In the previous section we completed a discussion on coordinate geometry. In this chapter we will see polynomials. We have seen the basics about polynomials in a previous chapter 20. This discussion is a continuation of the discussion that we had in that chapter.
• On many occasions, we have seen 'difference of squares of two numbers'.
• To explain it using algebra, we can use 'x' and 'y'.
♦ Let x be any number. Then square of that number is x2
♦ Let y be any number. Then square of that number is y2
• Difference between the squares is (x2 - y2)
• We have seen that this difference is the product of two quantities:
(i) Sum of the two numbers, which is (x+y)
(ii) Difference of the two numbers, which is (x-y)
That is., (x2-y2) = (x+y) (x-y)
• We have learned the above equation in the form of an identity.
♦ It helps to avoid lengthy calculations while solving many problems.
• In the identity, let us put some real values for 'y'.
♦ When y = 1, We get: (x2-1) = (x2-12) = (x+1)(x-1)
♦ When y = 2, We get: (x2-2) = [x2- (√2)2] = (x+√2)(x-√2)
♦ When y = 1⁄4, We get: (x2-1⁄4) = [x2- (1⁄2)2] = (x+1⁄2)(x-1⁄2)
Etc.,
• In the above examples, (x2-1), (x2-2), (x2-1/4) are all second degree polynomials
• (x+1), (x-1), (x+2), (x-2), (x+1⁄2), (x-1⁄2) are all first degree polynomials
♦ We have seen the conditions for polynomials here.
• So in each of the examples above, a second degree polynomial is written as the 'product of two first degree polynomials'
• Now consider the simple equation: 12 = 4×3
♦ Here, 4 and 3 are factors of 12
• In the same way, (x2-1) = (x+1)(x-1)
♦ (x+1) and (x-1) are factors of (x2-1)
In general we can write:
■ If a polynomial p(x) is the product of the polynomials q(x) and r(x)
THEN:
q(x) and r(x) are said to be factors of p(x)
Another example:
• Consider the product of (x-1) and (x-2). We have:
(x-1)×(x-2) = x2 - 3x + 2
• That is.,
x2 - 3x + 2 = (x-1)(x-2)
• So (x-1) and (x-2) are factors of (x2 -3x +2)
• Note that:
♦ The product (x2 -3x +2) is a second degree polynomial
♦ Both the factors (x-1) and (x-2) are first degree polynomials
• We can write: p(x) = x2 - 3x + 2
Then what is p(1) ?
♦ We get: p(1) = (12 - 3×1 + 2) = (1 - 3 + 2) = (-2+2) = 0
♦ Similarly p(2) = (22 - 3×2 + 2) = (4 - 6 + 2) = (-2+2) = 0
• But we can write (x2 - 3x + 2) as the product of two factors also:
p(x) = x2 - 3x + 2 = (x-1) (x-2)
♦ Then p(1) = (1-1)(1-2) = (0×-1) = 0
♦ Similarly p(2) = (2-1)(2-2) = (1×0) = 0
• From the factor (x-1), we get a number 1
• From the factor (x-2), we get a number 2
• Both these numbers can be used to make (x2 - 3x + 2) equal to zero
• Is there any other method to obtain the numbers '1' and '2'?
Let us try:
1. We want to make (x2 - 3x + 2) equal to zero.
• So let us write: x2 - 3x + 2 = 0
■ We get the same numbers
Let us try the same steps for another example:
• Consider the product of (x-4) and (x-7). We have:
(x-4)×(x-7) = x2 - 11x + 28
• That is.,
x2 - 11x + 28 = (x-4)(x-7)
• So (x-4) and (x-7) are factors of (x2 -11x +28)
• Note that:
♦ The product (x2 -11x +28) is a second degree polynomial
♦ Both the factors (x-4) and (x-7) are first degree polynomials
• We can write: p(x) = x2 - 11x + 28
Then what is p(4) ?
♦ We get: p(4) = (42 - 11×4 + 28) = (16 - 44 + 28) = (-28+28) = 0
♦ Similarly p(7) = (72 - 11×7 + 28) = (49 - 77 + 28) = (-28+28) = 0
• But we can write (x2 - 11x + 28) as the product of two factors also:
p(x) = x2 - 11x + 28 = (x-4) (x-7)
♦ Then p(4) = (4-4)(4-7) = (0×-3) = 0
♦ Similarly p(7) = (7-4)(7-7) = (3×0) = 0
• From the factor (x-4), we get a number 4
• From the factor (x-7), we get a number 7
• Both these numbers can be used to make (x2 - 11x + 28) equal to zero
• Is there any other method to obtain the numbers '4' and '7'?
