Question

How do you find all rational roots for ${{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=0?$

Answer 1

Hint: We will factorize the given algebraic equation or polynomial of degree $4$ to reduce it into the equations of third, second and then first degree polynomials. That is, we factorize the given fourth degree polynomial to get a third degree polynomial and a first degree polynomial in the first step. Then, it will further be factorized to get only first degree polynomials.

Complete step by step solution:
Consider the polynomial ${{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=0$
This equation is factored as follows:
Let us find a factor of the given polynomial,
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15={{x}^{4}}+{{x}^{3}}-5{{x}^{2}}-5x-3{{x}^{3}}-3{{x}^{2}}+15x+15$
We are taking the common factors outside as,
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=x\left( {{x}^{3}}+{{x}^{2}}-5x-5 \right)-3\left( {{x}^{3}}+{{x}^{2}}-5x-5 \right)$
 Hence, by taking the common factors from this we get
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( {{x}^{3}}+{{x}^{2}}-5x-5 \right)$
Now we will repeat the above process again to factorize $\left( {{x}^{3}}+{{x}^{2}}-5x-5 \right)$, $\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( {{x}^{3}}+{{x}^{2}}-5x-5 \right)$
And we get the following,
 $\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( \left( x+1 \right){{x}^{2}}-5\left( x+1 \right) \right)$
Since we got the previous step, we can see that there is another common factor in this product on the right-hand side.
We are taking the common factor $x+1$,
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( \left( x+1 \right){{x}^{2}}-5\left( x+1 \right) \right)$
We will get,
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( \left( x+1 \right)\left( {{x}^{2}}-5 \right) \right)$
We can rewrite it as,
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( x+1 \right)\left( {{x}^{2}}-5 \right).$
From this we can find the solutions.
Recall our equation,
${{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=0$
Also, we have
$\Rightarrow {{x}^{4}}-2{{x}^{3}}-8{{x}^{2}}+10x+15=\left( x-3 \right)\left( x+1 \right)\left( {{x}^{2}}-5 \right).$
From this we will get the following equation,
$\left( x-3 \right)\left( x+1 \right)\left( {{x}^{2}}-5 \right)=0$
This is true if $x-3=0,$ $x+1=0,$ or ${{x}^{2}}-5=0$
From this we will get the following equations,
$\Rightarrow x=3,$ $x=-1,$ or ${{x}^{2}}=5$
This implies that \[x=3,\] $x=-1$ or \[x=\pm \sqrt{5}\]
This can also be written as $x=3,$ $x=-1,$ $x=\sqrt{5},$ or $x=-\sqrt{5}$
There for the rational solutions of the above given polynomials are $x=3,$ $x=-1,$ $x=\sqrt{5},$ or $x=-\sqrt{5}.$

Note: We have done this procedure using factor by grouping.
This can be done using the following theorem which is called the rational root theorem:
Any rational roots of a polynomial $f\left( x \right)$ must be expressible in the form $\dfrac{p}{q}$ for integers $p,q$ with $p$ a divisor of the constant term and $q$ a divisor of the coefficient of the term with the highest power.