How do you simplify \[\left( {x - 5} \right)\left( {x + 2} \right)\]?

Hint: Here, we will multiply the given algebraic expression by using the FOIL method. Then by using the trigonometric identity, we will simplify the given trigonometric expression. FOIL method is a method of multiplying the binomials by multiplying the first terms, then the outer terms, then the inner terms and at last the last terms. It is possible to combine like terms.

Complete step-by-step answer:
In order to simplify the expression, we will multiply the brackets by multiplying each term present in the first bracket by each term present in the second bracket.
Therefore, we get
\[\left( {x - 5} \right)\left( {x + 2} \right) = x\left( {x + 2} \right) - 5\left( {x + 2} \right)\]
Now, using the distributive property of multiplication, we get
\[ \Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) = \left( {x \times x} \right) + \left( {x \times 2} \right) - \left( {5 \times x} \right) - \left( {5 \times 2} \right)\]
\[ \Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) = {x^2} + 2x - 5x - 10\]
Subtracting the like terms, we get

\[ \Rightarrow \left( {x - 5} \right)\left( {x + 2} \right) = {x^2} - 3x - 10\]
Therefore, this is the required answer.

Note:
An alternate way of simplifying the above brackets is that as we know the two brackets are a product of two linear polynomials and hence, they will simplify to form a quadratic polynomial.
Hence, after simplifying, they will be of the form \[a{x^2} + bx + c = 0\]
Now, in the brackets, \[\left( {x - 5} \right)\left( {x + 2} \right)\], substituting \[x = 0\], we get,
\[\left( {0 - 5} \right)\left( {0 + 2} \right) = \left( { - 5} \right)\left( 2 \right) = - 10\]
Hence, in the quadratic polynomial, \[a{x^2} + bx + c = 0\], when \[x = 0\], we get,
\[a\left( 0 \right) + b\left( 0 \right) + c = - 10\]
\[ \Rightarrow c = - 10\]
Now, again, substituting \[x = 1\] in \[\left( {x - 5} \right)\left( {x + 2} \right)\],
\[ \Rightarrow \left( {1 - 5} \right)\left( {1 + 2} \right) = \left( { - 4} \right)\left( 3 \right) = - 12\]
Hence, in the quadratic polynomial, \[a{x^2} + bx - 10\], when \[x = 1\], we get,
\[a\left( 1 \right) + b\left( 1 \right) - 10 = - 12\]
\[ \Rightarrow a + b = - 2\]………………………………\[\left( 1 \right)\]
Similarly, substituting \[x = 2\] in \[\left( {x - 5} \right)\left( {x + 2} \right)\],
\[ \Rightarrow \left( {2 - 5} \right)\left( {2 + 2} \right) = \left( { - 3} \right)\left( 4 \right) = - 12\]
Hence, in the quadratic polynomial, \[a{x^2} + bx - 10\], when \[x = 2\], we get,
\[a{\left( 2 \right)^2} + b\left( 2 \right) - 10 = - 12\]
\[ \Rightarrow 4a + 2b = - 2\]
Dividing both sides by 2,
\[ \Rightarrow 2a + b = - 1\]………………………………\[\left( 2 \right)\]
Now, we will use elimination method, thus, subtracting \[\left( 1 \right)\] from \[\left( 2 \right)\], we get,
\[ \Rightarrow 2a + b - a - b = - 1 - \left( { - 2} \right)\]
\[ \Rightarrow a = 1\]
Substituting the value \[a = 1\] in \[\left( 1 \right)\]
\[ \Rightarrow b = - 2 - 1 = - 3\]
Therefore, the required quadratic polynomial \[a{x^2} + bx + c = 0\] having \[a,b,c\] as \[1, - 3, - 10\] respectively is \[{x^2} - 3x - 10 = 0\]
$\therefore \left( x-5 \right)\left( x+2 \right)={{x}^{2}}-3x-10$
Hence, this is the required answer.