How do you factor ${{x}^{2}}+7x-60$?

Hint: Given polynomial is a second-degree polynomial and is also known as quadratic polynomial and the standard form of the second-degree polynomial is $a{{x}^{2}}+bx+c$.
Where, $a$ is the coefficient of the term ${{x}^{2}}$, $b$ is the coefficient of the term $x$ , and $c$ is constant.

Complete step by step solution:
The given polynomial is ${{x}^{2}}+7x-60$.
In the above polynomial $a=1,b=7$ and $c=-60$.
To factorize this polynomial find two factors whose sum will be equal to the coefficient of the middle term of the above polynomial which is 7 and if we multiply those factors will be equal to the constant term $c$ which is $-60$.
That is $-12$ and $5$
If we add both factors will get $7$ which is the coefficient of the middle term of the above polynomial.
$\Rightarrow -12+5=7$
And if we multiply both factors will get $-60$ which is equal to the constant of the polynomial.
$\Rightarrow -12\times 5=-60$
Now, rewrite the coefficient of the middle term using two factors $-12$ and $5$.
$\Rightarrow {{x}^{2}}+(-12+5)x-60$
$\Rightarrow {{x}^{2}}-12x+5x-60$
$\Rightarrow x(x-12)+5(x-12)$
$\Rightarrow (x-12)\times (x+5)$.

Hence, $(x-12)$ and $(x+5)$ are the factors of the polynomial ${{x}^{2}}+7x-60$.

Note: The shape of the quadratic polynomials is always parabolic.
The roots of quadratic polynomials are two real, zero, or one.
The degree is the highest power of the variables of the polynomial.
If the degree is one then the polynomial will be linear polynomial.