Question

How do you find the zeros of the polynomial function$f\left( x \right)={{x}^{3}}-{{x}^{2}}-42x$? \[\]

Answer 1

Hint: We recall that the zeros of any polynomial $p\left( x \right)$ are the value of $x$ for which $p\left( x \right)=0$. We find the zeroes of the given polynomial by factorization. We take $x$ common and find a quadratic polynomial of the type $a{{x}^{2}}+bx+c$ in the bracket. We use splitting the middle term method to factorize the quadratic polynomial. \[\]

Complete answer:
We know that the zeros or roots of any polynomial $p\left( x \right)$ for which $p\left( x \right)=0$. Linear factorization of polynomials is one of the methods to find zeros. We know from splitting the middle term method that we can factorize quadratic polynomial $a{{x}^{2}}+bx+c$ by splitting the middle term $b=p+q$such that $pq=c\times a$. \[\]
We are asked to find the zeros of the polynomial function
\[f\left( x \right)={{x}^{3}}-{{x}^{2}}-42x\]
We see that $x$ is multiplied with each of the terms of $f\left( x \right)$. We take $x$ common to have;
\[f\left( x \right)=x\left( {{x}^{2}}-x-42 \right)\]
We see that the expression in the bracket is quadratic polynomial. So we need to find $p,q$ to split the middle term $-x$such that $p+q=-1,pq=1\left( -42 \right)=-42$. We know that $6\times 7=42$. If we take $p=6,q=-7$ we shall get $pq=6\left( -7 \right)=-42,p+q=6+\left( -7 \right)=-1$. So we proceed to have
\[\begin{align}
  & \Rightarrow f\left( x \right)=x\left( {{x}^{2}}+\left( 6-7 \right)x-42 \right) \\
 & \Rightarrow f\left( x \right)=x\left( {{x}^{2}}+6x-7x-42 \right) \\
\end{align}\]
We take $x$ common in the first two terms and $-7$common in last two terms to have ;
\[\Rightarrow f\left( x \right)=x\left\{ x\left( x+6 \right)-7\left( x+6 \right) \right\}\]
We take $x+6$ common in the bracket to have;
\[\begin{align}
  & \Rightarrow f\left( x \right)=x\left\{ \left( x+6 \right)\left( x-7 \right) \right\} \\
 & \Rightarrow f\left( x \right)=x\left( x+6 \right)\left( x-7 \right) \\
\end{align}\]
Let us equate $f\left( x \right)$ to zero to find the zeros. We have
\[\begin{align}
  & f\left( x \right)=0 \\
 & \Rightarrow x\left( x+6 \right)\left( x-7 \right)=0 \\
\end{align}\]
If the product of three factors is zero at least one of them is zero. So we have;
\[\begin{align}
  & \Rightarrow x=0\text{ or }x+6=0\text{ or }x-7=0 \\
 & \Rightarrow x=0\text{ or }x=-6\text{ or }x=7 \\
\end{align}\]
So the zeros of $f\left( x \right)$ are $x=0,x=-6,x=7$.

Note: We can alternatively factorize using factor theorem which states that $x-a$ is a factor of $p\left( x \right)$ if and only if $p\left( a \right)=0$.We have here use trial and error to guess a root $x=0$ and then use polynomial division. We can solve the problem graphically by plotting $f\left( x \right)$ in a graph and see where it cuts the $x-$axis. We note that a polynomial with real coefficients of degree 3 has either 3 real roots or one real and two conjugate complex roots .