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Could someone please explain how to solve this : $x^4 - 10x^3 + 21x^2 + 40x - 100 = 0$ - not the answer only, but a step-by-step solution. I tried to solve it, with the help of khanacademy, but still I have no idea how to correctly solve it.

Thank you so much in advance!

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    $\begingroup$ How to solve a quartic equation? Three ways: (i) Numerically; or (ii) Use the Ferrari-Cardano procedure, or one of its kin; or (iii) Pray that by design it is one of the relatively few such equations that collapse, because there are some very simple roots. Prayer works for quartics in homework exercises. $\endgroup$ – André Nicolas Jan 30 '12 at 21:06
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    $\begingroup$ Actually, 4 ways: (iv) Use a computer. Mathematica solves this equation in 0.001764 seconds: {{x -> -2}, {x -> 2}, {x -> 5}, {x -> 5}}. $\endgroup$ – David G. Stork Sep 4 '15 at 16:21
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Since the polynomial has integer coefficients, the rational root theorem applies. Thus any rational root must be of the form $x=\pm p/q$, where $p$ divides the constant term 100 and $q$ divides the leading coefficient 1. In this case, the only possibility for $q$ is 1. This tells you that any rational root must be a divisor of $100=2^2*5^2$. It turns out that this polynomial does have rational roots, after which you find one you can perform polynomial division to get a complete factorization.

For instance, we have the potential rational roots $x=\pm2,\pm5,\pm10,\pm20\pm25,\pm50,\pm100$. We could plug in $x=5$ and verify that this is a root. Then, $$ \frac{x^4 - 10x^3 + 21x^2 + 40x - 100}{x-5} = x^3-5x^2-4x+20. $$ Since all the roots are rational, repeating this process will generate all of them. Not every polynomial with integer coefficients has rational roots (for instance $x^2-2=0$), so this won't always be the case.

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  • $\begingroup$ Thank You, it was very helpful! $\endgroup$ – Uzdawi Jan 30 '12 at 19:07
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$$x^4-10x^3 +25x^2-(4x^2-40x+100)=0$$

$$\Rightarrow x^2(x^2-10x+25)-4(x^2-10x+25)=0 $$

$$\Rightarrow (x^2-4)(x^2-10x+25)=0 $$

$$\Rightarrow (x-2)(x+2)(x-5)^2=0 $$

$$\Rightarrow x_1=-2 , x_2=2 , x_{3,4}=5$$

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    $\begingroup$ How did you even see that :O $\endgroup$ – Airdish Mar 12 '16 at 15:28
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First note that $x= \pm 2$ are the roots of the equation. Hence using the remainder theorem, we can proceed as follows:

$$ x^3(x-2) - 8x^2(x-2) +5x(x-2) + 50(x-2)=0$$ $$\qquad \Rightarrow (x-2)(x^3 -8x^2 +5x +50) =0$$ $$ (x-2)( x^2(x+2) -10x(x+2) +25(x+2)) =0 $$ $$\qquad \Rightarrow (x^2-4)(x^2-10x+25)=0 $$ Hence we can solve it as

$$ (x^2-4)(x-5)^2 =0 $$ or $$x=\{ -2,+2,5 \}$$

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The idea is to represent the polynomial $p(x) = x^4 - 10x^3 + 21x^2 + 40x - 100$ as a product of simple factors. By Vietta formula the free terms equals to the product of roots. Thus, if we confine ourselves to integer roots, they should divide $100$. Start with $x=2$, and verify that $p(2) = 16 - 10 \times 8 + 21 \times 4 + 40 \times 2 - 100 = 16 -80 + 84 - 80 + 100 = 0$.

Now applying long division we compute $$ p(x) = (x-2)\left( x^3 - 8 x^2 + 5 x + 50 \right) $$ Now try $x=5$. $5^2(5-8) + 5 ( 5+ 10) = 5( -15 + 15) = 0$. Apply long division again: $$ p(x) = (x-2)(x-5) \left(x^2 - 3 x - 10\right) = (x-2)(x+2)(x-5)^2 $$

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  • $\begingroup$ Thank You for explaining it so clearly! $\endgroup$ – Uzdawi Jan 30 '12 at 19:11
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    $\begingroup$ very last line should have $x^2-3x-10$? instead of $x^3$ $\endgroup$ – snulty Mar 16 '16 at 19:35

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