Math, asked by meharbatra, 9 months ago

Pls tell how to do this?

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Answered by Anonymous

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$\huge \tt {ANSWER:}$

$\large \tt {{x}^{3} + {13x}^{2} + 31x - 45}$

$\large \tt {= {x}^{3} - {x}^{2} + {14x}^{2} - 14x + 45x - 45}$

$\large \tt { = {x}^{2} (x - 1) + 14x(x - 1) + 45(x - 1)}$

$\large \tt { = (x - 1)( {x}^{2} + 14x + 45)}$

$\large \tt { = (x - 1)( {x}^{2} + 9x + 5x + 45)}$

$\large \tt { = (x - 1)(x(x + 9) + 5(x + 9))}$

$\large \tt { = (x - 1)(x + 9)(x + 5)}$

$\huge \tt {FOLLOW \: ME \: PLEASE}$

Answered by BrainlyAVYAM

Answer:

$x {}^{3} + 13x {}^{2} + 31x - 45 = 0 \\ = x {}^{3} - x {}^{2} + 14x {}^{2} - 14x+ 45x - 45 = 0 \\ = x {}^{2} (x - 1) + 14x(x - 1) + 45(x - 1) = 0 \\ = (x - 1)(x {}^{2} + 14x + 45) = 0 \\ = (x - 1)(x {}^{2} + 9x + 5x + 45) = 0 \\ = (x - 1)(x(x + 9) + 5(x + 9)) = 0 \\ = (x - 1)(x + 9)(x + 5) = 0 \\ = x = 1 \: \: and- 9 \: and- 5 \: \: answer$

Hey! Mate Here is your solution. Thanks

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