Question

How do you find the rest of the zeros given one of the zero is $\dfrac{1}{2}$ and the function $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$?

Answer 1

Hint: We take the help of long-term division. From the given root value of $\dfrac{1}{2}$ we find the factor as $2x-1$. Then we factorise the last quadratic equation using a quadratic solving process. We find the factored form and also the roots.

Complete step by step solution:
It is given that the zero is $\dfrac{1}{2}$ for the function $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$.
Therefore, $x=\dfrac{1}{2}$ is a root for $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$.
Simplifying we get $2x-1$ as the factor of $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$.
Now we divide $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$ by $2x-1$.
\[2x-1\overset{2{{x}^{3}}-13{{x}^{2}}+24x-9}{\overline{\left){\begin{align}
  & 4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9 \\
 & \underline{4{{x}^{4}}-2{{x}^{3}}} \\
 & -26{{x}^{3}}+61{{x}^{2}}-42x+9 \\
 & \underline{-26{{x}^{3}}+13{{x}^{2}}} \\
 & 48{{x}^{2}}-42x+9 \\
 & \underline{48{{x}^{2}}-24x} \\
 & -18x+9 \\
 & \underline{-18x+9} \\
 & 0 \\
\end{align}}\right.}}\]
The quotient is \[2{{x}^{3}}-13{{x}^{2}}+24x-9\] and we have to factorise it again.
We again find the value of $x=a$ for which the function $f\left( x \right)=2{{x}^{3}}-13{{x}^{2}}+24x-9=0$.
We take $x=a=3$.
We can see $f\left( 3 \right)=2\times {{3}^{3}}-13\times {{3}^{2}}+24\times 3-9=54-117+72-9=0$.
So, the root of the $f\left( x \right)=2{{x}^{3}}-13{{x}^{2}}+24x-9$ will be the function $\left( x-3 \right)$. 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 $\left( x-3 \right)$ is a factor of the polynomial \[2{{x}^{3}}-13{{x}^{2}}+24x-9\].
\[x-3\overset{2{{x}^{2}}-7x+3}{\overline{\left){\begin{align}
  & 2{{x}^{3}}-13{{x}^{2}}+24x-9 \\
 & \underline{2{{x}^{3}}-6{{x}^{2}}} \\
 & -7{{x}^{2}}+24x-9 \\
 & \underline{-7{{x}^{2}}+21x} \\
 & 3x-9 \\
 & \underline{3x-9} \\
 & 0 \\
\end{align}}\right.}}\]
Now we break the function \[2{{x}^{2}}-7x+3\] into a quadratic form.
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 $2{{x}^{2}}-7x+3=0$. The values of a, b, c are $2,-7,3$ respectively.
We put the values and get x as $x=\dfrac{7\pm \sqrt{{{7}^{2}}-4\times 2\times 3}}{2\times 2}=\dfrac{7\pm \sqrt{25}}{4}=\dfrac{7\pm 5}{4}=3,\dfrac{1}{2}$.
Therefore, the other roots of the function $f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9$ are $x=3,3,\dfrac{1}{2}$.
$f\left( x \right)=4{{x}^{4}}-28{{x}^{3}}+61{{x}^{2}}-42x+9={{\left( x-3 \right)}^{2}}{{\left( 2x-1 \right)}^{2}}$ is the factored form.

Note: In the factorisation we have two unique roots. They are in order of 2. Therefore, the total number of roots always will be 4 which is equal to the power value of the polynomial.