Question

$$n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) = 0$$

Answer

$$ \begin{matrix}n_1 = 0 & n_2 = 1 & n_3 = 2 \\[1 em] n_4 = 3 & n_5 = 4 & n_6 = 5 \\[1 em] n_7 = 6 & \\[1 em] \end{matrix} $$

Explanation

$$ \begin{aligned} n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6) &= 0&& \text{simplify left side} \\[1 em](1n^2-n)(n-2)(n-3)(n-4)(n-5)(n-6) &= 0&& \\[1 em](1n^3-2n^2-n^2+2n)(n-3)(n-4)(n-5)(n-6) &= 0&& \\[1 em](1n^3-3n^2+2n)(n-3)(n-4)(n-5)(n-6) &= 0&& \\[1 em](1n^4-3n^3-3n^3+9n^2+2n^2-6n)(n-4)(n-5)(n-6) &= 0&& \\[1 em](1n^4-6n^3+11n^2-6n)(n-4)(n-5)(n-6) &= 0&& \\[1 em](1n^5-10n^4+35n^3-50n^2+24n)(n-5)(n-6) &= 0&& \\[1 em](1n^6-15n^5+85n^4-225n^3+274n^2-120n)(n-6) &= 0&& \\[1 em]n^7-21n^6+175n^5-735n^4+1624n^3-1764n^2+720n &= 0&& \\[1 em] \end{aligned} $$

In order to solve $ \color{blue}{ x^{7}-21x^{6}+175x^{5}-735x^{4}+1624x^{3}-1764x^{2}+720x = 0 } $, first we need to factor our $ x $.

$$ x^{7}-21x^{6}+175x^{5}-735x^{4}+1624x^{3}-1764x^{2}+720x = x \left( x^{6}-21x^{5}+175x^{4}-735x^{3}+1624x^{2}-1764x+720 \right) $$

$ x = 0 $ is a root of multiplicity $ 1 $.

The remaining roots can be found by solving equation $ x^{6}-21x^{5}+175x^{4}-735x^{3}+1624x^{2}-1764x+720 = 0$.

$ \color{blue}{ x^{6}-21x^{5}+175x^{4}-735x^{3}+1624x^{2}-1764x+720 } $ is a polynomial of degree 6. To find zeros for polynomials of degree 3 or higher we use Rational Root Test.

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction $ \dfrac{p}{q} $, where p is a factor of the trailing constant and q is a factor of the leading coefficient.

The factor of the leading coefficient ( 1 ) is 1 .The factors of the constant term (720) are 1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 30 36 40 45 48 60 72 80 90 120 144 180 240 360 720 . Then the Rational Roots Tests yields the following possible solutions:

$$ \pm \frac{ 1 }{ 1 } , ~ \pm \frac{ 2 }{ 1 } , ~ \pm \frac{ 3 }{ 1 } , ~ \pm \frac{ 4 }{ 1 } , ~ \pm \frac{ 5 }{ 1 } , ~ \pm \frac{ 6 }{ 1 } , ~ \pm \frac{ 8 }{ 1 } , ~ \pm \frac{ 9 }{ 1 } , ~ \pm \frac{ 10 }{ 1 } , ~ \pm \frac{ 12 }{ 1 } , ~ \pm \frac{ 15 }{ 1 } , ~ \pm \frac{ 16 }{ 1 } , ~ \pm \frac{ 18 }{ 1 } , ~ \pm \frac{ 20 }{ 1 } , ~ \pm \frac{ 24 }{ 1 } , ~ \pm \frac{ 30 }{ 1 } , ~ \pm \frac{ 36 }{ 1 } , ~ \pm \frac{ 40 }{ 1 } , ~ \pm \frac{ 45 }{ 1 } , ~ \pm \frac{ 48 }{ 1 } , ~ \pm \frac{ 60 }{ 1 } , ~ \pm \frac{ 72 }{ 1 } , ~ \pm \frac{ 80 }{ 1 } , ~ \pm \frac{ 90 }{ 1 } , ~ \pm \frac{ 120 }{ 1 } , ~ \pm \frac{ 144 }{ 1 } , ~ \pm \frac{ 180 }{ 1 } , ~ \pm \frac{ 240 }{ 1 } , ~ \pm \frac{ 360 }{ 1 } , ~ \pm \frac{ 720 }{ 1 } ~ $$

Substitute the POSSIBLE roots one by one into the polynomial to find the actual roots. Start first with the whole numbers.

If we plug these values into the polynomial $ P(x) $, we obtain $ P(1) = 0 $.

To find remaining zeros we use Factor Theorem. This theorem states that if $\frac{p}{q}$ is root of the polynomial then this polynomial can be divided with $ \color{blue}{q x - p} $. In this example:

Divide $ P(x) $ with $ \color{blue}{x - 1} $

$$ \frac{ x^{6}-21x^{5}+175x^{4}-735x^{3}+1624x^{2}-1764x+720 }{ \color{blue}{ x - 1 } } = x^{5}-20x^{4}+155x^{3}-580x^{2}+1044x-720 $$

Polynomial $ x^{5}-20x^{4}+155x^{3}-580x^{2}+1044x-720 $ can be used to find the remaining roots.

Use the same procedure to find roots of $ x^{5}-20x^{4}+155x^{3}-580x^{2}+1044x-720 $

When you get second degree polynomial use step-by-step quadratic equation solver to find two remaining roots.

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