Question

How do you solve ${x^3} - 3{x^2} + 16x - 48 = 0$?

Answer 1

Hint: In this question we have to solve the given cubic polynomial, first factor the first two terms and then factor the second term, next rewrite the polynomial with the factored terms and equate each term to zero and then solve the equation to get the required values for $x$.

Complete step by step solution:
Given cubic polynomial is ${x^3} - 3{x^2} + 16x - 48 = 0$,
The given polynomial is a cubic polynomial and the degree of the polynomial will be 3 and so it will have three zeros.
Now observe that the first two terms in the cubic polynomial can be factored as ${x^3} - 3{x^2} = {x^2}\left( {x - 3} \right)$,
And now the second pair of terms can be factored as $16x - 48 = 16\left( {x - 3} \right)$,
Now rewriting the polynomial as,
$ \Rightarrow {x^2}\left( {x - 3} \right) + 16\left( {x - 3} \right) = 0$,
Now taking out the common terms , we get,
$ \Rightarrow \left( {{x^2} + 16} \right)\left( {x - 3} \right) = 0$,
Now equate each term to zero, we get, i.e., equate first term to zero, we get
$ \Rightarrow x - 3 = 0$,
Now add 3 to both sides of the equation we get,
$ \Rightarrow x - 3 + 3 = 0 + 3$,
Now simplifying we get,
$ \Rightarrow x = 0$,
Now equate second term to zero, we get,
$ \Rightarrow {x^2} + 16 = 0$,
Now subtract 16 to both sides of the equation we get,
$ \Rightarrow {x^2} + 16 - 16 = 0 - 16$,
Now simplifying we get,
$ \Rightarrow {x^2} = - 16$,
Now taking square root on both sides we get,
$ \Rightarrow \sqrt {{x^2}} = \sqrt { - 16} $,
Now simplifying we get,
$ \Rightarrow x = \sqrt { - 16} $,
Now simplifying we get,
$ \Rightarrow x = \sqrt {\left( { - 1} \right)16} $,
We know that .., substituting the value in the expression we get,
$ \Rightarrow x = \sqrt {{4^2} \cdot {i^2}} $,
Now further simplifying we get,
$ \Rightarrow x = \pm 4i$,
So, the values for $x$ are $3$ and $ \pm 4i$.

$\therefore $ The values of $x$ when the given polynomial i.e., ${x^3} - 3{x^2} + 16x - 48 = 0$ is solved will be equal to $3$ and $ \pm 4i$.

Note:
The traditional way of solving a cubic equation is to reduce it to a quadratic equation and then solve either by factoring or quadratic formula.
Like a quadratic equation has two real roots, a cubic equation may have possibly three real roots. But unlike quadratic equations which may have no real solution, a cubic equation has at least one real root. The other two roots might be real or imaginary. Whenever we are given a cubic equation, or any equation, you always have to arrange it in a standard form first.