# ICSE Revision Notes for Factorisation of Polynomials Class 10 Maths

 Chapter Name Factorisation of Polynomials Topics Covered Remainder Theorem and its Application Proof of the Remainder Theorem Factor Theorem and its Applications Proof of the Factor TheoremFactorisation of Quadratic Polynomials using Factor Theorem and SplittingFactorisation of Cubic Polynomial using Factor Theorem Related Study

### Remainder Theorem and Its Application

Remainder Theorem

Consider two polynomials p(x) and q(x), where p(x) = 5x4 − 4x2 − 50 and q(x) = x − 2. We know how to divide p(x) by q(x) using the long division method. The result of this division will give the quotient as 5x3 + 10x2 +16x + 32 and the remainder as 14.

The long division method of finding the remainder is quite tedious. There is a simpler way to find the above remainder. This method is generalized in the form of a theorem called the remainder theorem. This theorem helps us find the remainder when a polynomial is to be divided by a linear polynomial.

In this lesson, we will study the remainder theorem and some of its applications in the form of examples.

Understanding the Remainder Theorem

Consider the division of a polynomial p(x) by a polynomial q(x), where p(x) = 5x4 − 4x2 − 50 and q(x) = x − 2. In this case, we have:

Dividend = p(x) and divisor = q(x

On dividing p(x) by q(x) using the long division method, we get:

Quotient = 5x3 + 10x2 +16x + 32 and remainder = 14

Now, let us find the value of p(x) at x = 2.

p(2) = 5 × 24 − 4 × 22 − 50

= 5 × 16 − 4 × 4 − 50

= 80 − 16 − 50

= 14

Note how the value of p(2) is the same as the remainder obtained by the long division of p(x) by q(x). Also observe how x = 2 is a zero of the polynomial q(x).

Thus, if we replace x in the dividend with the zero (or root) of the divisor, then we  get the remainder.

This method of finding the remainder is called the remainder theorem. It can be stated as follows:

For a polynomial p(x) of a degree greater than or equal to 1 and for any real  number a, if p(x) is divided by a linear polynomial x a, then the remainder will be p(a).

### Proof of the Remainder Theorem

Statement

For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(x) is divided by a linear polynomial x a, then the remainder will be p(a).

Proof

Let p(x) be a polynomial of a degree greater than or equal to 1 and a be any real number. When divided by x a, let p(x) leave the remainder r(x). Let q(x) be the quotient obtained.

Then, p(x) = (x a) q(x) + r(x), where r(x) = 0 or degree r(x) < degree (x a

Now, x a is a polynomial of degree 1; so, either r(x) = 0 or r(x) = constant (since a polynomial of degree less than 1 is a constant).

Let r(x) = constant = r (say). Then, p(x) = (x a) q(x) +

On putting x = a, we get p(a) = (a a) q(a) + r = 0 × q(a) + r = r

Thus, if p(x) is divided by x a, then the remainder will be p(a).

Notes:

1. If p(x) is divided by x + a, then the remainder will be p(−a).
2. If p(x) is divided by ax b, then the remainder will be P(b/a)  .
3. If p(x) is divided by ax + b, then the remainder will be P(-b/a) .

### Solved Examples

Example 1Find the remainder when x3 x2a + 5xa is divided by x a

Let p(x) = x3 x2a + 5xa

According to the remainder theorem, if p(x) is divided by x a, then the remainder will be p(a).

On putting x a = 0, we get x = a

∴ Remainder = p(a

= a3 a2 × a + 5 × a ×

= a3 a3 + 5a2

= 5a2

Thus, when x3 x2a + 5xa is divided by x a, we get 5a2 as the remainder.

Example 2What is the remainder when 81x4 + 54x3− 9x2− 3x + 2 is divided by 3x + 2?

Let p(x) = 81x4 + 54x3 9x2 3x + 2

As per the remainder theorem, if p(x) is divided by ax + b, then the remainder will be p(-b/a).

On putting 3x + 2 = 0, we get x = -(2/3).

