Factor the polynomial x^3+2x^2-13x+10

Factor the polynomial x^3+2x^2-13x+10

f(x)= x^3+2x^2-13x+10

In order to factor the polynomial we need to know its roots. Using a guess (usually try 0, 1, -1, 2, -2, etc.) ,notice that f(1) = 0, since

1^3 + 2*1^2 - 13*1 + 10 = 0

Now, using synthetic division we can form factor out this root

f(x)/(x-1) = ( x^3+2x^2-13x+10 ) / (x-1)

The equivalent long division setup (using, e.g. Horner's method) is

1 2 -13 10
______________
-1|| -1 -3 10
______________
1 3 -10 0

The last line is the coefficients of the resulting quadratic equation, which is 1*x^2 + 3*x - 10 (plus remainder 0, which shows that 1 is a root). Thus,

f(x) = (x-1) (x^2 + 3x - 10).

The last quadratic term can be factored using standard algebraic methods; guessing at the roots, 5 and 2 are likely, since 5-2 = 3 for the middle coefficient, and 5*2 = 10 for the last one. In fact,

x^2 + 3x - 10 = (x+5)(x-2)

Thus,

f(x) = (x-1)(x+5)(x-2).

let f(x) = x^3+2x^2-13x+10

then f(1) = 1+2-13+10
so u get x=1 that is (x-1) is a factor.

divide so u get (x-1)(x^2+3x-10)

now u can see (x^2+3x-10) equals (x+2)(x-5)

therefore x^3+2x^2-13x+10 = (x-1)(x+2)(x-5)...

**** this is gonna be on my finals in 2 weeks and i cant even solve this.