# Miscellaneous Problems on Factorization

Here we will solve different types of Miscellaneous Problems on Factorization.

1. Factorize: x(2x + 5) – 3

Solution:

Given expression = x(2x + 5) – 3

= 2x2 + 5x – 3

= 2x2 + 6x – x – 3,

[Since, 2(-3) = - 6 = 6 × (-1), and 6 + (-1) = 5]

= 2x(x + 3) – 1(x + 3)

= (x + 3)(2x – 1).

2. Factorize: 4x2y – 44x2y + 112xy

Solution:

Given expression = 4x2y – 44x2y + 112xy

= 4xy(x2 – 11x + 28)

= 4xy(x2 – 7x – 4x + 28)

= 4xy{x(x – 7) – 4(x - 7)}

= 4xy(x - 7)(x - 4)

3. Factorize: (a – b)3 +(b – c)3 + (c – a)3.

Solution:

Let a – b = x, b – c = y, c – a = z. Adding, x + y + z = 0.

Therefore, the given expression = x3 + y3 + z3 = 3xyz (Since, x + y + z = 0).

Therefore, (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b)(b – c)(c –a).

4. Resolve into factors: x3 + x2 - $$\frac{1}{x^{2}}$$ + $$\frac{1}{x^{3}}$$

Solution:

Given expression = x3 + x2 - $$\frac{1}{x^{2}}$$ + $$\frac{1}{x^{3}}$$

= (x + $$\frac{1}{x}$$)(x2 – x ∙ $$\frac{1}{x}$$ + $$\frac{1}{x^{2}}$$) + (x + $$\frac{1}{x}$$)(x - $$\frac{1}{x}$$)

= (x + $$\frac{1}{x}$$){ x2 – x ∙ $$\frac{1}{x}$$ + $$\frac{1}{x^{2}}$$ + x - $$\frac{1}{x}$$}

= (x + $$\frac{1}{x}$$){ x2 – 1 + $$\frac{1}{x^{2}}$$ + x - $$\frac{1}{x}$$}

= (x + $$\frac{1}{x}$$)( x2 + x – 1 - $$\frac{1}{x}$$ + $$\frac{1}{x^{2}}$$)

5. Factorize: 27(a + 2b)3 + (a – 6b)3

Solution:

Given expression = 27(a + 2b)3 + (a – 6b)3

= {3(a + 2b)}3 + (a – 6b)3

= {3(a + 2b) + (a – 6b)}[{3(a + 2b)}2 – {3(a + 2b)}(a – 6b) + (a – 6b)2]

= (3a + 6b + a – 6b)[9(a2 + 4ab + 4b2) – (3a + 6b)(a – 6b) + a2 – 12ab + 36b2]

= 4a[9a2 + 36ab + 36b2 – {3a2 – 18ab + 6ba – 36b2} + a2 – 12ab  + 36b2]

= 4a(7a2 + 36ab + 108b2).

6. If x + $$\frac{1}{x}$$ = $$\sqrt{3}$$, find x^3 + $$\frac{1}{x^{3}}$$.

Solution:

x3 + $$\frac{1}{x^{3}}$$ = (x + $$\frac{1}{x}$$)(x2 – x ∙ $$\frac{1}{x}$$ + $$\frac{1}{x^{2}}$$)

= (x + $$\frac{1}{x}$$)[x2 + $$\frac{1}{x^{2}}$$ – 1]

= (x + $$\frac{1}{x}$$)[(x + $$\frac{1}{x}$$)2 – 3]

= $$\sqrt{3}$$ ∙ [($$\sqrt{3}$$)2 – 3]

= $$\sqrt{3}$$ × 0

= 0.

7. Evaluate: $$\frac{128^{3} + 272^{3}}{128^{2} - 128 \times 272 + 272^{2}}$$

Solution:

The given expression = $$\frac{128^{3} + 272^{3}}{128^{2} - 128 \times 272 + 272^{2}}$$

= $$\frac{(128 + 272)(128^{2} - 128 \times 272 + 272^{2})}{128^{2} - 128 \times 272 + 272^{2}}$$

= 128 + 272

= 400.

8. If a + b + c = 10, a2 + b2 + c2 = 38 and a3 + b3 + c3 = 160, find the value of abc.

Solution:

We know, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – bc – ca – ab).

Therefore, 160 – 3abc = 10(38 – bc – ca – ab).......................... (i)

Now, (a + b + c)2 = a2 + b2 + c2 + 2bc + 2ca + 2ab

Therefore, 102 = 38 + 2(bc + ca + ab).

⟹ 2(bc + ca + ab) = 102 – 38

⟹ 2(bc + ca + ab) = 100 – 38

⟹ 2(bc + ca + ab) = 62

Therefore, bc + ca + ab = $$\frac{62}{2}$$ = 31.

Putting in (i), we get,

160 – 3abc = 10(38 – 31)

⟹ 160 – 3abc = 70

⟹ 3abc = 160 - 70

⟹ 3abc = 90.

Therefore, abc = $$\frac{90}{3}$$ = 30.

9. Find the LCM and HCF of x2 – 2x – 3 and x2 + 3x + 2.

Solution:

Here, x2 – 2x – 3 = x2 – 3x + x – 3

= x(x – 3) + 1(x – 3)

= (x – 3)(x + 1).

And x2 + 3x + 2 = x2 + 2x + x + 2.

= x(x + 2) + 1(x + 2)

= (x + 2)(x + 1).

Therefore, by the definition of LCM, the required LCM = (x – 3)(x + 1)(x + 2).

Again, by definition of HCF, the required HCF = x + 1.

10. (i) Find the LCM and HCF of x3 + 27 and x2 – 9.

(ii) Find the LCM and HCF of x3 – 8, x2 - 4 and x2 + 4x + 4.

Solution:

(i) x3 + 27 = x3 + 33

= (x + 3)(x2 – x ∙ 3 + 32}

= (x + 3)(x2 – 3x + 9).

x2 – 9 = x2 – 32

= (x + 3)(x – 3).

Therefore, by definition of LCM,

the required LCM = (x + 3)(x2 – 3x + 9)(x – 3)

= (x2 – 9)(x2 – 3x + 9).

Again, by definition of HCF, the required HCF = x + 3.

(ii) x3 – 8 = x3 – 23

= (x – 2)(x2 + x ∙ 2 + 22)

= (x – 2)(x2 + 2x + 4).

x2 – 4 = x2 – 22

= (x + 2)(x - 2).

x2 + 4x + 4 = (x + 2)2.

Therefore, by the definition of LCM, the required LCM = (x – 2)(x + 2)2(x2 + 2x + 4).