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Question

Question: How do you factor \({{x}^{3}}-{{x}^{2}}-2x+2=0\)?...

How do you factor x3x22x+2=0{{x}^{3}}-{{x}^{2}}-2x+2=0?

Explanation

Solution

In order to factor first we will take common terms out from the given polynomial by grouping the terms. Then we will further simplify the obtained equation to get the factors of the given polynomial.

Complete step-by-step solution:
We have been given a polynomial x3x22x+2=0{{x}^{3}}-{{x}^{2}}-2x+2=0.
We have to find the factors of the given polynomial.
Now, we know that there is no suitable method to factorize the polynomial of degree 3. So we need to solve it numerically.
Now, we can rewrite the given polynomial by taking common terms out. Then we will get
x3x22x+2=0 x2(x1)2(x1)=0 \begin{aligned} & \Rightarrow {{x}^{3}}-{{x}^{2}}-2x+2=0 \\\ & \Rightarrow {{x}^{2}}\left( x-1 \right)-2\left( x-1 \right)=0 \\\ \end{aligned}
As (x1)\left( x-1 \right) is the common factor so taking the common factor out we will get
(x1)(x22)=0\Rightarrow \left( x-1 \right)\left( {{x}^{2}}-2 \right)=0
Now, equating each factor to zero we will get
x1=0\Rightarrow x-1=0 and x22=0\Rightarrow {{x}^{2}}-2=0
Simplifying each equation we will get
x=1\Rightarrow x=1
Let us consider x22=0\Rightarrow {{x}^{2}}-2=0
Now, we can rewrite the equation as x2(2)2=0\Rightarrow {{x}^{2}}-{{\left( \sqrt{2} \right)}^{2}}=0
Now, we know that a2b2=(ab)(a+b){{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)
Applying the formula to the above equation we will get
(x2)(x+2)=0\Rightarrow \left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right)=0
So we get x=2\Rightarrow x=\sqrt{2} and x=2\Rightarrow x=-\sqrt{2}
Hence we get the factors as (x1)(x2)(x+2)\left( x-1 \right)\left( x-\sqrt{2} \right)\left( x+\sqrt{2} \right).

Note: As the given polynomial is of the degree 3 so it is difficult to factor it by using a method like split middle term or completing the square method. Students may give the consider the factors of the given polynomial as (x1)(x22)\left( x-1 \right)\left( {{x}^{2}}-2 \right) which is also not incorrect but we can further factorize the expression (x22)\left( {{x}^{2}}-2 \right) so it is necessary to solve onward.