| Column - I | Column - II |
| P. \[9{{x}^{2}}+24x+16\] | (i) \[(2x-4)\] |
| Q. \[25{{x}^{2}}+30x+9\] | (ii) \[(4x+1)\] |
| R. \[40{{x}^{2}}+14x+1\] | (iii) \[(5x+3)\] |
| S. \[4{{x}^{2}}-16x+16\] | (iv) \[(3x+4)\] |
A) P\[\to \](iv); Q\[\to \](iii); R\[\to \](ii); S\[\to \](i)
B) P\[\to \](iii): Q\[\to \](i); R\[\to \](iv); S\[\to \](ii)
C) P\[\to \](ii); Q\[\to \](i); R\[\to \](iv): S\[\to \](iii)
D) P\[\to \](iv); Q\[\to \](iii); R\[\to \](i); S\[\to \](ii)
Correct Answer: A
Solution :
P. We have, \[9{{x}^{2}}+24x+16={{(3x)}^{2}}+2(3x)(4)+{{(4)}^{2}}\] \[={{(3x+4)}^{2}}=(3x+4)(3x+4)\] Q. We have, \[25{{x}^{2}}+30x+9={{(5x)}^{2}}+2(5x)(3)+{{(3)}^{2}}\] \[={{(5x+3)}^{2}}=(5x+3)(5x+3)\] R. We have, \[40{{x}^{2}}+14x+1=40{{x}^{2}}+10x+4x+1\] \[=10x(4x+1)+1(4x+1)=(10x+1)(4x+1)\] S. We have, \[4{{x}^{2}}-16x+16={{(2x)}^{2}}-2(2x)(4)+{{(4)}^{2}}\] \[={{(2x-4)}^{2}}=(2x-4)(2x-4)\]
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