Question

How do you factor ${x^3} - 2x + 1$ ?

Answer 1

Hint: In this question, we will use the concept of the formula for the quadratic equation. First, we will split the terms such that we will obtain some common factor which will lead to the linear and quadratic equation, then use the concept of the formula to obtain the factor of the quadratic equation.

Complete step by step solution:
In this question, we have given a function ${x^3} - 2x + 1$, we need to provide the factor of the given function.
In the given function, first we will split $2x$ as $\left( {x + x} \right)$ and the expression become,
$ \Rightarrow h\left( x \right) = {x^3} - \left( {x + x} \right) + 1$
Now, as we know that the product of positive and the negative number is always a negative number while the product of two negative number is always a positive number, so apply this concept in the above expression and obtain,
$ \Rightarrow h\left( x \right) = {x^3} - x - x + 1$
Now, we will make the group of two terms like,
$ \Rightarrow h\left( x \right) = \left( {{x^3} - x} \right) - \left( {x - 1} \right)$
From the above expression we can see that we can take $x$ as common from the first term and $1$ from the second term combination, so the expression become,
$ \Rightarrow h\left( x \right) = x\left( {{x^2} - 1} \right) - 1\left( {x - 1} \right)$
Now we rewrite the above expression as,
$ \Rightarrow h\left( x \right) = x\left( {{x^2} - {1^2}} \right) - 1\left( {x - 1} \right)$
As we know that, if the expression is in the form $\left( {{a^2} - {b^2}} \right)$ then we can write it as,
$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$
Now, we will use the above identity to the simplified expression as,
$ \Rightarrow h\left( x \right) = x\left( {x - 1} \right)\left( {x + 1} \right) - 1\left( {x - 1} \right)$
Now, we will take $\left( {x - 1} \right)$ as the common from each term,
$ \Rightarrow h\left( x \right) = \left( {x - 1} \right)\left( {x\left( {x + 1} \right) - 1} \right)$
Now we will simplify the second factor as,
$ \Rightarrow h\left( x \right) = \left( {x - 1} \right)\left( {{x^2} + x - 1} \right).....\left( 1 \right)$
As we know that if the quadratic equation is in the form $a{x^2} + bx + c$, then the roots of that equation will be,
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, we will apply this formula in the simplified expression’s factor $\left( {{x^2} + x - 1} \right)$ and obtain,
$ \Rightarrow {x_{1,2}} = \dfrac{{ - 1 \pm \sqrt {{{\left( 1 \right)}^2} - 4\left( 1 \right)\left( { - 1} \right)} }}{{2\left( 1 \right)}}$
After simplification we will get,
$ \Rightarrow {x_1} = \dfrac{{ - 1 + \sqrt 5 }}{2}$
$ \Rightarrow {x_2} = \dfrac{{ - 1 - \sqrt 5 }}{2}$
So, we will get the factor of the expression as,
$\left( {{x^2} + x - 1} \right) = \left( {x - \dfrac{{ - 1 + \sqrt 5 }}{2}} \right)\left( {x - \dfrac{{ - 1 - \sqrt 5 }}{2}} \right)$
From equation (1) we will get the factor of the expression as,
\[\therefore h\left( x \right) = \left( {x - 1} \right)\left( {x - \dfrac{{ - 1 + \sqrt 5 }}{2}} \right)\left( {x - \dfrac{{ - 1 - \sqrt 5 }}{2}} \right)\]

Therefore, the factors of the given expression are \[\left( {x - 1} \right)\left( {x - \dfrac{{ - 1 + \sqrt 5 }}{2}} \right)\left( {x - \dfrac{{ - 1 - \sqrt 5 }}{2}} \right)\].

Note:
As we know that the factors of the quadratic equation can be calculated by splitting the middle term such that the sum of the numbers is equal to the coefficient of the $x$ while the product of the number should be equal to the coefficients of ${x^2}$ and the constant.