How to factorize x² 3√3x 6
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Correction in question :-
Between x^2 and 3root3x and 6, there is plus sign.
x^2 + 3root3 x + 6
![= {x}^{2} + 3 \sqrt{3} x + 6 \\ \\ = {x}^{2} + 2\sqrt{3} x + \sqrt{3} x + 6 \\ \\ = x(x + 2 \sqrt{3} ) + \sqrt{3} (x + 2 \sqrt{3} ) \\ \\ = (x + 2 \sqrt{3} )(x + \sqrt{3} ) = {x}^{2} + 3 \sqrt{3} x + 6 \\ \\ = {x}^{2} + 2\sqrt{3} x + \sqrt{3} x + 6 \\ \\ = x(x + 2 \sqrt{3} ) + \sqrt{3} (x + 2 \sqrt{3} ) \\ \\ = (x + 2 \sqrt{3} )(x + \sqrt{3} )](https://tex.z-dn.net/?f=+%3D++%7Bx%7D%5E%7B2%7D+++%2B+3++%5Csqrt%7B3%7D+x+%2B+6+%5C%5C++%5C%5C++%3D++%7Bx%7D%5E%7B2%7D++%2B+2%5Csqrt%7B3%7D+x+%2B++%5Csqrt%7B3%7D+x++%2B+6+%5C%5C++%5C%5C++%3D+x%28x+%2B+2+%5Csqrt%7B3%7D+%29++%2B+%5Csqrt%7B3%7D+%28x+%2B+2+%5Csqrt%7B3%7D+%29+%5C%5C++%5C%5C++%3D+%28x+%2B+2+%5Csqrt%7B3%7D+%29%28x+%2B++%5Csqrt%7B3%7D+%29)
Between x^2 and 3root3x and 6, there is plus sign.
x^2 + 3root3 x + 6
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