∴ Remainder = p(-2/3)

= 81(-2/3)4 + 54(-2/3)3 – 9(-2/3)2 – 3(-2/3) + 2

= 81 × 16/81 – 54 × 8/27 - 9 × 4/9 + 3 × 2/3 + 2

= 16 – 16 – 4 + 2 + 2

= 0

Thus, when 81x4 + 54x3 − 9x2 − 3x + 2 is divided by 3x + 2, we get zero as the remainder.

Example 3Verify the remainder theorem for the division of 2x3 − 3x2 + 4 by x − 3.

Let p(x) = 2x3 − 3x2 + 4

Let us divide p(x) by x − 3 using the long division method.

Thus, the division of 2x3 − 3x2 + 4 by x − 3 yields the remainder 31.

Let us now find the remainder using the remainder theorem. According to this theorem, if p(x) is divided by x a, then the remainder will be p(a).

On putting x − 3 = 0, we get x = 3.

∴ Remainder = p(3)

= 2 × 33 − 3 × 32 + 4

= 54 − 27 + 4

= 31

Clearly, the remainder obtained by using the remainder theorem is the same as that obtained via the long division method. Hence, the remainder theorem is verified.

Example 4: For what value of m is p(x) = mx3 + 17x2− 31− 2m completely divisible by 3x + 1.

It is given that p(x) = mx 3 + 17x2 − 31x − 2

If p(x) is completely divisible by ax + b, then the remainder will be zero, i.e., p(- b/a) = 0.

On putting 3x + 1 = 0, we get x = -1/3 .

Using the remainder theorem, we can find the value of m as follows:

p(-1/3) = 0

⇒ m(-1/3)3 + 17(-1/3)2 – 31(-1/3) – 2m = 0

⇒ m(-1/27) + 17 × 1/9 + 31/3 – 2m = 0

⇒ - m/27 – 2m + 17/9 + 31/3 = 0

⇒ (-m – 54m)/27 + (17 + 93)/9 = 0

⇒ - (55m/27) + (110/9) = 0

⇒ - 55m/27 = - (110/9)

⇒ m = - 110/9 × (-27/55)

m = 6

Thus, when m = 6, mx3 + 17x2 − 31x − 2m is completely divisible by 3x + 1.

Example 5: Find the value of k for which p(x) = 4kx3 − 13x − 3k + 2

(i) is exactly divisible by 2x − 1.

(ii) leaves 3 as the remainder when divided by 2x + 3.

(i) We have p(x) = 4kx3 − 13x − 3k + 2

As per the remainder theorem, if p(x) is divided by ax b, then the remainder will be p(b/a).

On putting 2x − 1 = 0, we get x = 1/2.

Now, if 4kx3 − 13x − 3k + 2 is exactly divisible by 2x − 1, then the remainder will be zero, i.e., p(1/2) = 0.

Using the remainder theorem, we can find the value of k as follows:

p(1/2) = 0

⇒ 4k(1/2)3 – 13(1/2) – 3k + 2 = 0

⇒ 4k × 1/8 – 13/2 – 3k + 2 = 0

⇒ k/2 – 13/2 – 3k + 2 = 0

⇒ -5k/2 – 9/2 = 0

⇒ - 5k/2 = 9/2

⇒ k = -9/5

Thus, when k = -9/5, 4kx3 − 13x − 3k + 2 is exactly divisible by 2x − 1.

(ii) As per the remainder theorem, if p(x) is divided by ax + b, then the remainder will be p(-b/a).

On putting 2x + 3 = 0, we get x = - 3/2.

It is given that division of 4kx3 – 13x – 3k + 2 by 2x + 3 yields the remainder 3.

i.e., p(- 3/2) = 3

Using the remainder theorem, we can find the value of k as follows:

p(-3/2) = 3

⇒ 4k (-3/2)3 – 13(-3/2) – 3k + 2 = 3

⇒ 4k × (-27/8) + 39/2 – 3k + 2 = 3

⇒ - 27k/2 + 39/2 – 3k + 2 = 3

⇒ - (33/2)k + 43/2 = 3

⇒ - (33/2)k = 3 – 43/2

⇒ - (33/2)k = - (37/2)

⇒ k = 37/33

Thus, when k = 37/33, the division of 4kx3 – 13x – 3k + 2 by 2x + 3 leaves 3 as the remainder.

