Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.
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).
From Miscellaneous Problems on Factorization to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.