Question

How do you solve $4{{x}^{3}}+3{{x}^{2}}+x+2=0$?

Answer 1

Hint: We factor the given equation with the help of the vanishing method. In this method we find a number $a$ such that for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We assume $f\left( x \right)=4{{x}^{3}}+3{{x}^{2}}+x+2$ and take the value of $a$ as $-1$.

Complete step by step solution:
We find the value of $x=a$ for which the function $f\left( x \right)=4{{x}^{3}}+3{{x}^{2}}+x+2=0$.
We take $x=a=-1$.
We can see $f\left( -1 \right)=4{{\left( -1 \right)}^{3}}+3{{\left( -1 \right)}^{2}}+\left( -1 \right)+2=-4+3-1+2=0$.
So, the root of the $f\left( x \right)=4{{x}^{3}}+3{{x}^{2}}+x+2$ will be the function $\left( x+1 \right)$. This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x+1 \right)$ is a factor of the polynomial $4{{x}^{3}}+3{{x}^{2}}+x+2$.
We can now divide the polynomial $4{{x}^{3}}+3{{x}^{2}}+x+2$ by $\left( x+1 \right)$.
\[x+1\overset{4{{x}^{2}}-x+1}{\overline{\left){\begin{align}
  & 4{{x}^{3}}+3{{x}^{2}}+x+2 \\
 & \underline{4{{x}^{3}}+4{{x}^{2}}} \\
 & -{{x}^{2}}+x+2 \\
 & \underline{-{{x}^{2}}-x} \\
 & 2x+2 \\
 & \underline{2x+2} \\
 & 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with $4{{x}^{2}}$. We get \[4{{x}^{3}}+4{{x}^{2}}\]. We subtract it to get \[-{{x}^{2}}+x+2\]. We again equate with the highest power of the remaining terms. We multiply with $-x$ and subtract to get \[2x+2\]. At the end we had to multiply with 2 to complete the division. The quotient is \[4{{x}^{2}}-x+1\].
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $4{{x}^{2}}-x+1=0$. The values of a, b, c are $4,-1,1$ respectively.
We put the values and get x as
$x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\times 4\times 1}}{2\times 4}\\
=\dfrac{1\pm \sqrt{-15}}{8}\\
=\dfrac{1\pm i\sqrt{15}}{8}$.
The equation $4{{x}^{3}}+3{{x}^{2}}+x+2=0$ has two imaginary roots and one real root $x=-1$.

Note: We find the value of x for which the function $f\left( x \right)=4{{x}^{3}}+3{{x}^{2}}+x+2=0$. The discriminant value being negative square, we get the imaginary numbers root values.
In this case the value of $D=\sqrt{{{b}^{2}}-4ac}$ is non-square ${{b}^{2}}-4ac={{\left( -1 \right)}^{2}}-4\times 4\times 1=-15$.
This is a negative square value. That’s why the roots are imaginary.