Factor theorem and polynomial solution

Question

Find the value of $m$ if $(x-m)$ is a factor of $x^2-m^2 x+x+2$. I know if $(x-m)$ is a factor of $f (x)$ then $f(m)$ must be zero. But I could not reduce it.

Answer 1

When you divide $x^2-m^2x+x+2$ by $(x-m)$, you get-

the quotient part is $=x-m^2+1+m$, and the remainder part is $=2-m^3+m^2+m$.

Since, $(x-m)$ is a factor, the remainder part has to be zero.

The equation, $m^3-m^2-m-2$, has $3$ roots, namely- $m_1=2, \ m_2=-0.5(1+\sqrt 3i) \text{ and } m_3 =-0.5(1-\sqrt 3i)$.

Knowing the values of $m$, you can write the expression of the factor $(x-m)$ using the appropriate numerical value e.g. $(x-2)$.

Answer 2

HINT: Notice, $(x-m)$ is a factor of $x^2-m^2\cdot x+x+2$ then substituting $x=m$, the value of the polynomial should be zero, hence we get $$(m)^2-m^2(m)+m+2=0$$ $$m^3-m^2-m-2=0$$ You can solve above cubic equation for real value of $m$ using Newton's Method which gives $m=2$ real root hence factorizing the polynomial $$(m-2)(m^2+m+1)=0$$

Answer 3

You have that $f(m)={m}^{2}-{m}^{3}+m+2=- \left( m-2 \right) \left( {m}^{2}+m+1 \right)=0.$ Thus $m=2$ or $m$ is a root of the equation ${m}^{2}+m+1=0.$