Let us try:
1. We want to make (x2 - 11x + 28) equal to zero.
• So let us write: x2 - 11x + 28 = 0
■ We get the same numbers
Let us write a summary of what we have seen in the above two examples:
Example 1:
• (x-1) and (x-2) are the two factors of (x2 - 3x + 2)
♦ If we put x = 1 in (x2 - 3x + 2), it will become zero
♦ If we put x = 2 in (x2 - 3x + 2), it will become zero
• So we can write:
1 and 2 are the solutions of the polynomial equation:
x2 - 3x + 2 = 0
Example 2:
• (x-4) and (x-7) are the two factors of (x2 - 11x + 28)
♦ If we put x = 4 in (x2 - 11x + 28), it will become zero
♦ If we put x = 7 in (x2 - 11x + 28), it will become zero
• So we can write:
4 and 7 are the solutions of the polynomial equation:
x2 - 11x + 28 = 0
In the above two examples, Two first degree polynomials are multiplied to get a second degree polynomials. Let us now see a third degree polynomial:
Example 3:
1. Consider the three first degree polynomials: (x-3), (x-7) and (x-2)
2. Let us multiply them: (x-3) × (x-7) × (x-2)
= (x2 - 10x + 21) × (x-2)
= x3 - 10x2 + 21x - 2x2 + 20x - 42
= x3 - 12x2 + 41x - 42
3. We can write: p(x) = x3 - 12x2 + 41x - 42 = (x-3) (x-7) (x-2)
4. p(3) = (33 - 12×32 + 41×3 - 42) = (27 - 108 + 123 - 42) = 0
• p(7) = (73 - 12×72 + 41×7 - 42) = (343 - 588 + 287 - 42) = 0
• p(2) = (23 - 12×22 + 41×2 - 42) = (8 - 48 + 82 - 42) = 0
5. Thus we get:
3, 7 and 2 are the solutions of the polynomial equation:
x3 - 12x2 + 41x - 42 = 0
We can put the above findings in the form of two theorems:
Theorem 35.1
If the first degree polynomial (x-a) is a factor of the polynomial p(x), THEN p(a) = 0
Theorem 35.2
If the polynomial p(x) can be split in the form:
p(x) = (x - a1) (x - a2) . . . (x - an)
That is., as a product of first degree polynomials,
THEN a1, a2, . . . , an are the solutions of the equation p(x) = 0
• Note that, all the factors must be of the first degree
• The above theorems gives us a method to find solutions of polynomial equations.
• All we need to do is:
Split the given polynomial into factors such that, all the factors are first degree polynomials.
• We will analyse it using an example:
1. Consider the polynomial equation: x2 - 5x + 6 = 0
2. We want to split (x2 - 5x + 6) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - 5x + 6)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - 5x + 6) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - 5x + 6) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - 5x + 6), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - 5x + 6) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - 5x + 6 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -5
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 6
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = -5
(ii) b1b2 = 6
(iii) From (ii) we get: b2 = 6⁄b1
(iv) Substituting this in (i) we get: (b1 + 6⁄b1) = -5
• Simplifying we get: [(b1)2 + 5b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = -2 or -3
• Substituting these values in (i) we get: b2 = -2 or -3
(v) So we can write:
b1 = -2 and b2 = -3
10. Then the result in step (5) becomes:
(x2 - 5x + 6) = (x + b1) (x + b2)
⟹ (x2 - 5x + 6) = (x-2) (x-3)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - 5x + 6) = 0
are: 2 and 3
■ The reader may now be wondering:
'Why go through all the trouble, when (x2 - 5x + 6) = 0 can be easily solved as a quadratic equation?'
