Question

Find the roots of this equation by factorisation.
⟹ x³ + x + 2 = 0​

Answer 1

Step-by-step explanation:

${x}^{3} + x + 2 = 0 \\ \\ = > {x}^{3} + {x}^{2} - {x}^{2} - x + 2x + 2 = 0 \\ \\ = > {x}^{2} (x + 1) - x(x + 1) + 2(x + 1) = 0 \\ \\ = > (x + 1)( {x}^{2} - x + 2) = 0 \\ \\ = > x + 1 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > x = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ = > {x}^{2} - x + 2 = 0 \\ \\ = > x(x - 1) = - 2 \\ \\ = > x = - 2 \\ \\ = > x - 1 = - 2 \\ \\ = > x = - 1 \\ \\ \\ x = - 1 \: \: \: \: and \: \: \: \: - 2$

Answer 2

$\sf{x^3+x+2=0}$

Use the Rational Root Theorem (Rational Root Theorem : For a polynomial equation with integer coefficients $\sf{a_{n} x^{n}+a_{n-1} x^{n-1}+\ldots+a_{0}}$ . If $\sf{a_0}$ and $\sf{a_n}$ are integers. Then if there is a rational solution it could be found by checking all the numbers produced for ± Dividers of $\sf{a_0}$ ÷ Dividers of $\sf{a_n}$)

• Here, $\sf{a_0}$ = 24 and $\sf{a_n}$ = 1

• The Dividers of $\sf{a_0 : 1,2}$

• The Dividers of $\sf{a_n:1}$

Therefore Check the following rational numbers $\sf{\pm \dfrac{1,\:2}{1}}$

• -1/1 is a root of the expression, so factor out x + 1

$\to\sf{\displaystyle\left(x+1\right)\frac{x^3+x+2}{x+1}}$

$\to\sf{\displaystyle\left(x+1\right)\left(x^2+\frac{-x^2+x+2}{x+1}\right)}$

$\to\sf{\displaystyle\left(x+1\right)\left(x^2-x+\frac{2x+2}{x+1}\right)}$

$\to\sf{\displaystyle\left(x+1\right)\left(x^2-x+2\right)}$

Use Zero Factor Principle

• The Principle of Zero Products states that if the product of two numbers is 0, If ab = 0 then a = 0 or b = 0

$\boxed{\sf{x+1=0\quad \mathrm{or}\quad \:x^2-x+2=0}}$

$---\longrightarrow$ Solve x + 1 = 0

$\bigstar\:\:\:\:\sf{x+1=0}\quad\implies\sf{x+1-1=0-1}\quad\implies\underline{\underline{\sf{x=-1}}}$

$---\longrightarrow$ Solve x² - x + 2 = 0

We can Solve it with Quadratic Formula || Quadratic Equation Formula : For a Quadratic Equation of the form $\sf{ax^2+bx+c=0}$ the solutions are $\sf{x_{1,\:2}=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}$

• We have $\sf{a = 1, b = -1, c = 2}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\cdot \:2}}{2}}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-8}}{2}}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{1-8}}{2}}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{-7}}{2}}$

• Apply Radical Rule : $\sf{\sqrt{-a}=\sqrt{-1} \sqrt{a}}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{-1}\sqrt{7}}{2}}$

• Apply the Imaginary Rule : $\sf{\sqrt{-1}=i}$

$\to\rm{\displaystyle x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{7}i}{2}}$

• Separate The Solutions

$\to\rm{\displaystyle x_1=\frac{-\left(-1\right)+\sqrt{7}i}{2\cdot \:1},\:x_2=\frac{-\left(-1\right)-\sqrt{7}i}{2\cdot \:1}}$

• The Solutions to the Quadratic Equations are

$\to\rm{\displaystyle x=\frac{1}{2}+i\frac{\sqrt{7}}{2},\:x=\frac{1}{2}-i\frac{\sqrt{7}}{2}}$

$\rule{315}{1}$

The Final Solutions Are,

$\boxed{\boxed{\sf{\displaystyle x=-1,\:x=\frac{1}{2}+i\frac{\sqrt{7}}{2},\:x=\frac{1}{2}-i\frac{\sqrt{7}}{2}}}}$