Example 6Find the values of a and b for which p(x) = x3 + ax2 + bx − 20 leaves 0 and −2 as the remainders when divided by x − 5 and x − 3 respectively.

We have p(x) = x3 + ax2 + bx − 20

As per the remainder theorem, if p(x) is divided by x a, then the remainder will be p(a). On putting x − 5 = 0, we get x = 5.

On putting x − 3 = 0, we get x = 3.

Now, if the division of x3 + ax2 + bx − 20 by x − 5 leaves 0 as the remainder, then p(5) = 0.

⇒ 53 + a × 52 + b × 5 – 20 = 0

⇒ 125 + 25a + 5b – 20 = 0

⇒ 25a + 5b + 105 = 0

⇒ 5a + b = - 21 …(1)

Also, if the division of x3 + ax2 + bx − 20 by x − 3 leaves −2 as the remainder, then p(3) = −2.

⇒ 33 + a × 32 + b × 3 – 20 = - 2

⇒ 27 + 9a + 3b = - 2 + 20

⇒ 9a + 3b = 18 – 27

⇒ 9a + 3b = - 9

⇒ 3a + b = - 3 …(2)

On solving equations 1 and 2, we get:

5a + (−3 − 3a) = −21 (∵ b = −3 − 3a

⇒ 5a − 3a − 3 = −21

⇒ 2a = −21 + 3

⇒ 2a = −18

a = −9

Now, b = −3 − 3

b = − 3 − 3 × (−9)

b = − 3 + 27

b = 24

Thus, when a = −9 and b = 24, the divisions of x3 + ax2 + bx − 20 by x − 5 and x −3 leave 0 and −2 respectively as the remainders.

### Factor Theorem and its Applications

Factor Theorem

We know the relation between a number and its factor. If we divide 91 by 7, then we get 13 as the quotient and zero as the remainder. In this case, we say that 7 is a factor of

91 as the remainder is zero. Now, if we divide 107 by 9, then we get 11 as the quotient and 8 as the remainder. In this case, we say that 9 is not a factor of 107 as the remainder is not zero.

Thus, the relation between a number and its factor is given as follows:

If a number is completely divisible by another number, i.e., the remainder is zero, then the second number is a factor of the first number.

Similarly, a polynomial p(x) is said to be completely divisible by a polynomial q(x) if we get zero as the remainder on dividing p(x) by q(x). In this case, we say that q(x) is a factor of p(x).

We have studied the remainder theorem that helps us to find the remainder. Similarly, we have a factor theorem that helps us to determine whether or not a polynomial is a factor of another polynomial, without actually performing the division.

In this lesson, we will study the factor theorem and solve some problems based on it.

Understanding the Factor Theorem

We can easily determine whether a polynomial q(x) is a factor of a polynomial p(x) without performing the division. This can be done by using the factor theorem, which can be stated as follows:

For a polynomial p(x) of a degree greater than or equal to 1 and for any real number c

i) if p(c) = 0, then x c will be a factor of p(x) and

ii) if x c is a factor of p(x), then p(c) will be equal to zero.

Consider the polynomial, p(x) = x2 − 3x + 2.

On putting x = 2 in p(x), we get:

p(2) = 22 − 3 × 2 + 2

= 4 − 6 + 2

= 0

Thus, we can say that x − 2 is a factor of p(x), where 2 is a real number.

### Proof of the Factor Theorem

Statement

For a polynomial p(x) of a degree greater than or equal to 1 and for any real number c,

1. if p(c) = 0, then x c will be a factor of p(x) and
2. if x c is a factor of p(x), then p(c) will be equal to zero.

Proof

Let p(x) be a polynomial of a degree greater than or equal to 1 and c be any real number such that p(c) = 0. Let quotient q(x) be obtained when p(x) is divided by x c

(i) p(c) = 0

By the remainder theorem, the remainder obtained is p(c).

p(x) = (x c) q(x) + p(c

p(x) = (x c) q(x) [∵ p(c) = 0]

x c is a factor of p(x).

