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My teacher told me that I can factor any polynomial function using this formula: $a(x-x_1)(x-x_2)$, where $x_1$ and $x_2$ are the roots of the quadratic equation $ax^2+bx+c$. Though I couldn't find this formula on any website. Example: $x^2+7x-6$ is factored in $(x-\frac{5+\sqrt{69}}{2})(x-\frac{5-\sqrt{69}}{2})$

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    For "nice" polynomials, yes. For higher order polynomials, you have to find (sufficiently many) linear factors until the given polynomial reduces to a quadratic function. – 2017-02-17
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    It's not entirely clear what you mean, if you allow complex numbers to appear it is possible by fundamental theorem of algebra, but if not, you will have trouble with something like $x^2+1$ – 2017-02-17
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    and by the way $x^2+7x-6$ does not factor into $(x-\frac{5+\sqrt{69}}{2})((x-\frac{5-\sqrt{69}}{2})$ – 2017-02-17
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    @user160738 in fact a website said that this polynomial is prime so it can't be factored – 2017-02-17
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    They mean it can't be factored over the rational numbers. It can be factored as $\left(x + \frac{7+\sqrt{73}}{2}\right)\left(x + \frac{7-\sqrt{73}}{2}\right)$, but $\sqrt{73}$ is irrational. – 2017-02-17
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    Your teacher probably said "_any **quadratic** polynomial_", not just "any polynomial". And then an interpretation of this claim depends on what numbers we are allowed to use: all complex, all reals, or only rationals, or only integers. This statement is true over complex numbers, but necessarily over reals. – 2017-02-17

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