• The answer is that, It is essential to learn this new method because, it is the basis for 'the method of solving polynomial equations of higher degrees'
Another example:
1. Consider the polynomial equation: x2 + 2x - 15 = 0
2. We want to split (x2 + 2x - 15) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 + 2x - 15)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 + 2x - 15) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 + 2x - 15) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 + 2x - 15), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 + 2x - 15) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 + 2x - 15 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is 2
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is -15
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = 2
(ii) b1b2 = -15
(iii) From (ii) we get: b2 = -15⁄b1
(iv) Substituting this in (i) we get: (b1 + -15⁄b1) = 2
• Simplifying we get: [(b1)2 - 2b1 - 15] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = 5 or -3
• Substituting these values in (i) we get: b2 = -3 or 5
(v) So we can write:
b1 = 5 and b2 = -3
10. Then the result in step (5) becomes:
(x2 + 2x - 15) = (x + b1) (x + b2)
⟹ (x2 + 2x - 15) = (x+5) (x-3)
⟹ (x2 + 2x - 15) = (x-(-5)) (x-3)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 + 2x - 15) = 0
are: -5 and 3
One more example:
1. Consider the polynomial equation: x2 - x - 1 = 0
2. We want to split (x2 - x - 1) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - x - 1)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - x - 1) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - x - 1) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - x - 1), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - x - 1) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - x - 1 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -1
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is -1
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = -1
(ii) b1b2 = -1
(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 = [(√5)-1]⁄2 or -[(√5)+1]⁄2
• Substituting these values in (i) we get: b2 = -[(√5)+1]⁄2 or [(√5)-1]⁄2
(v) So we can write:
b1 = [(√5)-1]⁄2 and b2 = -[(√5)+1]⁄2
10. Then the result in step (5) becomes:
(x2 - x - 1) = (x + b1) (x + b2)
⟹ (x2 - x - 1) = (x + [(√5)-1]⁄2) (x - [(√5)+1]⁄2)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - x - 1) = 0
are: -[(√5)-1]⁄2 and [(√5)+1]⁄2
A different type:
• Consider the polynomial equation: 2x2 - 7x + 6 = 0
♦ In this case, the coefficient of x2 is not '1'.
• So we have to modify the equation. We will write (2x2 - 7x + 6) as 2[x2 - (7⁄2)x + 3]
• We can use the usual steps for the portion within the square brackets. After obtaining it's results, we will do the necessary changes.
1. Consider the polynomial equation: x2 - (7⁄2)x + 3 = 0
2. We want to split (x2 - (7⁄2)x + 3) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - (7⁄2)x + 3)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - (7⁄2)x + 3) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - (7⁄2)x + 3) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - (7⁄2)x + 3), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - (7⁄2)x + 3) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - (7⁄2)x + 3 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -(7⁄2)
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 3
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = - (7⁄2)
(ii) b1b2 = 3
(iii) From (ii) we get: b2 = 3⁄b1
(iv) Substituting this in (i) we get: (b1 + 3⁄b1) = - (7⁄2)
• Simplifying we get: [2(b1)2 + 7b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = -2 or -3⁄2
• Substituting these values in (i) we get: b2 = -3⁄2 or -2
(v) So we can write:
b1 = -2 and b2 = -3⁄2
10. Then the result in step (5) becomes:
(x2 - (7⁄2)x + 3) = (x + b1) (x + b2)
⟹ (x2 - (7⁄2)x + 3) = (x -2) (x - 3⁄2)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - (7⁄2)x + 3) = 0
are: 2 and 3⁄2.
12. Now let us analyse the situation:
• What we wanted originally is the solution of the equation:
2x2 - 7x + 6 = 0
13. This can be written as:
We want the solution of the equation:
2[x2 - (7⁄2)x + 3] = 0
14. When x = 2, the equation in (13) will become zero
• When x = 3⁄2, the equation in (13) will become zero
• So 2 and 3⁄2 are the solutions of the equation in (13)
Consequently, 2 and 3⁄2 are the solutions of the equation in (12) also
In the above examples, we wrote the given second degree polynomials as the product of two first degree polynomials.
But all second degree polynomials cannot be split like this. An example is shown below. We will try the same procedure:
1. Consider the polynomial equation: x2 + 1 = 0
2. We want to split (x2 + 1) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 + 1)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 + 1) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 + 1) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 + 1), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 + 1) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 + 1 = x2 + (b1+b2)x + b1b2
⟹ x2 + 0x + 1 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is 0
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 1
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = 0
(ii) b1b2 = 1
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = 0
• Simplifying we get: [(b1)2 + 1] = 0
⟹ [(b1)2] = -1
• But we cannot take the square root of a negative number.