(ii) x c is a factor of p(x

⇒ When divided by x c, p(x) leaves zero as the remainder.

However, by the remainder theorem, the remainder obtained is p(c). ⇒ p(c) = 0

Notes

1. x + c will be a factor of p(x) if p(−c) = 0
2. cx d will be a factor of p(x) if p(d/c) = 0
3. cx + d will be a factor of p(x) if p(-d/c) = 0
4. (x c) (x d) will be a factor of p(x) if p(c) = 0 and p(d) = 0

### Example Based on the Theorem

Solved Examples

Example 1Check whether or not x − 1 is a factor of x3 − 2x2 x + 2.

Let p(x) = x3 − 2x2 x + 2

According to the factor theorem, x − 1 will be a factor of p(x) if p(1) = 0.

p(1)= 13 − 2 × 12 − 1 + 2

= 1 − 2 − 1 + 2

= 0

Thus, x − 1 is a factor of x3 − 2x2 x + 2.

Example 2Using the factor theorem, show that 2x + 1 is a factor of 2x3 + 3x2− 11x− 6.

Let p(x) = 2x3 + 3x2 − 11x − 6

According to the factor theorem, 2x + 1 will be a factor of p(x) if p(-1/2) = 0.

p(-1/2) = 2(-1/2)3 + 3(-1/2)2 – 11(-1/2) – 6

= 1/4 + 3/4 + 11/2 – 6

= (-1 + 3 + 22 – 24)/4

= 0

Thus, 2x + 1 is a factor of 2x3 + 3x2 − 11x − 6.

Example 3For what value of m is x − 3 a factor of 3x2 − 3x + m

Let p(x) = 3x2 − 3x +

According to the factor theorem, x − 3 will be a factor of p(x) if p(3) = 0. p(3) = 3 × 32 − 3 × 3 +

So, 3 × 32 − 3 × 3 + m = 0

⇒ 3 × 9 − 9 + m = 0

⇒ 27 − 9 + m = 0

⇒ 18 + m = 0

m = −18

Thus, x − 3 is a factor of 3x2 − 3x + m when m = −18.

Example 4Check whether or not 2x2 − 11x + 25 is exactly divisible by 2x − 3.

Let p(x) = 2x2 − 11x + 25 and q(x) = 2x − 3

We know that p(x) will be exactly divisible by q(x) if q(x) is a factor of p(x).

On putting 2x − 3 = 0, we get x = 3/2.

On using the factor theorem, we get:

p(3/2) = 2(3/2)2 – 11(3/2) + 25

= 9/2 – 33/2 + 25

= 13 ≠ 0

Thus, q(x) is not a factor of p(x).

Hence, 2x2 − 11x + 25 is not exactly divisible by 2x − 3.

Example 5: Using the factor theorem, determine whether or not g(x) is a factor of f(x), where f(x) = 7x2 − 2√8x − 6 and g(x) = x − √2.

It is given that f(x) = 7x2 − 2√8x − 6 and g(x) = x − √2

According to the factor theorem, g(x) will be a factor of f(x) if f(√2) = 0 .

f(√2)= 7(√2)2 − 2√8 × √2− 6

= 7 × 2 − 2√16 − 6

= 14 − 8 − 6

= 0

Therefore, g(x) is a factor of f(x).

Example 6Using the factor theorem, show that ab, bc and ca are factors of a (b2 c2) + b (c2a2) + c (a2b2).

We have the given expression as a (b2 c2) + b (c2 a2) + c (a2 b2).

As per the factor theorem, x k will be a factor of a polynomial p(x) if p(x) = 0 when x = k

Let us consider p(a) = a (b2 c2) + b (c2 a2) + c (a2 b2) to be a polynomial in variable ‘a’. Take b and c as constants for the time being.

Now, as per the factor theorem, a b will be a factor of p(a) if p(a) = 0 when a = b On putting a = b in p(a), we get:

b (b2 c2) + b (c2 b2) + c (b2 b2

= b3 bc2 + bc2 b3 + c × 0

= 0

Thus, a b is a factor of a (b2 c2) + b (c2 a2) + c (a2 b2).

