Here we will solve different types of Problems on Factorization using a\(^{2}\) – b\(^{2}\) = (a + b)(a – b).
1. Factorize: 4a\(^{2}\) – b\(^{2}\) + 2a + b
Solution:
Given expression = 4a\(^{2}\) – b\(^{2}\) + 2a + b
= (4a\(^{2}\) – b\(^{2}\)) + 2a + b
= {(2a)\(^{2}\) – b\(^{2}\)} + 2a + b
= (2a + b)(2a – b) + 1(2a + b)
= (2a + b)(2a – b + 1)
2. Factorize: x\(^{3}\) – 3x\(^{2}\) – x + 3
Solution:
Given expression = x\(^{3}\) – 3x\(^{2}\) – x + 3
= (x\(^{3}\) – 3x\(^{2}\)) – x + 3
= x\(^{2}\)(x – 3) – 1(x – 3)
= (x – 3)(x\(^{2}\) – 1)
= (x – 3)(x\(^{2}\) – 1\(^{2}\))
= (x – 3)(x + 1)(x - 1)
3. Factorize: 4x\(^{2}\) – y\(^{2}\) + 2x – 2y – 3xy
Solution:
Given expression = 4x\(^{2}\) – y\(^{2}\) + 2x – 2y – 3xy
= x\(^{2}\) – y\(^{2}\) + 2x – 2y + 3x\(^{2}\) – 3xy
= (x + y)(x – y) + 2(x – y) + 3x(x – y)
= (x – y)(x + y + 2 + 3x)
= (x – y)(4x + y + 2)
4. Factorize: a\(^{4}\) + a\(^{2}\)b\(^{2}\) + b\(^{4}\)
Solution:
Given expression = a\(^{4}\) + a\(^{2}\)b\(^{2}\) + b\(^{4}\)
= a\(^{4}\) + 2a\(^{2}\)b\(^{2}\) + b\(^{4}\) - a\(^{2}\)b\(^{2}\)
= (a\(^{2}\))\(^{2}\) + 2 ∙ a\(^{2}\) ∙ b\(^{2}\) + (b\(^{2}\))\(^{2}\) - a\(^{2}\)b\(^{2}\)
= (a\(^{2}\) + b\(^{2}\))\(^{2}\) – (ab)\(^{2}\)
= (a\(^{2}\) + b\(^{2}\) + ab)( a\(^{2}\) + b\(^{2}\) – ab)
5. Factorize: x\(^{2}\) – 3x - 28
Solution:
Given expression = x\(^{2}\) – 3x - 28
= {x\(^{2}\) – 2 ∙ x ∙ \(\frac{3}{2}\) + (\(\frac{3}{2}\))\(^{2}\)} – (\(\frac{3}{2}\))\(^{2}\) - 28
= (x - \(\frac{3}{2}\))\(^{2}\) – (\(\frac{9}{4}\) + 28)
= (x - \(\frac{3}{2}\))\(^{2}\) – \(\frac{121}{4}\)
= (x - \(\frac{3}{2}\))\(^{2}\) – (\(\frac{11}{2}\))\(^{2}\)
= (x - \(\frac{3}{2}\) + \(\frac{11}{2}\))(x - \(\frac{3}{2}\) - \(\frac{11}{2}\))
= (x + 4)(x – 7)
6. Factorize: x\(^{2}\) + 5x + 5y – y\(^{2}\)
Solution:
Given expression = x\(^{2}\) + 5x + 5y – y\(^{2}\)
= (x\(^{2}\) – y\(^{2}\)) + 5x + 5y
= (x + y)(x – y) + 5(x + y)
= (x + y)(x – y + 5)
7. Factorize: x\(^{2}\) + xy – 2y - 4
Solution:
Given expression = x\(^{2}\) + xy – 2y – 4
= (x\(^{2}\) – 4) + xy – 2y
= (x\(^{2}\) – 2\(^{2}\)) + y(x – 2)
= (x + 2)(x – 2) + y(x – 2)
= (x - 2)(x + 2 + y)
= (x - 2)(x + y + 2)
8. Factorize: a\(^{2}\) – b\(^{2}\) – 10a + 25
Solution:
Given expression = a\(^{2}\) – b\(^{2}\) – 10a + 25
= (a\(^{2}\) – 10a + 25) – b\(^{2}\)
= (a\(^{2}\) – 2 ∙ a ∙ 5 + 5\(^{2}\)) – b\(^{2}\)
= (a – 5)\(^{2}\)– b\(^{2}\)
= (a – 5 + b)(a – 5 – b)
= (a + b – 5)(a – b – 5)
9. Factorize: x(x – 2) – y(y – 2)
Solution:
Given expression = x(x – 2) – y(y – 2)
= x\(^{2}\) – 2x – y\(^{2}\) + 2y
= (x\(^{2}\) – y\(^{2}\)) – 2x + 2y
= (x + y)(x – y) – 2(x – y)
= (x – y)(x + y – 2).