• So there are no numbers which will satisfy (b1+b2) = 0 and b1b2 = 1
10. That means we cannot split (x2 + 1) into two first degree polynomials (x + b1) and (x + b2)
We have seen a number of examples. We are now in a position to find the results using a lesser number of steps. A summary is given below:
1. b1 and b2 are the solutions when p(x) becomes equal to zero
2. (b1+b2) is equal to the coefficient of the x term in the given polynomial
• b1b2 is the constant term in the given polynomial
3. Using this information b1 and b2 can be calculated.
• In this calculation, we will encounter a quadratic equation
In the next section, we will see some solved examples.
• On many occasions, we have seen 'difference of squares of two numbers'.
• To explain it using algebra, we can use 'x' and 'y'.
♦ Let x be any number. Then square of that number is x2
♦ Let y be any number. Then square of that number is y2
• Difference between the squares is (x2 - y2)
• We have seen that this difference is the product of two quantities:
(i) Sum of the two numbers, which is (x+y)
(ii) Difference of the two numbers, which is (x-y)
That is., (x2-y2) = (x+y) (x-y)
• We have learned the above equation in the form of an identity.
♦ It helps to avoid lengthy calculations while solving many problems.
• In the identity, let us put some real values for 'y'.
♦ When y = 1, We get: (x2-1) = (x2-12) = (x+1)(x-1)
♦ When y = 2, We get: (x2-2) = [x2- (√2)2] = (x+√2)(x-√2)
♦ When y = 1⁄4, We get: (x2-1⁄4) = [x2- (1⁄2)2] = (x+1⁄2)(x-1⁄2)
Etc.,
• In the above examples, (x2-1), (x2-2), (x2-1/4) are all second degree polynomials
• (x+1), (x-1), (x+2), (x-2), (x+1⁄2), (x-1⁄2) are all first degree polynomials
♦ We have seen the conditions for polynomials here.
• So in each of the examples above, a second degree polynomial is written as the 'product of two first degree polynomials'
• Now consider the simple equation: 12 = 4×3
♦ Here, 4 and 3 are factors of 12
• In the same way, (x2-1) = (x+1)(x-1)
♦ (x+1) and (x-1) are factors of (x2-1)
In general we can write:
■ If a polynomial p(x) is the product of the polynomials q(x) and r(x)
THEN:
q(x) and r(x) are said to be factors of p(x)
Another example:
• Consider the product of (x-1) and (x-2). We have:
(x-1)×(x-2) = x2 - 3x + 2
• That is.,
x2 - 3x + 2 = (x-1)(x-2)
• So (x-1) and (x-2) are factors of (x2 -3x +2)
• Note that:
♦ The product (x2 -3x +2) is a second degree polynomial
♦ Both the factors (x-1) and (x-2) are first degree polynomials
• We can write: p(x) = x2 - 3x + 2
Then what is p(1) ?
♦ We get: p(1) = (12 - 3×1 + 2) = (1 - 3 + 2) = (-2+2) = 0
♦ Similarly p(2) = (22 - 3×2 + 2) = (4 - 6 + 2) = (-2+2) = 0
• But we can write (x2 - 3x + 2) as the product of two factors also:
p(x) = x2 - 3x + 2 = (x-1) (x-2)
♦ Then p(1) = (1-1)(1-2) = (0×-1) = 0
♦ Similarly p(2) = (2-1)(2-2) = (1×0) = 0
• From the factor (x-1), we get a number 1
• From the factor (x-2), we get a number 2
• Both these numbers can be used to make (x2 - 3x + 2) equal to zero
• Is there any other method to obtain the numbers '1' and '2'?
