In this part, we will look at the Factor Theorem, which uses the remainder theorem and learn how to factorise polynomials. Further, we will be covering the splitting method and the factor theorem method.

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## Factor Theorem

If p(x) is a polynomial of degree n > 1 and a is any real number, then

- x – a is a factor of p(x), if p(a) = 0, and
- p(a) = 0, if x – a is a factor of p(x).

Let’s look at an example to understand this theorem better.

**Browse more Topics under Polynomials**

- Polynomial and its Types
- Value of Polynomial and Division Algorithm
- Degree of Polynomial
- Factorisation of Polynomials
- Remainder Theorem
- Zeroes of Polynomial
- Geometrical Representation of Zeroes of a Polynomial

### Example:

Examine whether x + 2 is a factor of x^{3} + 3x^{2} + 5x + 6.

Solution: To begin with, we know that the zero of the polynomial (x + 2) is –2. Let p(x) = x^{3} + 3x^{2} + 5x + 6

Then, p(–2) = (–2)^{3} + 3(–2)^{2} + 5(–2) + 6 = –8 + 12 – 10 + 6 = 0

According to the factor theorem, if p(a) = 0, then (x – a) is a factor of p(x). In this example, p(a) = p(- 2) = 0

Therefore, (x – a) = {x – (-2)} = (x + 2) is a factor of ‘x^{3} + 3x^{2} + 5x + 6’ or p(x).

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## Factorisation of polynomials

You can factorise polynomials by splitting the middle term as follows: to begin with, consider a polynomial ax^{2} + bx + c with factors (px + q) and (rx + s). Therefore, we have ax^{2} + bx + c = (px + q) (rx + s). So, ax^{2} + bx + c = prx^{2} + (ps + qr) x + qs

If we compare the coefficients of x^{2}, we get a = pr. Also, on comparing the coefficients of x, we get b = ps + qr. Finally, on comparing the constants, we get c = qs. Hence, b is the sum of two numbers ‘ps’ and ‘qr’, whose product is (ps)(qr) = (pr)(qs) = ac.

Therefore, to factorise ax^{2} + bx + c, we have to write b as the sum of two numbers whose product is ‘ac’. Let’s look at an example to understand this clearly.

### Example

Factorise 6×2 + 17x + 5 by splitting the middle term.

Solution 1 (By splitting method): As explained above, if we can find two numbers, ‘p’ and ‘q’ such that, p + q = 17 and pq = 6 x 5 = 30, then we can get the factors.

After looking at the factors of 30, we find that numbers ‘2’ and ‘15’ satisfy both the conditions, i.e. p + q = 2 + 15 = 17 and pq = 2 x 15 = 30. So,

6x^{2} + 17x + 5 = 6x^{2} + (2 + 15)x + 5

= 6x^{2} + 2x + 15x + 5

= 2x(3x + 1) + 5(3x + 1)

= (3x + 1) (2x + 5)

Therefore, the factors of (6x^{2} + 17x + 5) are (3x + 1) and (2x + 5) with a remainder, zero.

### Example

Factorise y^{2} – 5y + 6 by using the Factor Theorem.

Solution: Let p(y) = y^{2} – 5y + 6. Now, if p(y) = (y – a) (y – b), the constant term will be ab as can be seen below,

p(y) = (y – a)(y – b)

= y^{2} – by – ay + ab

On comparing the constants, we get ab = 6. Next, the factors of 6 are 1, 2 and 3. Now, p(2) = 2^{2}– (5 × 2) + 6 = 4 – 10 + 6 = 0. So, (y – 2) is a factor of p(y). Also, p(3) = 3^{2} – (5 × 3) + 6 = 9 – 15 + 6 = 0. So, (y – 3) is also a factor of y^{2} – 5y + 6. Therefore, y^{2} – 5y + 6 = (y – 2)(y – 3)

## More Solved Examples for You

**Question 1:** **Factorise x ^{3} – 23x^{2} + 142x – 120**

**Answer :** Let p(x) = x^{3} – 23x^{2} + 142x – 120. To begin with, we will start finding the factors of the constant ‘– 120’, which are:

±1, ±2, ±3, ±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±40, ±60 and ±120

Further, by trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x). Also, we see that

[x^{3} – 23x^{2} + 142x – 120] = x^{3} – x^{2} – 22x^{2} + 22x + 120x – 120

So, by removing the common factors, we have x^{3} – 23x^{2} + 142x – 120 = x^{2}(x –1) – 22x(x – 1) + 120(x – 1)

Further, taking ‘x – 1’ common, we get x^{3} – 23x^{2} + 142x – 120 = (x – 1) (x^{2} – 22x + 120)

Therefore, x^{3} – 23x^{2} + 142x – 120 = (x – 1) (x^{2} – 22x + 120)

Also, note that if we divide p(x) by ‘x – 1’, then the result will be (x^{2} – 22x + 120)

Going on, x^{2} – 22x + 120 can be factorised further. So, by splitting the middle term, we get:

x^{2} – 22x + 120 = x^{2} – 12x – 10x + 120 … [ (– 12 – 10 = – 22) and {(–12)( –10) = 120}]

= x(x – 12) – 10(x – 12)

= (x – 12) (x – 10)

Therefore, we have x^{3} – 23x^{2} – 142x – 120 = (x – 1)(x – 10)(x – 12)

**Question 2:** **Factorise x ^{3} – 2x^{2} – x + 2**

**Answer :** Let p(x) = x^{3} – 2x^{2} – x + 2. To begin with, we will start finding the factors of the constant ‘2’, which are: 1, 2

By trial, we find that p(1) = 0. Hence, we conclude that (x – 1) is a factor of p(x).

So, by removing the common factors, we have x^{3} – 2x^{2} – x + 2 = x^{2}(x – 2) – (x – 2) = (x^{2} – 1)(x – 2)

= (x + 1)(x – 1)(x – 2) … [Using the identity (x^{2} – 1) = (x + 1)(x – 1)]

Therefore, the factors of x^{3} – 2x^{2} – x + 2 are (x + 1), (x – 1) and (x – 2)

**Question 3: Explain factor theorem with example?**

**Answer:** An example of factor theorem can be the factorization of 6×2 + 17x + 5 by splitting the middle term. In this example, one can find two numbers, ‘p’ and ‘q’ in a way such that, p + q = 17 and pq = 6 x 5 = 30. After that one can get the factors.

**Question 4: What is meant by a polynomial factor?**

**Answer:** A factor of polynomial P(x) refers to any polynomial whose division takes place evenly into P(x). For example, x + 2 is a factor belonging to the polynomial x^{2} – 4. The polynomial’s factorization is its representation as a product of its various factors. A good example can be the factorization of x^{2} – 4 is (x – 2)(x + 2).

**Question 5: Explain the formula of factor theorem?**

**Answer: **The Factor Theorem explain us that if the remainder f(r) = R = 0, then (x − r) happens to be a factor of f(x). The Factor Theorem is quite important because of its usefulness to find roots of polynomial equations.

**Question 6: Is it possible for a remainder to be negative?**

**Answer:** No, a remainder can never be a negative number.