Now, suppose p(b) = a (b2 c2) + b (c2 a2) + c (a2 b2) is a polynomial in variable ‘b’ and a and c are constants. Then, b c will be a factor of p(b) if p(b) = 0 when b = c

On substituting b = c in p(b), we find that the result is zero.

Similarly, we can take p(c) = a (b2 c2) + b (c2 a2) + c (a2 b2) to be a polynomial in variable ‘c’ and a and b as constants. Then, c a will be a factor of p(c) if p(c) = 0 when c = a. On substituting c = a in p(c), we find that the result is zero.

Hence, b c and c a are also factors of a (b2 c2) + b (c2 a2) + c (a2 b2).

### Factorisation of Quadratic Polynomials u sing Factor Theorem and Splitting

Factorisation of Quadratic Polynomials

We know that 7 × 6 = 42. Here, 7 and 6 are factors of 42. Now, consider the linear polynomials x − 2 and x + 1. On multiplying the two, we get: x (x + 1) − 2 (x + 1) = x2 + x − 2x − 2 = x2 x − 2, which is a quadratic polynomial. So, x − 2 and x + 1 are

factors of the quadratic polynomial

x2 x − 2. A quadratic polynomial can have a maximum of two factors.

In the above example, we found the quadratic polynomial from its two factors. We can also find the factors from the quadratic polynomial. This process of decomposing a polynomial into a product of its factors (which when multiplied give the original expression) is called factorisation

There are two ways of finding the factors of quadratic polynomials viz., by applying the factor theorem and by splitting the middle term. We will discuss these methods of factorisation in this lesson and also solve some examples based on them.

Factorisation of Quadratic Polynomials Using the Factor Theorem

The factor theorem states that: For a polynomial p(x) of a degree greater than or equal to 1 and for any real number a, if p(a) = 0, then x a will be a factor of p(x).

Consider the quadratic polynomial, p(x) = x2 − 5x + 6. To find its factors, we need to ascertain the value of x for which the value of the polynomial comes out to be zero. For this, we first determine the factors of the constant term in the polynomial, and then check the value of the polynomial at these points.

In the given polynomial, the constant term is 6 and its factors are ±1, ±2, ±3 and ±6.

Let us now check the value of the polynomial for each of these factors of 6.

p(1) = 12 − 5 × 1 + 6 = 1 − 5 + 6 = 2 ≠ 0

Hence, x − 1 is not a factor of p(x).

p(2) = 22 − 5 × 2 + 6 = 4 − 10 + 6 = 0

Hence, x − 2 is a factor of p(x).

p(3) = 32 − 5 × 3 + 6 = 9 − 15 + 6 = 0

Hence, x − 3 is also a factor of p(x).

We know that a quadratic polynomial can have a maximum two factors which are already obtained as: (x − 2) and (x − 3).

Thus, the given polynomial = p(x) = x2 − 5x + 6 = (x − 2) (x − 3)

Solved Examples

Example 1Factorise x2 − 7x + 10 using the factor theorem.

Let p(x) = x2 − 7x + 10

The constant term is 10 and its factors are ±1, ±2, ±5 and ±10.

Let us check the value of the polynomial for each of these factors of 10. p(1) = 12 − 7 × 1 + 10 = 1 − 7 + 10 = 4 ≠ 0

Hence, x − 1 is not a factor of p(x).

p(2) = 22 − 7 × 2 + 10 = 4 − 14 + 10 = 0

Hence, x − 2 is a factor of p(x).

p(5) = 52 − 7 × 5 + 10 = 25 − 35 + 10 = 0

Hence, x − 5 is a factor of p(x).

We know that a quadratic polynomial can have a maximum of two factors. We have obtained the two factors of the given polynomial, which are x − 2 and x − 5.

