**Chapter 6: Factorization of Polynomials Exercise – 6.5**

**Question: 1**

Using factor theorem, factorize of the polynomials:

x^{3 }+ 6x^{2 }+ 11x + 6

**Solution:**

Given polynomial, f(x) = x^{3 }+ 6x^{2 }+ 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)

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, it cannot have more than three linear factors.

∴ f(x) = k(x + 1)(x + 2)(x + 3)

=> x^{3 }+ 6x^{2 }+ 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)

∴ x^{3 }+ 6x^{2 }+ 11x + 6 = (x + 1)(x + 2)(x + 3)

**Question: 2**

Using factor theorem, factorize of the polynomials:

x^{3 }+ 2x^{2 }– x – 2

**Solution:**

Given, f(x) = x^{3}+ 2x^{2 }– 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.

∴ f(x) = k(x – 1)(x + 2)(x + 1 )

x^{3 }+ 2x^{2} – 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)

So, x^{3} + 2x^{2} – x – 2 = (x – 1)(x + 2)(x + 1)

**Question: 3**

Using factor theorem, factorize of the polynomials:

x^{3} – 6x^{2} + 3x + 10

**Solution:**

Let, f(x) = x^{3} – 6x^{2} + 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.

∴ f(x) = k(x + 1)(x – 2)(x – 5)

Substitute x = 0 on both sides

=> x^{3}– 6x^{2} + 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 )

so, x^{3} – 6x^{2} + 3x + 10 = (x + 1)(x – 2)(x – 5)

**Question: 4**

Using factor theorem, factorize of the polynomials:

x^{4} –7x^{3} + 9x^{2 }+ 7x –10

**Solution:**

Given, f(x) = x^{4}–7x^{3} + 9x^{2} + 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.

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

=> x^{4 }–7x^{3} + 9x^{2 }+ 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)

So, x^{4} – 7x^{3} + 9x^{2} + 7x – 10

= (x – 1)(x + 1)(x – 2)(x – 5)

**Question: 5**

Using factor theorem, factorize of the polynomials:

x^{4} – 2x^{3} – 7x^{2} + 8x + 12

**Solution:**

Given,

f(x) = x^{4} – 2x^{3} –7x^{2} + 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

So, 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)

=> x^{4 }– 2x^{3 }–7x^{2 }+ 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)

so, x^{4} – 2x^{3 }– 7x^{2} + 8x + 12 = (x – 2)(x + 1)(x + 2)(x – 3)

**Question: 6**

Using factor theorem, factorize of the polynomials:

x^{4} + 10x^{3 }+ 35x^{2} + 50x + 24

**Solution:**

Given, f(x) = x^{4 }+ 10x^{3 }+ 35x^{2} + 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)

=> x^{4 }+ 10x^{3} + 35x^{2 }+ 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, x^{4} + 10x^{3} + 35x^{2} + 50x + 24 = (x + 1)(x + 2)(x + 3)(x + 4)

**Question: 7**

Using factor theorem, factorize of the polynomials:

2x^{4}–7x^{3}–13x^{2 }+ 63x – 45

**Solution:**

Given, f(x) = 2x^{4}–7x^{3}–13x^{2 }+ 63x – 45

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

The factors of the coefficient of x^{4} 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

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

=> x^{2} – 4x + 3 is the factor of f(x)

Divide f(x) with x^{2} – 4x + 3 to get other three factors

By long division,

2x^{2} + x – 15

x^{2} – 4x + 3 2x^{4} – 7x^{3} – 13x^{2 } + 63x – 45

2x^{4} – 8x^{3 }+ 6x^{2}

(-) (+) (-)

x^{3} – 19x^{2 } + 63x

x^{3 }– 4x^{2} + 3x

(-) (+) (-)

-15x^{2} + 60x - 45

-15x^{2} + 60x - 45

(+) (-) (+)

0

=> 2x^{4 }– 7x^{3 }– 13x^{2} + 63x – 45 = (x^{2}– 4x + 3) (2x^{2}+ x – 15)

=> 2x^{4 }- 7x^{3}– 13x^{2} + 63x – 45 = (x – 1) (x – 3) (2x^{2}+ x – 15)

Now,

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

= 2x(x + 3) – 5 (x + 3)

= (2x – 5) (x + 3)

So, 2x^{4} – 7x^{3} – 13x^{2} + 63x – 45 = (x – 1)(x – 3)(x + 3)(2x – 5)

