Question

Factor $ - 5{x^3} + 18x + 8$ ?

Answer 1

Hint: A polynomial having degree 2 in one variable in form of $f\left( x \right) = a{x^2} + bx + c$ where $a,b,c \in R$ and $a \ne 0$, when equated to zero becomes quadratic equation of degree 2. To solve the given quadratic equation we will compare the given equation with the general form of the quadratic equation and then, we will find the root of the equation. The general form of quadratic equation is $f\left( x \right) = a{x^2} + bx + c$. $a$ is called as the leading coefficient and $c$ is called absolute term of quadratic equation. The value satisfying the quadratic equation is called the roots of the quadratic equation. The quadratic equation will always have two roots. The nature of the root may be either real or imaginary.
Formula for finding the root of the quadratic equation is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.

Complete step by step answer:
The given quadratic equation is,
$ - 5{x^3} + 18x + 8$
Compare the given equation with general form of quadratic equation,
$f\left( x \right) = a{x^2} + bx + c$.
Therefore, $a = - 5, b = 18$ and $c = 8$.
Suppose the roots of the quadratic equation are $\left( {\alpha ,\beta } \right)$.

Substitute the value of $a,b,$and $c$ in the formula,
$
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow x = \dfrac{{ - 18 \pm \sqrt {{{\left( {18} \right)}^2} - 4 \times \left( { - 5} \right) \times 8} }}{{2 \times \left( { - 5} \right)}} \\
   \Rightarrow x = \dfrac{{ - 18 \pm \sqrt {324 + 160} }}{{ - 10}} \\
 $

Solve the equation to find the value of $x$.
$
  x = \dfrac{{ - 18 \pm \sqrt {484} }}{{ - 10}} \\
   \Rightarrow x = \dfrac{{ - 18 \pm 22}}{{ - 10}} \\
 $
Consider the positive value of $x$.
 $
   \Rightarrow x = \dfrac{{ - 18 + 22}}{{ - 10}} \\
   \Rightarrow x = \dfrac{2}{{ - 10}} \\
   \Rightarrow x = - 5 \\
 $
Consider the negative value of $x$.
$
   \Rightarrow x = \dfrac{{ - 18 - 22}}{{ - 10}} \\
   \Rightarrow x = \dfrac{{ - 40}}{{ - 10}} \\
   \Rightarrow x = 4 \\
 $

Therefore the roots of the equation are $ - 5$ and $4$ .
The factors of the given equation are $\left( {x + 5} \right)$ and $\left( {x - 4} \right)$.

Note: The roots are the factor of the equation when they will be equated to the zero. While solving the problems students are advised to use the formula of quadratic equation $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ . Compare the value of given quadratic equation with general equation of quadratic equation $f\left( x \right) = a{x^2} + bx + c$ to find the coefficient of leading term and absolute term.