Answer:
[tex](x-2)(x-5)(x+1)[/tex]
Step-by-step explanation:
We are factoring the cubic expression:
[tex]x^3-6x^2+3x+10[/tex]
First, we can list out the factors of the constant term:
10 → 1, 2, 5, -1, -2, -5
and figure out which ones result in a zero:
(1)
[tex](1)^3-6(1)^2+3(1)+10 \stackrel{?}{=} 0[/tex]
[tex]1-6+3+10 \stackrel{?}{=} 0[/tex]
[tex]8\ne 0[/tex] ❌
(2)
[tex](2)^3-6(2)^2+3(2)+10 \stackrel{?}{=} 0[/tex]
[tex]8-24+6+10 \stackrel{?}{=} 0[/tex]
[tex]0=0[/tex] ✅
(5)
[tex](5)^3-6(5)^2+3(5)+10 \stackrel{?}{=} 0[/tex]
[tex]125-150+15+10 \stackrel{?}{=} 0[/tex]
[tex]0=0[/tex] ✅
(-1)
[tex](-1)^3-6(-1)^2+3(-1)+10 \stackrel{?}{=} 0[/tex]
[tex]-1 -6 -3 + 10 \stackrel{?}{=} 0[/tex]
[tex]0=0[/tex] ✅
(-2)
[tex](-2)^3-6(-2)^2+3(-2)+10 \stackrel{?}{=} 0[/tex]
[tex]-8-24-6+10 \stackrel{?}{=} 0[/tex]
[tex]-38\ne 0[/tex] ❌
(-5)
[tex](-5)^3-6(-5)^2+3(-5)+10 \stackrel{?}{=} 0[/tex]
[tex]-125-150-15+10 \stackrel{?}{=} 0[/tex]
[tex]-280 \ne 0[/tex] ❌
Now, we can multiply together factors with the same zeros as we found for the cubic:
[tex]\boxed{(x-2)(x-5)(x+1)}[/tex]
Multiplying this out, we can see that no coefficient is needed because this factored expression is equal to the cubic expression:
[tex](x-2)(x-5)(x+1) = x^3 -6x^2 + 3x + 10[/tex]