Solve x^5 + 30x^4 + 75x^3 - 3250x^2 - 15000x = 0.

Step-by-step solution

1. 1

   Take out the obvious common factor. Every term contains $x$:

   $x\left(x^4+30x^3+75x^2-3250x-15000\right)=0$

   so $x=0$ is a root immediately, and the remaining work is a quartic.

2. 2

   Search the divisors of 15000 for an integer root. Testing $x=10$ in the quartic:

   $10000+30000+7500-32500-15000=0\;\checkmark$

   so $(x-10)$ is a factor. Testing is cheap and decisive; guessing large roots such as $\pm 25$ without evaluating is exactly how wrong factorizations get published.

3. 3

   Divide and keep factoring. Synthetic division of the quartic by $x-10$ leaves

   $x^3+40x^2+475x+1500$

   and $x=-5$ satisfies this cubic: $-125+1000-2375+1500=0\;\checkmark$. Dividing again by $x+5$ gives the quadratic

   $x^2+35x+300$

4. 4

   Factor the final quadratic. Two numbers with product $300$ and sum $35$ are $15$ and $20$:

   $x^2+35x+300=(x+15)(x+20)$

   so the complete factorization is

   $x^5+30x^4+75x^3-3250x^2-15000x=x(x-10)(x+5)(x+15)(x+20)$

5. 5

   List the roots and verify them all.

   $x=0,\;10,\;-5,\;-15,\;-20$

   Two independent checks confirm the set. Vieta: the roots must sum to $-30$, and $0+10-5-15-20=-30\;\checkmark$; the product of the four nonzero roots must equal the quartic’s constant $-15000$ divided by its leading coefficient $1$, and $10\cdot(-5)\cdot(-15)\cdot(-20)=-15000\;\checkmark$. Numerically solving the quintic returns exactly $10,-20,-15,-5,0$ as well.