**Question: 8**

Using factor theorem, factorize of the polynomials:

3x^{3} - x^{2} – 3x + 1

**Solution:**

Given, f(x) = 3x^{3} - x^{2} – 3x + 1

The factors of constant term 1 is ± 1

The factors of the coefficient of x^{2} = 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

So, x – 1 is the factor of f(x)

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

By long division method,

3x^{2} + 2x – 1

x – 1 3x^{3} – x^{2 }– 3x + 1

3x^{3} – x^{2 }

(-) (+)

2x^{2} – 3x

2x^{2} – 2x

(-) (+)

- x + 1

- x + 1

(+) (-)

0

=> 3x^{3}- x^{2} – 3x + 1 = (x – 1)( 3x^{2} + 2x – 1)

Now,

3x^{2} + 2x -1 = 3x^{2} + 3x – x - 1

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

= (3x – 1)(x + 1)

Hence, 3x^{3}- x^{2}- 3x + 1= (x – 1) (3x – 1)(x + 1)

**Question: 9**

Using factor theorem, factorize of the polynomials:

x^{3}- 23x^{2} + 142x - 120

**Solution:**

Let, f(x) = x^{3}- 23x^{2} + 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

So, (x – 1) is the factor of f(x)

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

By long division,

x^{2} – 22x + 120

x – 1 x^{3} – 23x^{2} + 142x – 120

x^{3} – x^{2}

(-) (+)

- 22x^{2} + 142x

- 22x^{2} + 22x

(+) (-)

120x – 120

120x – 120

(-) (+)

0

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

Now,

x^{2} – 22x + 120 = x^{2} – 10x – 12x + 120

= x(x – 10) – 12(x – 10)

= (x – 10) (x – 12)

Hence, x^{3} – 23x^{2} + 142x – 120 = (x – 1) (x – 10) (x – 12)

**Question: 10**

Using factor theorem, factorize of the polynomials:

y^{3} – 7y + 6

**Solution:**

Given, f(y) = y^{3} – 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

So, (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)

=> y^{3 }– 7y + 6 = k(y – 1)( y – 2)(y + 3) —– 1

Substitute k = 0 in eq 1

=> 0 – 0 + 6 = k(-1)(-2)(3)

=> 6 = 6k

=> k = 1

y^{3 }– 7y + 6 = (1)(y – 1)( y – 2)(y + 3)

y^{3} – 7y + 6 = (y – 1)( y – 2)(y + 3)

Hence, y^{3}–7y + 6 = (y – 1)( y – 2)(y + 3)

**Question: 11**

Using factor theorem, factorize of the polynomials:

x^{3 }– 10x^{2 }– 53x – 42

**Solution:**

Given,

f(x) = x^{3}–10x^{2 }– 53x – 42

The constant in f(x) is - 42

The factors of - 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, x + 1 = 0

=> x = - 1

f(-1) = (−1)^{3 }–10(−1)^{2 }– 53(−1) – 42

= -1 – 10 + 53 – 42

= 0

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

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

By long division,

x^{2} – 11x – 42

x + 1 x^{3} – 10x^{2} – 53x – 42

x^{3 } + x^{2}

(-) (-)

-11x^{2} – 53x

-11x^{2} – 11x

(+) (+)

- 42x – 42

- 42x – 42

(+) (+)

0

=> x^{3} – 10x^{2} – 53x – 42 = (x + 1) (x^{2} – 11x – 42)

Now,

x^{2} – 11x – 42 = x^{2} – 14x + 3x – 42

= x(x – 14) + 3(x – 14)

= (x + 3)(x – 14)

Hence, x^{3} – 10x^{2} – 53x – 42 = (x + 1) (x + 3)(x – 14)

**Question: 12**

Using factor theorem, factorize of the polynomials:

y^{3} – 2y^{2} – 29y – 42

**Solution:**

Given, f(x) = y^{3} – 2y^{2} – 29y – 42

The constant in f(x) is - 42

The factors of -42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, y + 2 = 0

=> y = – 2

f(-2) = (−2)^{3} – 2(−2)^{2}–29(−2) – 42

= -8 -8 + 58 – 42

= 0

So, (y + 2) is the factor of f(y)

Now, divide f(y) with (y + 2) to get other factors

By, long division

y^{2} – 4y – 21

y + 2 y^{3} – 2y^{2} – 29y – 42

y^{3} + 2y^{2}

(-) (-)

