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Find the roots of the equation using the Quadratic Formula
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$\frac{3}{\frac{2x^3-5x^2-2x-3}{4x^3-13x^2+4x-3}}=0$
Learn how to solve problems step by step online. Find the roots of 3/((2x^3-5x^2-2x+-3)/(4x^3-13x^24x+-3)). Find the roots of the equation using the Quadratic Formula. Divide fractions \frac{3}{\frac{2x^3-5x^2-2x-3}{4x^3-13x^2+4x-3}} with Keep, Change, Flip: a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}. Multiply both sides of the equation by 2x^3-5x^2-2x-3. We can factor the polynomial \left(4x^3-13x^2+4x-3\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 -3.