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).

9th Grade Math

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