- 4y^{2} – 29y

- 4y^{2} – 8y

(+) (+)

- 21y – 42

- 21y – 42

(+) (+)

0

=> y^{3} – 2y^{2} – 29y – 42 = (y + 2) (y^{2} – 4y – 21)

Now,

y^{2} – 4y – 21 = y^{2} – 7y + 3y – 21

= y(y – 7) +3(y – 7)

= (y – 7)(y + 3)

Hence, y^{3} – 2y^{2} – 29y – 42 = (y + 2) (y – 7)(y + 3)

**Question: 13**

Using factor theorem, factorize of the polynomials:

2y^{3 }– 5y^{2 }– 19y + 42

**Solution:**

Given, f(x) = 2y^{3 }– 5y^{2 }– 19y + 42

The constant in f(x) is + 42

The factors of 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, y – 2 = 0

=> y = 2

f(2) = 2(2)^{3 }– 5(2)^{2 }– 19(2) + 42

= 16 – 20 – 38 + 42

= 0

So, (y – 2) is the factor of f(y)

Now, divide f(y) with (y – 2) to get other factors

By, long division method

2y^{2} – y – 21

y – 2 2y^{3} – 5y^{2} -19y + 42

2y^{3} – 4y^{2}

(-) (+)

- y^{2} – 19y

- y^{2} + 2y

(+) (-)

- 21y + 42

- 21y + 42

(+) (-)

0

=> 2y^{3} – 5y^{2} - 19y + 42 = (y – 2) (2y^{2} – y – 21)

Now,

2y^{2} – y – 21

The factors are (y + 3) (2y – 7)

Hence, 2y^{3} – 5y^{2} -19y + 42 = (y – 2) (y + 3) (2y – 7)

**Question: 14**

Using factor theorem, factorize of the polynomials:

x^{3 }+ 13x^{2 }+ 32x + 20

**Solution:**

Given, f(x) = x^{3 }+ 13x^{2 }+ 32x + 20

The constant in f(x) is 20

The factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10, ± 20

Let, x + 1 = 0

=> x = -1

f(-1) = (−1)^{3}+13(−1)^{2} + 32(−1) + 20

= -1 + 13 – 32 + 20

= 0

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

Divide f(x) with (x + 1) to get other factors

By, long division

x^{2} + 12x + 20

x + 1, x^{3} + 13x^{2} + 32x + 20

x^{3} + x^{2}

(-) (-)

12x^{2} + 32x

12x^{2} + 12x

(-) (-)

20x – 20

20x – 20

(-) (-)

0

=> x^{3} + 13x^{2} +32x + 20 = (x + 1)( x^{2} + 12x + 20)

Now,

x^{2} + 12x + 20 = x^{2} + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

The factors are (x + 10) and (x + 2)

Hence, x^{3} + 13x^{2} + 32x + 20 = (x + 1)(x + 10)(x + 2)

**Question: 15**

Using factor theorem, factorize of the polynomials:

x^{3 }– 3x^{2} – 9x – 5

**Solution:**

Given, f(x) = x^{3 }– 3x^{2} – 9x – 5

The constant in f(x) is -5

The factors of -5 are ±1, ±5

Let, x + 1 = 0

=> x = -1

f(-1) = (−1)^{3} - 3(−1)^{2} - 9(-1) - 5

= -1 – 3 + 9 – 5

= 0

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

Divide f(x) with (x + 1) to get other factors

By, long division

x^{2} – 4x – 5

x + 1 x^{3} – 3x^{2} – 9x – 5

x^{3} + x^{2}

(-) (-)

- 4x^{2} – 9x

- 4x^{2} – 4x

(+) (+)

- 5x – 5

- 5x – 5

(+) (+)

0

=> x^{3} – 3x^{2} – 9x – 5 = (x + 1)( x^{2} – 4x – 5)

Now,

x^{2} – 4x – 5 = x^{2} – 5x + x – 5

= x(x – 5) + 1(x – 5)

The factors are (x – 5) and (x + 1)

Hence, x^{3} – 3x^{2} – 9x – 5 = (x + 1)(x – 5)(x + 1)

**Question: 16**

Using factor theorem, factorize of the polynomials:

2y^{3} + y^{2} – 2y – 1

**Solution:**

Given, f(y) = 2y^{3} + y^{2} – 2y – 1

The constant term is 2

The factors of 2 are ± 1, ± 1/2

Let, y – 1= 0

=> y = 1

f(1) = 2(1)^{3} +(1)^{2} – 2(1) – 1

= 2 + 1 – 2 – 1

= 0

So, (y – 1) is the factor of f(y)

