Question

factorise x³-2x²-5x+6

Answer 1

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Answer 2

So the solution is $(x-1)(x+2)(x-3)$

Step-by-step explanation:

Given;

$f(x)=x^{3}-2x^{2} -5x+6$

Putting factor of $-6$ to get value of $f(x)$. If $f(x)=0$ then $x$ is the root.

Factor of $6$ are; $1,-1,2,-2,3,-3,6,-6$

$f(1)=1^{3}-2\times1^{2} -5\times1+6=0$

∴ $1$ is the root.

∴$(x-1)$ is the factor.

$f(-1)=-1^{3}-2\times-1^{2} -5\times-1+6=8$

∴$-1$ is not root.

∴$(x+1)$ is not factor.

$f(2)=2^{3}-2\times2^{2} -5\times2+6=-4$

∴$2$ is not root.

∴$(x-2)$  is not factor.

$f(-2)=-2^{3}-2\times-2^{2} -5\times-2+6=0$

∴$-2$ is root.

∴$(x+2)$  is factor.

$f(3)=3^{3}-2\times3^{2} -5\times3+6=0$

∴$3$ is root.

∴$(x-3)$  is factor.

$f(-3)=-3^{3}-2\times-3^{2} -5\times-3+6=-24$

∴$-3$ is not root.

∴$(x+3)$  is not factor.

∴ $f(x)=(x-1)(x+2)(x-3)$

∴The solution is $(x-1)(x+2)(x-3)$