Let us try:
1. We want to make (x2 - 3x + 2) equal to zero.
• So let us write: x2 - 3x + 2 = 0
3. This is a second degree equation of the form ax2 + bx + c = 0 (Details here)
Where: a = 1, b = (-3) and c = 2
4. So we can use the general formula to solve the equation
5. b2-4ac = (-3)2-4×1×2 = 9 - 8 = 1
• So √[b2-4ac] = √1 = ±1
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-3)±1 = 3 ±1 = (3 +1) or (3 -1)
⟹ numerator = 4 or 2
• The denominator = 2a = 2×1 = 2
• Thus x = 4⁄2 or 2⁄2
⟹ x = 2 or 1• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-3)±1 = 3 ±1 = (3 +1) or (3 -1)
⟹ numerator = 4 or 2
• The denominator = 2a = 2×1 = 2
• Thus x = 4⁄2 or 2⁄2
■ We get the same numbers
Let us try the same steps for another example:
• Consider the product of (x-4) and (x-7). We have:
(x-4)×(x-7) = x2 - 11x + 28
• That is.,
x2 - 11x + 28 = (x-4)(x-7)
• So (x-4) and (x-7) are factors of (x2 -11x +28)
• Note that:
♦ The product (x2 -11x +28) is a second degree polynomial
♦ Both the factors (x-4) and (x-7) are first degree polynomials
• We can write: p(x) = x2 - 11x + 28
Then what is p(4) ?
♦ We get: p(4) = (42 - 11×4 + 28) = (16 - 44 + 28) = (-28+28) = 0
♦ Similarly p(7) = (72 - 11×7 + 28) = (49 - 77 + 28) = (-28+28) = 0
• But we can write (x2 - 11x + 28) as the product of two factors also:
p(x) = x2 - 11x + 28 = (x-4) (x-7)
♦ Then p(4) = (4-4)(4-7) = (0×-3) = 0
♦ Similarly p(7) = (7-4)(7-7) = (3×0) = 0
• From the factor (x-4), we get a number 4
• From the factor (x-7), we get a number 7
• Both these numbers can be used to make (x2 - 11x + 28) equal to zero
• Is there any other method to obtain the numbers '4' and '7'?
Let us try:
1. We want to make (x2 - 11x + 28) equal to zero.
• So let us write: x2 - 11x + 28 = 0
3. This is a second degree equation of the form ax2 + bx + c = 0 (Details here)
Where: a = 1, b = (-11) and c = 28
4. So we can use the general formula to solve the equation
5. b2-4ac = (-11)2-4×1×28 = 121 - 112 = 9
• So √[b2-4ac] = √9 = ±3
• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-11)±3 = 11 ±3 = (11 +3) or (11 -3)
⟹ numerator = 14 or 8
• The denominator = 2a = 2×1 = 2
• Thus x = 14⁄2 or 8⁄2
⟹ x = 7 or 4• This '±' sign is already present in the numerator in the formula
• The numerator is: -b±√[b2-4ac] = -(-11)±3 = 11 ±3 = (11 +3) or (11 -3)
⟹ numerator = 14 or 8
• The denominator = 2a = 2×1 = 2
• Thus x = 14⁄2 or 8⁄2
■ We get the same numbers
Let us write a summary of what we have seen in the above two examples:
Example 1:
• (x-1) and (x-2) are the two factors of (x2 - 3x + 2)
♦ If we put x = 1 in (x2 - 3x + 2), it will become zero
♦ If we put x = 2 in (x2 - 3x + 2), it will become zero
• So we can write:
1 and 2 are the solutions of the polynomial equation:
x2 - 3x + 2 = 0
Example 2:
• (x-4) and (x-7) are the two factors of (x2 - 11x + 28)
♦ If we put x = 4 in (x2 - 11x + 28), it will become zero
♦ If we put x = 7 in (x2 - 11x + 28), it will become zero
• So we can write:
4 and 7 are the solutions of the polynomial equation:
x2 - 11x + 28 = 0
In the above two examples, Two first degree polynomials are multiplied to get a second degree polynomials. Let us now see a third degree polynomial:
Example 3:
1. Consider the three first degree polynomials: (x-3), (x-7) and (x-2)
2. Let us multiply them: (x-3) × (x-7) × (x-2)
= (x2 - 10x + 21) × (x-2)
= x3 - 10x2 + 21x - 2x2 + 20x - 42
= x3 - 12x2 + 41x - 42
3. We can write: p(x) = x3 - 12x2 + 41x - 42 = (x-3) (x-7) (x-2)
4. p(3) = (33 - 12×32 + 41×3 - 42) = (27 - 108 + 123 - 42) = 0
• p(7) = (73 - 12×72 + 41×7 - 42) = (343 - 588 + 287 - 42) = 0
• p(2) = (23 - 12×22 + 41×2 - 42) = (8 - 48 + 82 - 42) = 0
5. Thus we get:
3, 7 and 2 are the solutions of the polynomial equation:
x3 - 12x2 + 41x - 42 = 0
We can put the above findings in the form of two theorems:
Theorem 35.1
If the first degree polynomial (x-a) is a factor of the polynomial p(x), THEN p(a) = 0
Theorem 35.2
If the polynomial p(x) can be split in the form:
p(x) = (x - a1) (x - a2) . . . (x - an)
That is., as a product of first degree polynomials,
THEN a1, a2, . . . , an are the solutions of the equation p(x) = 0
• Note that, all the factors must be of the first degree
• The above theorems gives us a method to find solutions of polynomial equations.
