$$ 125 x^{3}-\frac{1}{125 x^{3}}=234 $$ factorize by factor theorem: $$ x^{3}-2 x^{2}-x+2 $$. (4) $$ P(x)=x^{3}-2 x^{2}-x+2 $$ $$ y=2 $$ and possible factors of 2 are $$ \pm 1, \pm 2 $$ $$ y=1 $$ and possible factor of 1 are 1 possible factors

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Chen Mckenzie · Accepted Answer

The solutions for $$ x $$ are approximately $$ x \approx 1.23 $$ and $$ x \approx -0.1 $$. The factorization of $$ P(x) $$ is $$ P(x) = (x - 1)(x - 2)(x + 1) $$.

\[ 125 x^{3}-\frac{1}{125 x^{3}}=234 \] factorize by factor theorem: \( x^{3}-2 x^{2}-x+2 \). (4) \( P(x)=x^{3}-2 x^{2}-x+2 \) \( y=2 \) and possible factors of 2 are \( \pm 1, \pm 2 \) \( y=1 \) and possible factor of 1 are 1 possible factors

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\[ 125 x^{3}-\frac{1}{125 x^{3}}=234 \] factorize by factor theorem: \( x^{3}-2 x^{2}-x+2 \). (4) \( P(x)=x^{3}-2 x^{2}-x+2 \) \( y=2 \) and possible factors of 2 are \( \pm 1, \pm 2 \) \( y=1 \) and possible factor of 1 are 1 possible factors

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