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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