• All we need to do is:
Split the given polynomial into factors such that, all the factors are first degree polynomials.
• We will analyse it using an example:
1. Consider the polynomial equation: x2 - 5x + 6 = 0
2. We want to split (x2 - 5x + 6) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - 5x + 6)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - 5x + 6) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - 5x + 6) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - 5x + 6), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - 5x + 6) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - 5x + 6 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -5
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 6
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = -5
(ii) b1b2 = 6
(iii) From (ii) we get: b2 = 6⁄b1
(iv) Substituting this in (i) we get: (b1 + 6⁄b1) = -5
• Simplifying we get: [(b1)2 + 5b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = -2 or -3
• Substituting these values in (i) we get: b2 = -2 or -3
(v) So we can write:
b1 = -2 and b2 = -3
10. Then the result in step (5) becomes:
(x2 - 5x + 6) = (x + b1) (x + b2)
⟹ (x2 - 5x + 6) = (x-2) (x-3)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - 5x + 6) = 0
are: 2 and 3
■ The reader may now be wondering:
'Why go through all the trouble, when (x2 - 5x + 6) = 0 can be easily solved as a quadratic equation?'
• The answer is that, It is essential to learn this new method because, it is the basis for 'the method of solving polynomial equations of higher degrees'
Another example:
1. Consider the polynomial equation: x2 + 2x - 15 = 0
2. We want to split (x2 + 2x - 15) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 + 2x - 15)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 + 2x - 15) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 + 2x - 15) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 + 2x - 15), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 + 2x - 15) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 + 2x - 15 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is 2
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is -15
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = 2
(ii) b1b2 = -15
(iii) From (ii) we get: b2 = -15⁄b1
(iv) Substituting this in (i) we get: (b1 + -15⁄b1) = 2
• Simplifying we get: [(b1)2 - 2b1 - 15] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = 5 or -3
• Substituting these values in (i) we get: b2 = -3 or 5
(v) So we can write:
b1 = 5 and b2 = -3
10. Then the result in step (5) becomes:
(x2 + 2x - 15) = (x + b1) (x + b2)
⟹ (x2 + 2x - 15) = (x+5) (x-3)
⟹ (x2 + 2x - 15) = (x-(-5)) (x-3)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 + 2x - 15) = 0
are: -5 and 3
One more example:
1. Consider the polynomial equation: x2 - x - 1 = 0
2. We want to split (x2 - x - 1) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - x - 1)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - x - 1) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - x - 1) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - x - 1), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - x - 1) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - x - 1 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -1
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is -1
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = -1
(ii) b1b2 = -1
(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 = [(√5)-1]⁄2 or -[(√5)+1]⁄2
• Substituting these values in (i) we get: b2 = -[(√5)+1]⁄2 or [(√5)-1]⁄2
(v) So we can write:
b1 = [(√5)-1]⁄2 and b2 = -[(√5)+1]⁄2
10. Then the result in step (5) becomes:
(x2 - x - 1) = (x + b1) (x + b2)
⟹ (x2 - x - 1) = (x + [(√5)-1]⁄2) (x - [(√5)+1]⁄2)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - x - 1) = 0
are: -[(√5)-1]⁄2 and [(√5)+1]⁄2
A different type:
• Consider the polynomial equation: 2x2 - 7x + 6 = 0
♦ In this case, the coefficient of x2 is not '1'.
• So we have to modify the equation. We will write (2x2 - 7x + 6) as 2[x2 - (7⁄2)x + 3]
• We can use the usual steps for the portion within the square brackets. After obtaining it's results, we will do the necessary changes.
