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# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.5 | Set 1

• Last Updated : 30 Apr, 2021

### Question 1. Using factor theorem, factorize of the polynomials: x3 + 6x2 + 11x + 6

Solution:

Given that, polynomial eqn., f(x) = x3 + 6x2 + 11x + 6

The constant term in f(x) is 6,

The factors of 6 are ± 1, ± 2, ± 3, ± 6

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x) and we get,

f(-1) = (−1)3 + 6(−1)2 + 11(−1) + 6

= – 1 + 6 – 11 + 6 = 12 – 12 = 0

So, (x + 1) is the factor of f(x)

Similarly, (x + 2) and (x + 3) are also the factors of f(x)

Since, f(x) is a polynomial having a degree 3, therefore it cannot have more than three linear factors.

Hence, f(x) = k(x + 1)(x + 2)(x + 3)

x3 + 6x2 + 11x + 6 = k(x + 1)(x + 2)(x + 3)

Substitute x = 0 on both the sides

0 + 0 + 0 + 6 = k(0 +1)(0 + 2)(0 + 3)

6 = k(1*2*3)

6 = 6k

k = 1

Substitute k value in f(x) = k(x + 1)(x + 2)(x + 3)

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

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

Hence, x3 + 6x2 + 11x + 6 = (x + 1)(x + 2)(x + 3)

### Question 2. Using factor theorem, factorize of the polynomials: x3 + 2x2 – x – 2

Solution:

Given that, f(x) = x3+ 2x2 – x – 2

The constant term in f(x) is -2,

The factors of (-2) are ±1, ± 2,

Let, x – 1 = 0

x = 1

Substitute the value of x in f(x)

f(1) = (1)3 + 2(1)2 – 1 – 2

1 + 2 – 1 – 2 = 0

Similarly, the other factors (x + 1) and (x + 2) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

therefore, f(x) = k(x – 1)(x + 2)(x + 1 )

x3 + 2x2 – x – 2 = k(x – 1)(x + 2)(x + 1 )

Substitute x = 0 on both the sides

0 + 0 – 0 – 2 = k(-1)(1)(2)

– 2 = – 2k

k = 1

Substitute k value in f(x) = k(x – 1)(x + 2)(x + 1)

f(x) = (1)(x – 1)(x + 2)(x + 1)

f(x) =  (x – 1)(x + 2)(x + 1)

therefore, x3 + 2x2 – x – 2 = (x – 1)(x + 2)(x + 1)

### Question 3. Using factor theorem, factorize of the polynomials : x3 – 6x2 + 3x + 10

Solution:

Given that, f(x) = x3 – 6x2 + 3x + 10

The constant term in f(x) is 10,

The factors of 10 are ± 1, ± 2, ± 5, ± 10,

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (−1)3– 6(−1)2 + 3(−1) + 10

-1 – 6 – 3 + 10 = 0

Similarly, the other factors (x – 2) and (x – 5) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

therefore, f(x) = k(x + 1)(x – 2)(x – 5)

Substitute x = 0 on both sides

x3– 6x2 + 3x + 10 = k(x + 1)(x – 2)(x – 5)

0 – 0 + 0 + 10 = k(1)(-2)(-5)

10 = k(10)

k = 1

Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)

f(x) = (1)(x + 1)(x – 2)(x – 5 )

therefore, x3 – 6x2 + 3x + 10 = (x + 1)(x – 2)(x – 5)

### Question 4. Using factor theorem, factorize of the polynomials : x4 –7x3 + 9x2 + 7x –10

Solution:

Given that, f(x) = x4–7x3 + 9x2 + 7x – 10

The constant term in f(x) is 10,

The factors of 10 are ± 1, ± 2, ± 5, ±10,

Let, x – 1 = 0

x = 1

Substitute the value of x in f(x)

f(x) = 14 – 7(1)3 + 9(1)2 + 7(1) – 10

1 – 7 + 9 + 7 – 10

10 – 10 = 0

(x – 1) is the factor of f(x)

Similarly, the other factors are (x + 1), (x – 2), (x – 5)

Since, f(x) is a polynomial of degree 4. So, it cannot have more than four linear factor.

therefore, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

x4 –7x3 + 9x2 + 7x – 10 = k(x – 1)(x + 1)(x – 2)(x – 5)

Put x = 0 on both sides

0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5)

– 10 = k(-10)

k = 1

Substitute k = 1 in f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5)

(x – 1)(x + 1)(x – 2)(x – 5)

therefore, x4 – 7x3 + 9x2 + 7x – 10 = (x – 1)(x + 1)(x – 2)(x – 5)

### Question 5. Using factor theorem, factorize of the polynomials : x4 – 2x3 – 7x2 + 8x + 12

Solution:

Given that,

f(x) = x4 – 2x3 –7x2 + 8x + 12

The constant term f(x) is equal is 12,

The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12,

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (−1)4 – 2(−1)3–7(−1)2 + 8(−1)+12

1 + 2 – 7 – 8 + 12 = 0

therefore, x + 1 is factor of f(x)

Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)

x4 – 2x3 –7x2 + 8x + 12 = k(x + 1)(x + 2)(x – 3)(x – 2)

Substitute x = 0 on both sides,

0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3)

12 = 12K

k = 1

Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)

f(x) = (x – 2)(x + 1)(x + 2)(x – 3)

Hence, x4 – 2x3 – 7x2 + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3)

### Question 6. Using factor theorem, factorize of the polynomials : x4 + 10x3 + 35x2 + 50x + 24

Solution:

Given that, f(x) = x4 + 10x3 + 35x2 + 50x + 24

The constant term in f(x) is equal to 24,

The factors of 24 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 8, ± 12, ± 24,

