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

How do you find the zeros of the polynomial functionf(x)=x3−x2−42xf\left( x \right)={{x}^{3}}-{{x}^{2}}-42x? $$$$

Explanation

Solution

We recall that the zeros of any polynomial p(x)p\left( x \right) are the value of xx for which p(x)=0p\left( x \right)=0. We find the zeroes of the given polynomial by factorization. We take xx common and find a quadratic polynomial of the type ax2+bx+ca{{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(x)p\left( x \right) for which p(x)=0p\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 ax2+bx+ca{{x}^{2}}+bx+c by splitting the middle term b=p+qb=p+qsuch that pq=c×apq=c\times a. $$$$
We are asked to find the zeros of the polynomial function
f(x)=x3−x2−42xf\left( x \right)={{x}^{3}}-{{x}^{2}}-42x
We see that xx is multiplied with each of the terms of f(x)f\left( x \right). We take xx common to have;
f(x)=x(x2−x−42)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,qp,q to split the middle term −x-xsuch that p+q=−1,pq=1(−42)=−42p+q=-1,pq=1\left( -42 \right)=-42. We know that 6×7=426\times 7=42. If we take p=6,q=−7p=6,q=-7 we shall get pq=6(−7)=−42,p+q=6+(−7)=−1pq=6\left( -7 \right)=-42,p+q=6+\left( -7 \right)=-1. So we proceed to have

& \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{aligned}$$ We take $x$ common in the first two terms and $-7$common in last two terms to have ; $$\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{aligned} & \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{aligned}$$ Let us equate $f\left( x \right)$ to zero to find the zeros. We have $$\begin{aligned} & f\left( x \right)=0 \\\ & \Rightarrow x\left( x+6 \right)\left( x-7 \right)=0 \\\ \end{aligned}$$ If the product of three factors is zero at least one of them is zero. So we have; $$\begin{aligned} & \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{aligned}$$ 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 .