How do you factor in the expression \[{x^2} + 2x - 15\]?

Hint: First understand the general quadratic equation which contains a,b,c values.
Then find the two numbers that multiply to give ac and add to give b while comparing the given equation.
Then find numbers to rewrite the x term and now the question is in four parts then we find the first two-part factor and the last two-part separately. Then we get the solution.

Complete step by step answer:
A equation of the form \[f(x) = a{x^2} + bx + c\] where\[a\],\[b\] \[c\] are constants, real numbers and \[a \ne 0\] is called a quadratic equation.
In our question given, \[{x^2} + 2x - 15\]
here \[a = 1,\,b = 2,\,c = - 15\]
In the first step, we multiply the \[a\] and \[c\] values,
It means \[ac = (1)( - 15)\]
\[ = - 15\]
First, we need two numbers: their multiply gives \[ac\] and add gives \[b\].
It means, we need two numbers: their multiply gives \[ - 15\] and add gives \[2\].
Because \[ac = - 15,b = 2\]
 the positive factors of \[ - 15\] are \[3,5\] but one of the factors has to be negative then only to make \[ - 15\].
So we try to find that,
\[( - 3) \times 5 = - 15\] and \[( - 3) + 5 = 2\]
\[3 \times ( - 5) = - 15\] and \[3 + ( - 5) = - 2\]
Perfect factors are \[ - 3,5\].
Because we multiply the value is \[15\] and add to get \[2\].
Now rewrite \[2x\] with \[ - 3x\]and \[5x\]
\[{x^2} - 3x + 5x - 15\]
Now take x common from the first two terms and 5 from the last two terms.
Now we get,
\[x(x - 3) + 5(x - 3)\]
Take common to \[(x - 3)\]on both part then we get,
\[(x - 3)(x + 5)\]
\[{x^2} + 2x - 15\] factor is \[(x - 3)(x + 5)\]

Note:
If students try to solve this quadratic equation with the help of a quadratic formula then they will get factors instead of factored expression.