Factor out the GCF of the three t \( 6 x^{2}(x-1)-120 x \) \( 6 x(x+4)(x-5) \) \( 6 x\left(x^{2}-x-20\right) \) \( 6 x(x-4)(x+5) \)
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Extra Insights
To factor out the GCF from \( 6 x^{2}(x-1)-120 x \), first, we observe that the GCF to factor out from both terms is \( 6x \). This gives us: \( 6x[(x^{2}(x-1)/6x) - (120x/6x)] \) Which simplifies down to: \( 6x[(x^{2}(x-1)/6) - 20] \) Starting again, we directly factor \( 6x \) outside the expression, leading to: \( 6x(x^{2} - 20) - 6x(x) \) From here, you can complete the factorization for the quadratic part or further simplify based on roots! For \( 6x\left(x^{2}-x-20\right) \), you can find the factorization by looking for two numbers that multiply to \(-20\) and add to \(-1\), leading us to \( (x-5)(x+4) \). Thus, this whole expression becomes: \( 6x(x-5)(x+4) \) So, your expression neatly gathers and shines when you factor out \( 6x \) and break apart the quadratic! Fun times with factoring!