Thus, we can write the given polynomial as:

p(x) = x2 − 7x + 10 = (x − 2) (x − 5)

Example 2Factorise x4y2 − 5x2y2 + 6y2

x4y2 − 5x2y2 + 6y2 = y2(x4 − 5x2 + 6)

Let x2 = a

⇒ (x2)2 = a2

x4 = a2

x4y2 − 5x2y2 + 6y2 = y2(a2 − 5a + 6)

= y2 × f(a), where f(a) = a2 − 5a + 6

Here, f(a) is a quadratic polynomial and the factors of the constant term ‘6’ are ±1, ±2, ±3 and ±6.

f(1) = 12 − 5 × 1 + 6 = 1 − 5 + 6 = 2 ≠ 0

Thus, a − 1 is not a factor of f(a).

f(2) = 22 − 5 × 2 + 6 = 4 − 10 + 6 = 0

Thus, a − 2 is a factor of f(a).

f(3) = 32 − 5 × 3 + 6 = 9 − 15 + 6 = 0

Thus, a − 3 is a factor of f(a).

We know that a quadratic polynomial can have a maximum of two factors. We have obtained the two factors of the given polynomial, which are a − 2 and a − 3.

Thus, we can write the given polynomial as:

f(a) = a2 − 5a + 6 = (a − 2) (a − 3)

Hence, x4y2 − 5x2y2 + 6y2 = y2(a − 2) (a − 3)

= y2(x2 − 2) (x2 − 3)

Example 3Factorise 4x (y2 + x − 1 + 3/x) + y2(y2 − 2)

4x (y2 + x − 1 + 3/x) + y2(y2 − 2) − 20

= 4xy2 + 4x2 − 4x + 12 + (y2)2 − 2y2 − 20

= (2x)2 + (y2)2 + 2 × 2x × y2 − 4x − 2y2 + 12 − 20

= (2x + y2)2 − 2 (2x + y2) − 8

= a2 − 2a − 8

= f(a), where a = 2x + y2

Here, f(a) is a quadratic polynomial and the factors of the constant term ‘8’ are ±1, ±2, ±4 and ±8.

f(1) = 12 − 2 × 1 − 8 = 1 − 2 − 8 = −9 ≠ 0

Thus, a − 1 is not a factor of f(a).

f(−1) = (−1)2 − 2 × (−1) − 8 = 1 + 2 − 8 = −5 ≠ 0

Thus, a + 1 is not a factor of f(a).

f(2) = 22 − 2 × 2 − 8 = 4 − 4 − 8 = −8 ≠ 0

Thus, a − 2 is not a factor of f(a).

f(−2) = (−2)2 − 2 × (−2) − 8 = 4 + 4 − 8 = 0

Thus, a + 2 is a factor of f(a).

f(4) = 42 − 2 × 4 − 8 = 16 − 8 − 8 = 0

Thus, a − 4 is a factor of f(a).

We know that a quadratic polynomial can have a maximum of two factors. We have obtained the two factors of the given polynomial, which are a + 2 and a − 4.

Thus, we can write the given polynomial as:

f(a) = a2 − 2a − 8 = (a + 2) (a − 4)

Hence, 4x (y2 + x − 1 + 3/x) + y2(y2 − 2) − 20 = (2x + y2 + 2) (2x + y2 − 4)

Example 4Factorise 12x2 − √2x − 12 by splitting the middle term.

The given polynomial is 12x2 − √2x − 12.

Here, ac = 12 × (−12) = −144. The middle term is- √2 .

Therefore, we have to split - √2 into two numbers such that their product is −144 and their sum is - √2.

These numbers are −9 √2 and 8 √2 (∵ −9 √2 + 8 √2 = - √2 and −9 √2 × 8 √2 = −144). Thus, we have:

12x2 − √2x − 12 = 12x2 − 9√2x + 8√2x − 12

= 3√2 x (2√2x − 3) + 4 (2√2 x − 3)

= (2√2x − 3) (3√2 x + 4)

Example 5Factorise 2x2 − 11x + 15 by splitting the middle term.

The given polynomial is 2x2 − 11x + 15.

Here, ac = 2 × 15 = 30. The middle term is −11. Therefore, we have to split −11 into two numbers such that their product is 30 and their sum is −11. These numbers are −5 and −6

[∵ (−5) + (−6) = −11 and (−5) × (−6) = 30].