Let, x + 1 = 0

x = -1

Substitute the value of x in f(x)

f(-1) = (-1)4 + 10(-1)3 + 35(-1)2 + 50(-1) + 24

1-10 + 35 – 50 + 24 = 0

(x + 1) is the factor of f(x)

Similarly, (x + 2), (x + 3), (x + 4) are also the factors of f(x)

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

x4 + 10x3 + 35x2 + 50x + 24 = k(x + 1)(x + 2)(x + 3)(x + 4)

Substitute x = 0 on both sides

0 + 0 + 0 + 0 + 24 = k(1)(2)(3)(4)

24 = k(24)

k = 1

Substitute k = 1 in f(x) = k(x + 1)(x + 2)(x + 3)(x + 4)

f(x) = (1)(x + 1)(x + 2)(x + 3)(x + 4)

f(x) = (x + 1)(x + 2)(x + 3)(x + 4)

Hence, x4 + 10x3 + 35x2 + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)

### Question 7. Using factor theorem, factorize of the polynomials : 2x4–7x3–13x2 + 63x – 45

Solution:

Given that, f(x) = 2x4–7x3–13x2 + 63x – 45

The factors of constant term – 45 are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45,

The factors of the coefficient of x4 is 2.

Hence possible rational roots of f(x) are ± 1, ± 3, ± 5, ± 9, ± 15, ± 45, ± 1/2, ± 3/2, ± 5/2, ± 9/2, ± 15/2, ± 45/2

Let, x – 1 = 0

x = 1

f(1) = 2(1)4 – 7(1)3 – 13(1)2 + 63(1) – 45

2 – 7 – 13 + 63 – 45 = 0

Let, x – 3 = 0

x = 3

f(3) = 2(3)4 – 7(3)3 – 13(3)2 + 63(3)  – 45

162 – 189 – 117 + 189 – 45 = 0

therefore, (x – 1) and (x – 3) are the roots of f(x)

x2 – 4x + 3 is the factor of f(x)

Divide f(x) with x2 – 4x + 3 to get other three factors,

By using long division we get,

2x4 – 7x3 – 13x2 + 63x – 45 = (x2– 4x + 3) (2x2+ x – 15)

2x4 – 7x3– 13x2 + 63x – 45 = (x – 1) (x – 3) (2x2+ x – 15)

Now,

2x2 + x – 15 = 2x2 + 6x – 5x –15

2x(x + 3) – 5 (x + 3)

(2x – 5) (x + 3)

Hence, 2x4 – 7x3 – 13x2 + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5)

### Question 8. Using factor theorem, factorize of the polynomials : 3x3 – x2 – 3x + 1

Solution:

Given that, f(x) = 3x3 – x2 – 3x + 1

The factors of constant term 1 is ± 1,

The factors of the coefficient of x2 = 3,

The possible rational roots are ±1, 1/3,

Let, x – 1 = 0

x = 1

f(1) = 3(1)3 – (1)2 – 3(1) + 1

3 – 1 – 3 + 1 = 0

therefore, x – 1 is the factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By using long division method we get,

3x3– x2 – 3x + 1 = (x – 1)( 3x2 + 2x – 1)

Now,

3x2 + 2x -1 = 3x2 + 3x – x – 1

3x(x + 1) -1(x + 1)

(3x – 1)(x + 1)

Hence, 3x3– x2– 3x + 1 = (x – 1) (3x – 1)(x + 1)

### Question 9. Using factor theorem, factorize of the polynomials : x3– 23x2 + 142x – 120

Solution:

Given that, f(x) = x3– 23x2 + 142x – 120

The constant term in f(x) is -120,

The factors of -120 are ±1, ± 2, ± 3, ± 4, ± 5, ± 6, ± 8, ± 10, ± 12, ± 15, ± 20, ± 24, ± 30, ± 40, ± 60, ± 120,

Let, x – 1 = 0

x = 1

f(1) = (1)3– 23(1)2 + 142(1) – 120

1 – 23 + 142 – 120 = 0

therefore, (x – 1) is the factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By using long division we get,

x3 – 23x2 + 142x – 120 = (x  – 1) (x2 – 22x + 120)

Now,

x2 – 22x + 120 = x2 – 10x – 12x + 120

x(x – 10) – 12(x – 10)

(x – 10) (x – 12)

Hence, x3 – 23x2 + 142x – 120 = (x – 1) (x – 10) (x – 12)

### Question 10. Using factor theorem, factorize of the polynomials : y3 – 7y + 6

Solution:

Given that, f(y) = y3 – 7y + 6

The constant term in f(y) is 6,

The factors are ± 1, ± 2, ± 3, ± 6,

Let, y – 1 = 0

y = 1

f(1) = (1)3 – 7(1) + 6

1 – 7 + 6 = 0

therefore, (y – 1) is the factor of f(y)

Similarly, (y – 2) and (y + 3) are also the factors

Since, f(y) is a polynomial which has degree 3, it cannot have more than 3 linear factors

f(y) = k(y – 1)(y – 2)(y + 3)

y3 – 7y + 6 = k(y – 1)( y – 2)(y + 3)      —————–(i)

Substitute k = 0 in eqn. 1

0 – 0 + 6 = k(-1)(-2)(3)

6 = 6k

k = 1

y3 – 7y + 6 = (1)(y – 1)( y – 2)(y + 3)

y3 – 7y + 6 = (y – 1)( y – 2)(y + 3)

Hence, y3–7y + 6 = (y – 1)( y – 2)(y + 3)

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