10. Factorize: a\(^{3}\) + 2a\(^{2}\) – a - 2
Solution:
Given expression = a\(^{3}\) + 2a\(^{2}\) – a - 2
= a\(^{2}\)(a + 2) – 1(a + 2)
= (a + 2)(a\(^{2}\) – 1)
= (a + 2)(a\(^{2}\) – 1\(^{2}\))
= (a + 2)(a + 1)(a – 1)
11. Factorize: a\(^{4}\) + 64
Solution:
Given expression = a\(^{4}\) + 64
= (a\(^{2}\))\(^{2}\) + 8\(^{2}\)
= (a\(^{2}\))\(^{2}\) + 2 ∙ a\(^{2}\) ∙ 8 + 8\(^{2}\) - 2 ∙ a\(^{2}\) ∙ 8
= (a\(^{2}\) + 8)\(^{2}\) – 16a\(^{2}\)
= (a\(^{2}\) + 8)\(^{2}\) – (4a)\(^{2}\)
= (a\(^{2}\) + 8 + 4a)(a\(^{2}\) + 8 - 4a)
= (a\(^{2}\) + 4a + 8)(a\(^{2}\) - 4a + 8)
11. Factorize: x\(^{4}\) + 4
Solution:
Given expression = x\(^{4}\) + 4
= (x\(^{2}\))\(^{2}\) + 2\(^{2}\)
= (x\(^{2}\))\(^{2}\) + 2 ∙ x\(^{2}\) ∙ 2 + 2\(^{2}\) - 2 ∙ x\(^{2}\) ∙ 2
= (x\(^{2}\) + 2)\(^{2}\) – 4x\(^{2}\)
= (x\(^{2}\) + 2)\(^{2}\) – (2x)\(^{2}\)
= (x\(^{2}\) + 2 + 2x) (x\(^{2}\) + 2 – 2x)
= (x\(^{2}\) + 2x + 2) (x\(^{2}\) – 2x + 2)
12. Express x\(^{2}\) – 5x + 6 as the difference of two squares and then factorize.
Solution:
Given expression = x\(^{2}\) – 5x + 6
= x\(^{2}\) – 2 ∙ x ∙ \(\frac{5}{2}\) + (\(\frac{5}{2}\))\(^{2}\) + 6 - (\(\frac{5}{2}\))\(^{2}\)
= (x - \(\frac{5}{2}\))\(^{2}\) + 6 - \(\frac{25}{4}\)
= (x - \(\frac{5}{2}\))\(^{2}\) - \(\frac{1}{4}\)
= (x - \(\frac{5}{2}\))\(^{2}\) – (\(\frac{1}{2}\))\(^{2}\), [Difference of two squares]
= (x - \(\frac{5}{2}\) + \(\frac{1}{2}\))(x - \(\frac{5}{2}\) - \(\frac{1}{2}\))
= (x – 2)(x - 3)
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