Question

(4x-2)² solve this integers
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Answer 1

${(4x - 2)}^{2} \\ \\ = {(4x)}^{2} + {(2)}^{2} - 2(4x)(2) \\ \\ = 16 {x}^{2} + 4 - 16x \\ \\ = 16 {x}^{2} - 16x + 4 \\ \\ = 4 {x}^{2} - 4x + 1$

Answer 2

Answer:

${(4x - 2)}^{2} \\ \\ using \: formula \\ \\ {(a - b)}^{2} \\ \\ = {a}^{2} - 2ab + {b}^{2} \\ \\ here \: a = 4x \\ \\ and \: b = 2 \\ \\ so \: {(4x)}^{2} - 2 \times 4x \times 2 + {(2)}^{2} \\ \\ = 4 {x}^{2} - 16x + 4 \\ \\ = 4( {x}^{2} - 4x + 1) \\ \\ = {x}^{2} - 4x + 1$

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