Question

Number of transformed equations of $x^3+2x^2+x+1=0$ by eliminating third term

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

To eliminate the third term from the equation $x^3+2x^2+x+1=0$......(1) let us use the substitution $x+h$ in place of $x$ in (1).

The we have,

$(x+h{)}^3+2(x+h{)}^2+(x+h)+1=0$

or, $(x^3+3x^{2}h+3x{h}^2+h^3)+2(x^2+2xh+h^2)+(x+h)+1=0$

or, $x^3+x^2(3h+2)+x(3h^2+4h+1)+(h^3+2h^2+h+1)=0$......(2)

To eliminate the third term from the equation we mus have $3h^2+4h+1=0\Rightarrow (h+1)(3h+1)\Rightarrow h=-1,-\frac{1}{3}$.

As we are getting two values of $h$ so there will be two transformed equation after putting the values of $h$ in equation (2).