Question

How do you find the roots of ${{x}^{3}}-5{{x}^{2}}+4x+10=0$?

Answer 1

Hint: In this question, we are given a cubic equation (equation of degree 3) and we need to find the roots of this equation. For this we will find the first root of the equation by hit and trial method. After that we will use the root to form a factor of the equation. Then we will divide the factor by the given equation to find a quadratic equation (other factor of equation). We will solve this quadratic equation using a quadratic formula to find the remaining two roots. Quadratic formula for an equation of the form $a{{x}^{2}}+bx+c=0$ is given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\].

Complete step by step solution:
Here we are given the equation $f\left( x \right)={{x}^{3}}-5{{x}^{2}}+4x+10=0$. Since the equation is of degree 3 (cubic equation) so we will first get three roots of this equation. Let us find the first root by hit and trial method. Taking x = 1 in the equation we see that,
${{\left( 1 \right)}^{3}}-5{{\left( 1 \right)}^{2}}+4\left( 1 \right)+10=0\Rightarrow 1-5+4+10=0\Rightarrow -4+4+10=0\Rightarrow 10\ne 0$.
So 1 is not the root of f(x) = 0.
Taking x = -1, in the equation f(x) we see that,
${{\left( -1 \right)}^{3}}-5{{\left( -1 \right)}^{2}}+4\left( -1 \right)+10=0$.
Simplifying we get, $-1-5-4+10=0\Rightarrow -6-4+10=0\Rightarrow -10+10=0\Rightarrow 0=0$
Hence -1 is one of the roots of f(x) = 0.
By factor theorem we know if c is a root of the equation g(x) = 0 then (x-c) is the factor of the equation g(x) = 0. So here, (x-(-1)) is the factor of f(x) = 0 i.e. (x+1) is the factor of f(x) = 0.
Let us divide f(x) by (x+1) to get the other factor in terms of quadratic function, we get,
\[\begin{align}
  & x+1\overset{{{x}^{2}}-6x+10}{\overline{\left){{{x}^{3}}-5{{x}^{2}}+4x+10}\right.}} \\
 & \,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+{{x}^{2}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\dfrac{-\,\,\,\,\,-\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{-6{{x}^{2}}+4x+10} \\
 & \,\,\,\,\,\,\,\,\,\,\,-6{{x}^{2}}-6x \\
 & \,\,\,\,\,\,\,\,\,\,\dfrac{\,\,+\,\,\,\,\,+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\begin{align}
  & 10x+10 \\
 & 10x+10 \\
\end{align}} \\
 & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{-\,\,\,\,\,-\,\,\,}{0} \\
 & \,\,\,\,\,\,\,\,\,\, \\
\end{align}\]
So by division algorithm we can write it as, ${{x}^{3}}-5{{x}^{2}}+4x+10=\left( x+1 \right)\left( {{x}^{2}}-6x+10 \right)$.
As ${{x}^{3}}-5{{x}^{2}}+4x+10=0$ so $\left( x+1 \right)\left( {{x}^{2}}-6x+10 \right)=0$.
Let us solve $\left( {{x}^{2}}-6x+10 \right)=0$.
Comparing $a{{x}^{2}}+bx+c=0$ we get a = 1, b = -6 and c = 10. Now let us apply the quadratic formula given by \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\] we get \[x=\dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( 10 \right)}}{2}\Rightarrow x=\dfrac{6\pm \sqrt{36-40}}{2}\Rightarrow x=\dfrac{6\pm \sqrt{-4}}{2}\].
We know that $\sqrt{-x}$ can be written as $\sqrt{x}i$ because $\sqrt{-1}=i$ so we get \[x=\dfrac{6\pm \sqrt{4}i}{2}\].
We know that ${{\left( 2 \right)}^{2}}=4$ so $\sqrt{4}=2$. Hence we get \[x=\dfrac{6\pm 2i}{2}\].
Value of x becomes \[x=\dfrac{6+2i}{2}\text{ and }x=\dfrac{6-2i}{2}\].
Taking 2 common from the numerator we get \[x=\dfrac{2\left( 3+i \right)}{2}\text{ and }x=\dfrac{2\left( 3-i \right)}{2}\].
Cancelling 2 from the numerator and the denominator we get x = 3+i and x = 3-i which are the required values of x.
Hence -1, 3+i, 3-i are the required roots of the given equation f(x) = 0 i.e. ${{x}^{3}}-5{{x}^{2}}+4x+10=0$.

Note: Students should take care of the signs while forming factors of the equation. Do not forget to use i for negative term in square root. Make sure remainder is zero while dividing f(x) by (x+1). For finding first roots, always try to put values as x = 1, -1, 2, -2 or students can use the rational roots theorem.