Question

How do you factor ${x^3} - 7x + 6 = 0$?

Answer 1

Hint: In the given problem, we have to factorise the given polynomial equation and represent it as a product of its factors. The given polynomial is of degree $3$ and is thus called a cubic polynomial. For factoring the polynomial, first we have to simplify the given polynomial and then find its factors. For finding the factors of a cubic polynomial, we first find out a root of the polynomial by hit and trial method. Hence, we get one factor using hit and trial and then find the remaining two roots by dividing the given polynomial by the factor already found.

Complete step by step solution:
Consider the given polynomial ${x^3} - 7x + 6 = 0$ as $p\left( x \right)$.
Now, using hit and trial method,
Putting $x = 1$in the polynomial $p\left( x \right)$,
\[p(x = 1) = {\left( 1 \right)^3} - 7\left( 1 \right) + 6\]
$ \Rightarrow $$p\left( {x = 1} \right) = 1 - 7 + 6$
$ \Rightarrow $$p\left( {x = 1} \right) = 0$
Hence, $p(1) = 0$
So, we conclude that $x = 1$is a root of the polynomial. Hence, by factor theorem, $\left( {x - 1} \right)$ is a factor of the polynomial.
Now, $p(x) = \left( {x - 1} \right)\left( {{x^2} + x - 6} \right)$
Now quadratic polynomials $\left( {{x^2} + x - 6} \right)$ can be factored further by making use of splitting the middle term method.
For factoring the quadratic polynomial $\left( {{x^2} + x - 6} \right)$ , we can use the splitting method in which the middle term is split into two terms such that the sum of the terms gives us the original middle term and product of the terms gives us the product of the constant term and coefficient of ${x^2}$.
So, $\left( {{x^2} + x - 6} \right)$
$ \Rightarrow {x^2} + 3x - 2x - 6$
We split the middle term $x$ into two terms $3x$ and $ - 2x$ since the product of these terms, $ - 6{x^2}$ is equal to the product of the constant term and coefficient of ${x^2}$ and sum of these terms gives us the original middle term, $x$.
$ \Rightarrow x\left( {x + 3} \right) - 2\left( {x + 3} \right)$
$ \Rightarrow \left( {x + 3} \right)\left( {x - 2} \right)$
Therefore,
${x^3} - 7x + 6 = \left( {x - 1} \right)\left( {{x^2} + x - 6} \right)$
$ \Rightarrow {x^3} - 7x + 6 = \left( {x - 1} \right)\left( {x + 3} \right)\left( {x - 2} \right)$
So, factored form of $\left[ {{x^3} - 7x + 6} \right]$ is $\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right)$.
So, ${x^3} - 7x + 6 = 0$ can also be written as $\left( {x - 1} \right)\left( {x - 2} \right)\left( {x + 3} \right) = 0$

Note: Such polynomials can be factored by using the hit and trial method and then using factor theorem finding the factors of the polynomials. Cubic polynomials are polynomials with degree of variable as $3$.