Divide f(y) with (y – 1) to get other factors

By, long division

2y^{2} + 3y + 1

y – 1, 2y^{3} + y^{2 }– 2y – 1

2y^{3} – 2y^{2}

(-) (+)

3y^{2} – 2y

3y^{2} – 3y

(-) (+)

y – 1

y – 1

(-) (+)

0

=> 2y^{3} + y^{2 }– 2y – 1 = (y – 1) (2y^{2} + 3y + 1)

Now,

2y^{2} + 3y + 1 = 2y^{2} + 2y + y + 1

= 2y(y + 1) + 1(y + 1)

= (2y + 1) (y + 1) are the factors

Hence, 2y^{3} + y^{2 }– 2y – 1 = (y – 1) (2y + 1) (y + 1)

**Question: 17**

Using factor theorem, factorize of the polynomials:

x^{3} – 2x^{2} – x + 2

**Solution:**

Let, f(x) = x^{3} – 2x^{2} – x + 2

The constant term is 2

The factors of 2 are ±1, ± 1/2

Let, x – 1= 0

=> x = 1

f(1) = (1)^{3 }– 2(1)^{2} – (1) + 2

= 1 – 2 – 1 + 2

= 0

So, (x – 1) is the factor of f(x)

Divide f(x) with (x – 1) to get other factors

By, long division

x^{2} – x – 2

x – 1, x^{3} – 2x^{2} – y + 2

x^{3} – x^{2}

(-) (+)

- x^{2} – x

- x^{2} + x

(+) (-)

– 2x + 2

– 2x + 2

(+) (-)

0

=> x^{3} – 2x^{2} – y + 2 = (x – 1) (x^{2} – x – 2)

Now,

x^{2} – x – 2 = x^{2} – 2x + x – 2

= x(x – 2) + 1(x – 2)

=(x – 2)(x + 1) are the factors

Hence, x^{3} – 2x^{2} – y + 2 = (x – 1)(x + 1)(x – 2)

**Question: 18**

Factorize each of the following polynomials:

1. x^{3 }+ 13x^{2} + 31x – 45 given that x + 9 is a factor

2. 4x^{3 }+ 20x^{2} + 33x + 18 given that 2x + 3 is a factor

**Solution:**

1. x^{3 }+ 13x^{2} + 31x – 45 given that x + 9 is a factor

let, f(x) = x^{3 }+ 13x^{2} + 31x – 45

given that (x + 9) is the factor

divide f(x) with (x + 9) to get other factors

by , long division

x^{2} + 4x – 5

x + 9 x^{3} + 13x^{2} + 31x – 45

x^{3 } + 9x^{2}

(-) (-)

4x^{2} +31x

4x^{2} +36x

(-) (-)

-5x – 45

-5x – 45

(+) (+)

0

=> x^{3} + 13x^{2} + 31x – 45 = (x + 9)( x^{2} + 4x – 5)

Now,

x^{2} + 4x – 5 = x^{2} + 5x – x – 5

= x(x + 5) -1(x + 5)

= (x + 5) (x – 1) are the factors

Hence, x^{3} + 13x^{2} + 31x – 45 = (x + 9)(x + 5)(x – 1)

2. 4x^{3 }+ 20x^{2 }+ 33x + 18 given that 2x + 3 is a factor

let, f(x) = 4x^{3 }+ 20x^{2 }+ 33x + 18

given that 2x + 3 is a factor

divide f(x) with (2x + 3) to get other factors

by, long division

2x^{2} + 7x + 6

2x + 3, 4x^{3 }+ 20x^{2} + 33x + 18

4x^{3} + 6x^{2}

(-) (-)

14x^{2} – 33x

14x^{2} – 21x

(-) (+)

12x + 18

12x + 18

(-) (-)

0

=> 4x^{3 }+ 20x^{2} + 33x + 18 = (2x + 3) (2x^{2} + 7x + 6)

Now,

2x^{2} + 7x + 6 = 2x^{2} + 4x + 3x + 6

= 2x(x + 2) + 3(x + 2)

= (2x + 3)(x + 2) are the factors

Hence, 4x^{3 }+ 20x^{2} + 33x + 18 = (2x + 3)(2x + 3)(x + 2)