Question

What are the possible rational zeros for \[f(x) = 4{x^3} - 9{x^2} + 6x - 1\] and how do you find them?

Answer 1

Hint: To solve this question, we first need to find one of the possible rational zeros using the trial-and-error method. Next, we need to divide the original polymer with the factor we obtained using the previously obtained zero. The quotient will be a quadratic equation. We would have to factorize that quadratic equation, obtain all factors of the original polynomial, and then finally find the zeros using the obtained factors.

Complete step-by-step answer:
The Given Polynomial is \[f(x) = 4{x^3} - 9{x^2} + 6x - 1\] .
We need to find out its rational zeros. For the first zero, we need to use the trial-and-error approach.
By trial-and-error approach, we get \[x = 1\] for \[f(x) = 0\] . Thus, \[(x - 1)\] is a factor of \[f(x)\] .
Now, for the rest of the zeroes, we first need to obtain a quadratic expression along with the above factor by dividing \[f(x)\] with \[(x - 1)\] .
On performing the division, we obtain the following expression.
  \[f(x) = (x - 1)(4{x^2} - 5x + 1)\] .
For obtaining the remaining zeros, we need to factorize the quadratic part of \[f(x)\] , i.e., \[4{x^2} - 5x + 1\] .
This can be done by splitting \[5x\] into \[x\] and \[4x\] , so that pairing and factorization can be performed.
Thus, we have
  \[
  f(x) = 4{x^3} - 9{x^2} + 6x - 1 \\
  f(x) = (x - 1)(4{x^2} - 4x - x + 1) \\
   \Rightarrow f(x) = (x - 1)\left( {4x(x - 1) - 1(x - 1)} \right) \\
   \Rightarrow f(x) = (x - 1)\left( {(x - 1)(4x - 1)} \right) \\
   \Rightarrow f(x) = (x - 1)(x - 1)(4x - 1) \\
   \Rightarrow x = 1,1,\dfrac{1}{4} \;
 \]
We obtain \[x = 1,1,\dfrac{1}{4}\] if \[f(x) = 0\] . At these values of \[x\] , the value of \[f(x)\] will be zero. Hence, these values are the rational zeros of \[f(x) = 4{x^3} - 9{x^2} + 6x - 1\] .
So, the correct answer is “ \[x = 1,1,\dfrac{1}{4}\] if \[f(x) = 0\] ”.

Note: In some questions like this, two or more zeros of a polynomial can have the exact same value. However, they are counted as different zeroes. This is because the factors are being repeated in the factored expression of the polynomial.