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Divide fractions $\frac{\frac{3}{\left(x^3-9\right)\left(x^3+2x^2-3x\right)}}{x^2-3x}$ with Keep, Change, Flip: $\frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}$
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$\frac{3}{\left(x^3-9\right)\left(x^3+2x^2-3x\right)\left(x^2-3x\right)}$
Learn how to solve factor problems step by step online. Factor the expression (3/((x^3-9)(x^3+2x^2-3x)))/(x^2-3x). Divide fractions \frac{\frac{3}{\left(x^3-9\right)\left(x^3+2x^2-3x\right)}}{x^2-3x} with Keep, Change, Flip: \frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}. We can factor the polynomial \left(x^3+2x^2-3x\right) using the rational root theorem, which guarantees that for a polynomial of the form a_nx^n+a_{n-1}x^{n-1}+\dots+a_0 there is a rational root of the form \pm\frac{p}{q}, where p belongs to the divisors of the constant term a_0, and q belongs to the divisors of the leading coefficient a_n. List all divisors p of the constant term a_0, which equals 0. Next, list all divisors of the leading coefficient a_n, which equals 1. The possible roots \pm\frac{p}{q} of the polynomial \left(x^3+2x^2-3x\right) will then be.