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Home / Algebra / Solving Equations and Inequalities / Quadratic Equations - Part I
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Section 2.5 : Quadratic Equations - Part I

8. Use factoring to solve the following equation.

\[{x^4} - 2{x^3} - 3{x^2} = 0\]

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Start Solution

Do not let the fact that this equation is not a quadratic equation convince you that you can’t do it! Note that we can factor an \({x^2}\)out of the equation. Doing that gives,

\[{x^2}\left( {{x^2} - 2x - 3} \right) = 0\]

The quantity in the parenthesis is a quadratic and we can factor it. The full factoring of the equation is then,

\[{x^2}\left( {x - 3} \right)\left( {x + 1} \right) = 0\] Show Step 2

Now all we need to do is use the zero factor property to get,

\[\begin{array}{*{20}{c}}{{x^2} = 0}\\{x = 0}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{x - 3 = 0}\\{x = 3}\end{array}\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}\begin{array}{*{20}{c}}{x + 1 = 0}\\{x = - 1}\end{array}\]

Therefore the three solutions are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 0,\,\,x = 3\,\,{\mbox{and }}x = - 1}}\)