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Find the break even points of the polynomial $\frac{x^2-3x-7}{\left(x+1\right)^2\left(2x+3\right)}$ by putting it in the form of an equation and then set it equal to zero
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$\frac{x^2-3x-7}{\left(x+1\right)^2\left(2x+3\right)}=0$
Learn how to solve classify algebraic expressions problems step by step online. Find the break even points of the expression (x^2-3x+-7)/((x+1)^2(2x+3)). Find the break even points of the polynomial \frac{x^2-3x-7}{\left(x+1\right)^2\left(2x+3\right)} by putting it in the form of an equation and then set it equal to zero. Multiply both sides of the equation by \left(x+1\right)^2\left(2x+3\right). To find the roots of a polynomial of the form ax^2+bx+c we use the quadratic formula, where in this case a=1, b=-3 and c=-7. Then substitute the values of the coefficients of the equation in the quadratic formula: \displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. Simplifying.