Question

How do I find all real and complex zeros of ${{x}^{3}}+4{{x}^{2}}+5x$?

Answer 1

Hint: In the given question we have to find all real and complex zeros of ${{x}^{3}}+4{{x}^{2}}+5x$. Real zeros are the values of $x$ when $y$ equals to zero. Complex zeros are values of $x$ when $y$ equals to zero but they cannot be seen on the graph. In the given example use the quadratic formula for deriving the factors of the equation.

Complete step by step solution:
The given equation in the question is,
$\Rightarrow {{x}^{3}}+4{{x}^{2}}+5x$
Now set the equation equal to $'0'$
$\Rightarrow $${{x}^{3}}+4{{x}^{2}}+5x$
Now take $'x'$ as common from the above equation
$\Rightarrow $$x\left( {{x}^{2}}+4x+5 \right)=0$
Now, by the product property we can write above equation as,
$x=0$ and ${{x}^{2}}+4x+5=0$
So, $x=0$ and ${{x}^{2}}+4x+5=0$
So, $x=0$ this is one of the roots,
Now, for factorization of ${{x}^{2}}+4x+5$ use the quadratic formula,
$\Rightarrow $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Put, $a=1,b=4$ and $c=5$ this value are get from comparing general quadratic equation $a{{x}^{2}}+bx+c$with ${{x}^{2}}+4x+5$
$\Rightarrow $$x=\dfrac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( 5 \right)}}{2\left( 1 \right)}$
Now, simplify the above expression.
$\Rightarrow $$x=\dfrac{-4\pm \sqrt{16-20}}{20}$
Subtracting $20$ from $16$ we get $-4$
$\therefore x=\dfrac{-4\pm \sqrt{-4}}{2}$
We can write $\sqrt{-4}$ as,
$\Rightarrow $$\sqrt{4\times \left( -1 \right)}=\sqrt{4}\sqrt{-1}$
We can write $\sqrt{-4}$ as,
$\Rightarrow $$\sqrt{4\times \left( -1 \right)}=\sqrt{4}\sqrt{-1}$
We know that $\sqrt{-1}=i$
Therefore, $\sqrt{-4}$ is equal to
$\Rightarrow $$\sqrt{-4}=\sqrt{4}i$
$\Rightarrow $$=2i$
Therefore expression will be,
$\Rightarrow $$x=\dfrac{-4\pm 2i}{2}$
Now dividing with $2$ we get,
$\Rightarrow $$x=-2\pm i$
$\Rightarrow $$x=-2+i$ and $x=-2-i$
Therefore, the function ${{x}^{3}}+4{{x}^{2}}+5x$ has $3$ roots one of the roots is real and other $2$ roots complex numbers.
The roots are $0,-2+i$ and $-2-i$

Therefore, the real and complex zeros of ${{x}^{3}}+4{{x}^{2}}+5x$ are $0,-2+i$ and $-2-i$

Additional Information:
Real zero are the value of $x$ when $y$ equals to zero and they represent the $x$-intercept of the graphs. Complex zeros are values of $x$ when $y$ equals to zero, but they cannot be seen on the graph. Complex zeros consist of imaginary numbers. An imaginary number $i$ is equal to the square root of negative one.
Also, zeros are nothing but the roots of the equation. A polynomial will have exactly as many roots as its degree. The degree is the highest exponent of the polynomial.

Note: Remember the formula for calculating of root. Also remember that $\sqrt{-1}$ is equal to the $'i'$ which is an imaginary number. Be careful while putting a sign before a number so that you get the correct answer. Also, give proper signs while performing calculations.