Question

How do you solve the inequality ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20<0$ $?$

Answer 1

Hint: This is a polynomial of $4$ degree. To solve this question we need to find the roots of the given polynomial. For solving the question the knowledge of Inequation should also be known to us. To find the roots of $4$ degree polynomial we need to divide the polynomial unless we get the polynomial of $2$ degree or is a quadratic polynomial.

Complete step-by-step solution:
For the polynomial with more than $2$ degree, we get one of the roots by hit and trial method. In the hit and trial method we are supposed to find the roots randomly, check the number and see for which number the polynomial turns to be $0$. So in this problem we will be using the same concept to find the roots until we get quadratic polynomials.
To find the root using hit and trial method for the given polynomial ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-
4x-20$, we get $1$ as the root. So the polynomial could be written as
${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20=(x-1)f(x)$
We need to find the value of the other root so we need to know the quotient which is $f(x)$
$f(x)=\dfrac{{{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20}{x-1}$
$= {{x}^{3}}+9{{x}^{2}}+24x+20$
Again we have another polynomial with $3$ as the degree. Here again we will use the hit and trial method to find the root of the polynomial ${{x}^{3}}+9{{x}^{2}}+24x+20$. Using this method the root for this equation is $2$.
$({{x}^{3}}+9{{x}^{2}}+24x+20)=(x+2)g(x)$
We need to find the value of the other root so we need to know the quotient which is $g(x)$
$g(x)=\dfrac{{{x}^{3}}+9{{x}^{2}}+24x+20}{x+2}$
$= {{x}^{2}}+7x+10$
Now we get quadratic polynomials, roots of this polynomial can be found by factorisation method.
On middle term factoring the polynomial ${{x}^{2}}+7x+10$ we get
$\Rightarrow {{x}^{2}}+2x+5x+10$
Taking $x$ and $5$ common we get
$\Rightarrow x(x+2)+5(x+2)$
On further calculating we get roots as
$\Rightarrow (x+2)(x+5)$
So, all the roots of the given polynomial ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20$ are $1,-2,-2$ and $-
5$ .
$\therefore $ The polynomial could be written as
\[{{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20=(x-1)(x+2)(x+2)(x+5)\]
Given a question, ask us to get the values of $x$ for which the polynomial becomes less than 0.
${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20<0$
$\Rightarrow (x-1)(x+2)(x+2)(x+5)<0$
We will now have found the range of $x$ where the polynomial turns to $0$. To do this we will use the number line and will check at what values of $x$ the polynomial turns to be negative.

We will check a value from each interval and see in which interval does the value of $x$ give polynomials.
On doing so we infer that for values of $x$ above $1$and below $-5$ the polynomial ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20$ becomes positive while for the values $-5,-2$ and $1$ the polynomial is zero.
$\therefore $ For $x\in (-5,1)-\left\{ -2 \right\}$ the given polynomial comes to be negative.

Note: To keep a check whether the solving is correct or not, keep in mind, the number of roots are the same as the degree of the polynomial. Always check a single value from an interval to check the sign of the polynomial. This problem do not include $-5$ and $-2$, because the question here is ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20<0$and not ${{x}^{4}}+8{{x}^{3}}+15{{x}^{2}}-4x-20\le 0$. So we need to avoid these types of errors.