2 = -1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = -1⁄b1
(iv) Substituting this in (i) we get: (b1 + -1⁄b1) = 1
• Simplifying we get: [(b1)2 - b1 - 1] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = [1-(√5)]⁄2 and b2 = [1+(√5)]⁄2
2. So we get:
• p(x) = [x2 + x - 1] = [(x+{[1-(√5)]⁄2})(x+{[1+(√5)]⁄2})]
= [(x-{-[(√5)-1]⁄2})(x-{[(√5)+1]⁄2})]
• We can write: (x-{-[(√5)-1]⁄2}) and (x-{[(√5)+1]⁄2}) are the factors of p(x) = [x2 + x - 1]
3. When p(x) = 0, the solutions are {-[(√5)-1]⁄2} and {[(√5)+1]⁄2}. That is:
p({-[(√5)-1]⁄2}) will be zero
p({[(√5)+1]⁄2}) will be zero
• In other words, we must put {-[(√5)-1]⁄2} or {[(√5)+1]⁄2} in the place of x, to make p(x) equal to zero
Part (vii): p(x) = 2x2 - 5x + 2
The question can be modified as: 2[x2 - (5⁄2)x + 1]
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -5⁄2
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = -5⁄2
• Simplifying we get: [2(b1)2 + 5b1 + 2] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2 and b2 = -1⁄2
2. So we get:
p(x) = [x2 - (5⁄2)x + 1] = [(x-2)(x-(1⁄2))]
• We can write: (x-2) and (x-(1⁄2)) are the factors of p(x) = [x2 - (5⁄2)x + 1]
3. When p(x) = 0, the solutions are 2 and (1⁄2). That is:
p(2) will be zero
p(1⁄2) will be zero
• In other words, we must put 2 or (1⁄2) in the place of x, to make p(x) equal to zero
• The geometric meaning of this situation can be seen in the fig.35.4 below:
 |
| Fig.35.4 |
• The inner magenta curve is a part of the plot of p(x) = [2x2 - 5x + 2]
• In both cases, 'y value' becomes zero at x = 2 and x = 1⁄2
Part (viii): p(x) = 6x2 - 7x + 2
The question can be modified as: 6[x2 - (7⁄6)x + 1⁄3]
1. From the coefficients in the given equation, we can write:
(i) (b1+b2) = -7⁄6
(ii) b1b2 = 1⁄3
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄3b1
(iv) Substituting this in (i) we get: (b1 + 1⁄3b1) = -7⁄6
• Simplifying we get: [18(b1)2 + 21b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get:
b1 = -2⁄3 and b2 = -1⁄2
2. So we get:
• p(x) = [x2 - (7⁄6)x + 1⁄3] = [(x-(2⁄3))(x-(1⁄2))]
• We can write: (x-(2⁄3)) and (x-(1⁄2)) are the factors of p(x) = [x2 - (7⁄6)x + 1⁄3]
3. When p(x) = 0, the solutions are (2⁄3) and (1⁄2). That is:
p(2⁄3) will be zero
p(1⁄2) will be zero
• In other words, we must put (2⁄3) or (1⁄2) in the place of x, to make p(x) equal to zero
Solved example 35.2
Find a second degree polynomial p(x) such that p(1) = 0 and p(-2) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put 1 in place of x, that polynomial will become zero
♦ So 1 is a solution of p(x)
♦ So (x-1) is a factor of p(x)
• If we put -2 in place of x, that polynomial will become zero
♦ So -2 is a solution of p(x)
♦ So (x-(-2)) is a factor of p(x)
⟹ (x+2) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = (x-1)(x+2)
= x2 + x - 2
Solved example 35.3
Find a second degree polynomial p(x) such that p(1+√3) = 0 and p(1-√3) = 0
Solution:
1. Suppose we have a second degree polynomial p(x)
• If we put (1+√3) in place of x, that polynomial will become zero
♦ So (1+√3) is a solution of p(x)
♦ So (x-(1+√3)) is a factor of p(x)
• If we put (1-√3) in place of x, that polynomial will become zero
♦ So (1-√3) is a solution of p(x)
♦ So (x-(1-√3)) is a factor of p(x)
2. We got two factors of the second degree polynomial p(x)
• Both the factors are first degree polynomials
3. Any second degree polynomial will have only two factors
• And each of those factors will be a first degree polynomial
• In our case we obtained both the factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-(1+√3)][(x-(1-√3)]
5. Let A = (1+√3) and B = (1-√3)
Then we get: p(x) = [x-A][x-B]
= x2 - (A+B)x + AB
6. A+B = (1+√3) + (1-√3) = 2
AB = (1+√3)(1-√3) = [12-(√3)2] = [1-3] = -2
7. Then the result in (5) becomes:
p(x) = x2 - (A+B)x + AB
= x2 -2x - 2
Solved example 35.4
Find a third degree polynomial p(x) such that p(1) = 0, p(√2) = 0 and p(-√2) = 0
Solution:
1. Suppose we have a third degree polynomial p(x)
• If we put (1) in place of x, that polynomial will become zero
♦ So 1 is a solution of p(x)
♦ So (x-1) is a factor of p(x)
• If we put (√2) in place of x, that polynomial will become zero
♦ So (√2) is a solution of p(x)
♦ So (x-(√2)) is a factor of p(x)
• If we put (-√2) in place of x, that polynomial will become zero
♦ So (-√2) is a solution of p(x)
♦ So (x-(-√2)) is a factor of p(x) same as (x+(√2)) is a factor of p(x)
2. We got three factors of the third degree polynomial p(x)
• All the three factors are first degree polynomials
3. Any third degree polynomial will have only three factors
• And each of those factors will be a first degree polynomial
• In our case we obtained all the three factors
• If we multiply those factors, we will get the original polynomial
4. So we can write:
p(x) = [x-1][x-(√2)][(x+(√2)]
5. Let us first multiply the last two factors:
[x-(√2)][(x+(√2)] = [x2-(√2)2] = [x2-2]
Now we get:
• p(x) = [x-1][x2-2] = x3 - x2 -2x + 2
• The geometric representation of p(x) can be seen in the fig.35.5 below:
 |
| Fig.35.5 |
• One point is on the negative side of the x axis.
♦ Between -1.5 and -1. It is closer to -1.5. This point is (x = -√2).
♦ We know that √2 = 1.414. That is why, the point is closer to 1.5
• Two points are on the positive side of the x axis
♦ One is at exact (x = 1)
♦ The other point is between 1 and 1.5, closer to 1.5. It is (x = √2)
Solved example 35.5
Prove that the polynomial x2 + x + 1 cannot be written as a product of first degree polynomials
Solution:
1. Let us first assume that the given polynomial can be written as a product of first degree polynomials
• Let those first degree polynomials be (x+b1) and (x+b2)
2. From the coefficients in the given equation, we can write:
(i) (b1+b2) = 1
(ii) b1b2 = 1
Where b1 and b2 are the required solutions when p(x) = 0
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = 1
• Simplifying we get: [(b1)2 + -b1 + 1] = 0
• This is a quadratic equation in (b1)
3. But there is no number that will satisfy [(b1)2 + -b1 + 1] = 0
• That means b1 does not exist
• So we cannot use 2(i) to obtain b2 either
• Both b1 and b2 does not exist.
• Consequently, there is no (x+b1) and (x+b2)
• Thus our assumption in (1) is wrong
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