Thus, we have:

2x2 − 11x + 15 = 2x2 − 5x − 6x + 15

= x (2x − 5) − 3 (2x − 5)

= (2x − 5) (x − 3)

Example 6: Factorise (3y − 1)2 − 6y + 2.

(3y − 1)2 − 6y + 2 = 9y2 + 1 − 6y − 6y + 2

= 9y2 − 12y + 3

= 3 (3y2 − 4y + 1)

Here, ac = 1 × 3 = 3. The middle term is −4. Therefore, we have to split −4 into two numbers such that their product is 3 and their sum is −4. These numbers are −1 and −3 [∵ (−3) + (−1) = −4 and (−3) × (−1) = 3].

Thus, we have:

3 (3y2 − 4y + 1) = 3 (3y2 − 3y y + 1)

= 3 [3y (y − 1) − 1 (y − 1)]

= 3 (y − 1) (3y − 1)

Example 7Find the dimensions of a rectangle whose area is given by the polynomial 20p2 + 69p + 54.

We know that area of a rectangle = Length × Breadth

Area of the rectangle is given by the polynomial 20p2 + 69p + 54. So, its factors will be the required dimensions of the rectangle.

In the given polynomial, ac = 20 × 54 = 1080. The middle term is 69. Therefore, we have to split 69 into two numbers such that their product is 1080 and their sum is 69. These numbers are 45 and 24 (∵ 45 + 24 = 69 and 45 × 24 = 1080).

Thus, we have:

20p2 + 69p + 54 = 20p2 + 45p + 24p + 54

= 5p (4p + 9) + 6 (4p + 9)

= (4p + 9) (5p + 6)

Hence, the dimensions of the rectangle are 5p + 6 and 4p + 9.

Example 8: Factorise: 2(3x + 4/5x)2 + 19(3x + 4/5x + 9/19).

2(3x + 4/5x)2 + 19(3x + 4/5x + 9/19)

= 2(3x + 4/5x)2 + 19(3x + 4/5x) + 9

= 2(3x + 4/5x)2 + 18(3x + 4/5x) + (3x + 4/5x) + 9

= 2(3x + 4/5x) [(3x + 4/5x) + 9] + 1[(3x + 4/5x) + 9]

= [(3x + 4/5x) + 9][2(3x + 4/5x) + 1]

= (3x + 4/5x + 9)(6x + 8/5x + 1)

### Factorisation of Cubic Polynomial Using Factor Theorem

Factorization of Cubic Polynomials

A cubic polynomial can be written as p(x) = ax3 + bx2 + cx + d, where a, b, c and d are real numbers. We cannot factorize a cubic polynomial in the manner in which we factorize a quadratic polynomial. We use a different approach for this purpose.

A cubic polynomial can have a maximum of three linear factors. By knowing one of these factors, we can reduce it to a quadratic polynomial. Thus, to factorize a cubic polynomial, we first find a factor by the hit and trial method or by using the factor theorem, and then reduce the cubic polynomial into a quadratic polynomial.

The resultant quadratic polynomial is solved by splitting its middle term or by using the factor theorem.

In this lesson, we will learn how to factorize a cubic polynomial and solve some examples related to the same.

Hit and trial method

Hit and trial method is used to find the factors or roots of a polynomial of degree more than two.

In this method, we put some value in the given polynomial to see if it satisfies the polynomial. If it does, then it is the zero of that polynomial. Using this method, we can reduce a polynomial of degree, say n, to a polynomial of degree n − 1.

Solved Examples

Example 1: Factorize x3 − 3x2 x + 3.

Let p(x) = x3 − 3x2 x + 3

The constant term is 3.

The factors of 3 are ±1 and ±3.

Let us take x = 1 and find the value of p(x).

p(1) = 13 − 3 × 12 − 1 + 3

= 1 − 3 − 1 + 3

= 0

Thus, x − 1 is a factor of p(x), using factor theorem.

Now, we have to group the terms of p(x) such that we can take x − 1 as common.

Thus, we have:

p(x) = x3 − 3x2 x + 3

= x3 x2 − 2x2 + 2x − 3x + 3

= x2(x − 1) − 2x (x − 1) − 3 (x − 1)

= (x − 1) (x2 − 2x − 3) … (1)

Next, we factorize x2 − 2x − 3 by splitting its middle term.

