How do you multiply \[\left( {x + 5} \right)\left( {x - 4} \right)\]?

Hint:In order to expand the above expression use the order of expanding as FOIL i.e. First,Outside,Inside,Last and later for simplification combine all the like terms to get the required answer.

Complete step by step solution:
We are given a polynomial having one variable $x$ in the term.
Let’s suppose the function given be $f\left( x \right)$
$f\left( x \right) = \left( {x + 5} \right)\left( {x - 4} \right)$
First we have to expand the above equation, to expand any binomial expression there is an order by which we can expand the expression easily. You can use the acronym FOIL which means First, Outside,inside,Last to remember the order of expanding .
$
f\left( x \right) = x\left( {x - 4} \right) + 5\left( {x - 4} \right) \\
= {x^2} - 4x + 5x - 20 \\
$
Now Simplification means to combine all the like term in the equation, so in our equation combine terms having $x$
Our equation becomes
\[f\left( x \right) = {x^2} + x - 20\]
Therefore, expansion and simplification of \[\left( {x + 5} \right)\left( {x - 4} \right)\]is \[{x^2} + x - 20\]

Note:
Quadratic Equation: A quadratic equation is an equation which can be represented in the form of $a{x^2} + bx + c$where $x$is the unknown variable and a,b,c are the numbers known where $a \ne 0$. If $a = 0$then the equation will become a linear equation and will no longer be quadratic .
The degree of the quadratic equation is of the order 2.
In order to determine the roots to a quadratic equation, there are couple of ways,
1.Using splitting up the middle term method:
let $a{x^2} + bx + c$ calculate the product of coefficient of ${x^2}$and the constant term and factorise it into two factors in a way that either addition or subtraction of the two gives the middle term and multiplication gives the product value.
2.You can also alternatively use a direct method which uses Quadratic Formula to find both roots of a quadratic equation as
$x1 = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}$ and $x2 = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
x1,x2 are root to quadratic equation $a{x^2} + bx + c$