Question

factorion x³-6x²+3x+10​

Answer 1

Answer:

$\underline{\huge{\texttt{(x - 2),\: (x + 1),\: (x - 5)}}}$

Step-by-step explanation:

Hey Readers,

$p(x)=x^3-6x^2+3x+10\\p(2)=(2)^3-6(2)^2+3(2)+10=8-24+6+10=0$

One factor is $(x-2)$

$g(x)=(x-2)$

Dividing p(x) by g(x) that is $(x^3-6x^2+3x+10)\div(x-2)$ we get $x^2-4x-5$

Splitting the middle term, we get,

$x^2-4x-5\\x^2-5x+1x-5\\(x^2-5x)(x-5)\\x(x-5)+1(x-5)\\(x+1)(x-5)$

So the factors are:-

$(x-2), (x+1), (x-5)$

Thanks!!