Question

Using factor theorem, factorise: $x^3-6x^2+3x+10$

Answer 1

For factoring $x^3-6x^2+3x+10,$ we should know atleast one zero of this polynomial. Once we know it, we can divide the polynomial by the factor to find the quotient and factor the quotient further to find other zeroes.

Keeping $x=1,{\left(1\right)}^3-6{\left(1\right)}^2+3\left(1\right)+10\ne 0$

$\Rightarrow x=2,{\left(2\right)}^3-6{\left(2\right)}^2+3\left(2\right)+10=0$

so, $(x-2)$ is a factor.

Now,

$\Rightarrow$ Quotient $=x^2-4x-5$

Using common factor theorem.,

$x^2+x-5x-5$

$x(x+1).5(x+1)$

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

To find zeroes $(x+1)(x-5)=0$

$\Rightarrow x=-1,x=5$

So, the zeroes of polynomial $x=-1,2,5.$

Hence, the answer should be $(x+1)(x-2)(x-5)$