1. Consider the polynomial equation: x2 - (7⁄2)x + 3 = 0
2. We want to split (x2 - (7⁄2)x + 3) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 - (7⁄2)x + 3)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 - (7⁄2)x + 3) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 - (7⁄2)x + 3) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 - (7⁄2)x + 3), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 - (7⁄2)x + 3) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 - (7⁄2)x + 3 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is -(7⁄2)
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 3
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = - (7⁄2)
(ii) b1b2 = 3
(iii) From (ii) we get: b2 = 3⁄b1
(iv) Substituting this in (i) we get: (b1 + 3⁄b1) = - (7⁄2)
• Simplifying we get: [2(b1)2 + 7b1 + 6] = 0
• This is a quadratic equation in (b1)
Solving it we get: b1 = -2 or -3⁄2
• Substituting these values in (i) we get: b2 = -3⁄2 or -2
(v) So we can write:
b1 = -2 and b2 = -3⁄2
10. Then the result in step (5) becomes:
(x2 - (7⁄2)x + 3) = (x + b1) (x + b2)
⟹ (x2 - (7⁄2)x + 3) = (x -2) (x - 3⁄2)
11. Based on theorem 35.2 that we saw above, the solutions of (x2 - (7⁄2)x + 3) = 0
are: 2 and 3⁄2.
12. Now let us analyse the situation:
• What we wanted originally is the solution of the equation:
2x2 - 7x + 6 = 0
13. This can be written as:
We want the solution of the equation:
2[x2 - (7⁄2)x + 3] = 0
14. When x = 2, the equation in (13) will become zero
• When x = 3⁄2, the equation in (13) will become zero
• So 2 and 3⁄2 are the solutions of the equation in (13)
Consequently, 2 and 3⁄2 are the solutions of the equation in (12) also
In the above examples, we wrote the given second degree polynomials as the product of two first degree polynomials.
But all second degree polynomials cannot be split like this. An example is shown below. We will try the same procedure:
1. Consider the polynomial equation: x2 + 1 = 0
2. We want to split (x2 + 1) into factors
• Each factor must be a first degree polynomial
3. How many such factors will there be for (x2 + 1)?
Ans: Only two factors
Because, if there are 'more than two', the degree of (x2 + 1) will also be 'more than two'
4. Recall how we write a first degree polynomial in a general form. It is:
ax+b (Details here)
5. So let the two factors of (x2 + 1) be:
(a1x + b1) and (a2x + b2)
• In the given polynomial (x2 + 1), the coefficient of x2 is 1
• So both a1 and a2 must be 1
♦ Other wise we will not get 1 as the coefficient of x2
• So the factors can be written as: (x + b1) and (x + b2)
• Thus we can write:
(x2 + 1) = (x + b1) (x + b2)
6. Their product is obtained as:
(x + b1) (x + b2) = x2 + (b1+b2)x + b1b2.
7. So we can write:
x2 + 1 = x2 + (b1+b2)x + b1b2
⟹ x2 + 0x + 1 = x2 + (b1+b2)x + b1b2
8. We have learned about terms, coefficients and variable parts. (Details here)
We now compare the terms on both sides:
• The coefficient of 'x2 term' on the left side is 1
♦ The coefficient of 'x2 term' on the right side is 1
• The coefficient of 'x term' on the left side is 0
♦ The coefficient of 'x term' on the right side is (b1+b2)
• The 'constant term' on the left side is 1
♦ The 'constant term' on the right side is b1b2
9. based on the comparison, we can write:
(i) (b1+b2) = 0
(ii) b1b2 = 1
(iii) From (ii) we get: b2 = 1⁄b1
(iv) Substituting this in (i) we get: (b1 + 1⁄b1) = 0
• Simplifying we get: [(b1)2 + 1] = 0
⟹ [(b1)2] = -1
• But we cannot take the square root of a negative number.
• So there are no numbers which will satisfy (b1+b2) = 0 and b1b2 = 1
10. That means we cannot split (x2 + 1) into two first degree polynomials (x + b1) and (x + b2)
1. b1 and b2 are the solutions when p(x) becomes equal to zero
2. (b1+b2) is equal to the coefficient of the x term in the given polynomial
• b1b2 is the constant term in the given polynomial
3. Using this information b1 and b2 can be calculated.
• In this calculation, we will encounter a quadratic equation
• For that, we can use the general formula
In the next section, we will see some solved examples.
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