Question

What are the rational zeros for $ f(x) = 5{x^3} - 31{x^2} + 31x - 5 $ and how do you find all zeros?

Answer 1

Hint: In this question, we are given a cubic equation and we have to find its rational zeros. Zeros of a polynomial are those values of the unknown variables at which the value of the function comes out to be zero. Rational numbers are of the form $ \dfrac{p}{q} $ where p and q are integers and q is not equal to zero.

Complete step-by-step answer:
We have to find the zeros of $ f(x) = 5{x^3} - 31{x^2} + 31x - 5 $
So, we will put this equation equal to zero.
 $
   \Rightarrow 5{x^3} - 31{x^2} + 31x - 5 = 0 \\
   \Rightarrow 5{x^3} - 5{x^2} - 26{x^2} + 26x + 5x - 5 = 0 \\
   \Rightarrow 5{x^2}(5x - 1) - 26x(x - 1) + 5(x - 1) = 0 \\
   \Rightarrow (x - 1)(5{x^2} - 26x + 5) = 0 \;
  $
Now, the obtained equation is a quadratic equation so we solve it by factorization too –
 $
   \Rightarrow (x - 1)(5{x^2} - 25x - x + 5) = 0 \\
   \Rightarrow (x - 1)[5x(x - 5) - 1(x - 5)] = 0 \\
   \Rightarrow (x - 1)(5x - 1)(x - 5) = 0 \\
   \Rightarrow x - 1 = 0,\,5x - 1 = 0,\,x - 5 = 0 \\
   \Rightarrow x = 1,\,x = \dfrac{1}{5},\,x = 5 \;
  $
We will reject the non-rational values of x, but we see that all the zeros obtained are rational so none of them is rejected.
Hence the zeros of $ f(x) = 5{x^3} - 31{x^2} + 31x - 5 $ is $ x = 1,\,x = \dfrac{1}{5},\,x = 5 $
So, the correct answer is “ $ x = 1,\,x = \dfrac{1}{5},\,x = 5 $ ”.

Note: A polynomial equation is an algebraic expression in which the unknown variable quantities are raised to some power. The highest exponent of the polynomial equation is known as its degree, so the degree of the equation is 3, a polynomial equation has exactly as many as zeros as the degree of the equation so it has 3 zeros. We find a few zeros of a polynomial by hit and trial method until we get a quadratic equation. After getting the quadratic equation, we solve it by using factorization but if we are not able to solve an equation using factorization then we use the quadratic formula to find the solutions.