The middle term is −2. We have to find two numbers such that their product is −3 and their sum is 2. These two numbers are 3 and −1.

Thus, we have:

x2 − 2x − 3 = x2 − (3 − 1)x − 3

= x2 − 3x + x − 3

= x (x − 3) + 1 (x − 3)

= (x − 3) (x + 1)

On substituting in equation 1, we get:

p(x) = (x − 1) (x − 3) ( x + 1)

Example 2: If x + 3 is a factor of the polynomial f(x) = x3 − 7x + 6, then factorize f(x).

We have x + 3 as a factor of the polynomial f(x) = x3 + 0x2 − 7x + 6.

Let us divide f(x) by x + 3.

f(x) = x3 − 7x + 6 = (x + 3) (x2 − 3x + 2)

= (x + 3) (x2 x − 2x + 2)

= (x + 3) [x (x − 1) − 2 (x − 1)]

= (x + 3) (x − 1) (x − 2)

Example 3Factorize 2x37x2 + 7x − 2.

Let p(x) = 2x3 7x2 + 7x − 2

Let us take x = 1 and find the value of p(x).

p(1) = 2 × 13 7 × 12 + 7 × 1 − 2

= 2 − 7 + 7 − 2

= 0

Thus, x − 1 is a factor of p(x).

Now, we have to group the terms of p(x) such that we can take x − 1 as common.

Thus, we have:

p(x) = 2x3 2x2 − 5x2 + 5x + 2x − 2

= 2x2(x − 1) − 5x (x − 1) + 2 (x − 1)

=(x − 1) (2x2 5x + 2) … (1)

Next, we factorize 2x2 5x + 2 by splitting its middle term. The middle term is −5. We have to find two numbers such that their product is 4 and their sum is 5. These two numbers are 4 and 1.

Thus, we have:

2x2 5x + 2 = 2x2 (4 + 1)x + 2

= 2x2 4x − x + 2

= 2x (x − 2) − 1 (x − 2)

= (2x − 1) (x − 2)

On substituting in equation 1, we get:

p(x) = (x − 1) (2x − 1) (x − 2)

Example 4Factorize x3 − 23x2 + 142x − 120.

Let p(x) = x3 − 23x2 + 142x − 120

Let us take x = 1 and find the value of p(x).

p(1)= 13 − 23 × 12 + 142 × 1 − 120

= 1 − 23 + 142 − 120

= 0

Thus, x − 1 is a factor of p(x).

Now, we have to group the terms of p(x) such that we can take x − 1 as common.

Thus, we have:

p(x)= x3 − 23x2 + 142x − 120

= x3 x2 − 22x2 + 22x + 120x − 120

= x2(x −1) − 22x (x − 1) + 120 (x − 1)

= (x − 1) (x2 − 22x + 120) … (1)

Next, we factorize x2 − 22x + 120 by splitting its middle term. The middle term is −22.

We have to find two numbers such that their product is 120 and their sum is 22. These two numbers are 12 and 10.

Thus, we have:

x2 − 22x + 120= x2 − 12x − 10x + 120

= x (x − 12) − 10 (x − 12)

= (x − 12) (x − 10)

On substituting in equation (1), we get:

x3 − 23x2 − 142x − 120 = (x − 1) (x − 12) (x − 10)

Example 5Factorize the cubic polynomial p(x) = 6x3 + 5x2 − 12x + 4.

We have p(x) = 6x3 + 5x2 − 12x + 4

Let us take x = −2 and then find the value of p(x).

p(−2)= 6 × (−2)3 + 5 × (−2)2 − 12 × (−2) + 4

= −48 + 20 + 24 + 4

= 0

Thus, x + 2 is a factor of p(x).

Now, let us divide p(x) by x + 2.

∴ 6x3 + 5x2 − 12x + 4 = (x + 2) (6x2 − 7x + 2)

= (x + 2) (6x2 − 4x − 3x + 2)

= (x + 2) [2x (3x − 2) − 1 (3x − 2)]

= (x + 2) (3x − 2) (2x − 1)