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When you start factoring a polynomial, a good first step is looking for a common factor (also known as the Greatest Common Factor, or GCF). A common factor is a number or expression that divides each term in the polynomial evenly. When we find this common factor, we can factor it out. This makes the polynomial simpler to work with.
In our exercise, the common factor is \(a+b\). Each term in \( (a+b)x^2 + (a+b)x - 12(a+b) \) has \(a+b\) as a factor. By factoring it out, we rewrite the expression into a simpler form: \[ (a+b)(x^2 + x - 12) \]
Once the common factor of \(a+b\) is factored out, we can then focus on the remaining polynomial inside the parentheses.

Quadratic expressions are polynomials of degree 2, usually written in the form \(ax^2 + bx + c\). Factoring these expressions can involve finding two binomials that multiply together to give the original quadratic expression.
In the equation \(x^2 + x - 12\), we are trying to find two numbers that, when multiplied, give \(-12\) and, when added, give \(1\). These two numbers are \(4\) and \(-3\).
Rewriting the expression using these numbers, we get: \[ x^2 + x - 12 = (x + 4)(x - 3) \]
This makes it easy to see the factors of the quadratic expression, simplifying the process of solving or simplifying polynomials.

Polynomial factorization is rewriting a polynomial as a product of simpler polynomials. These steps involve breaking down the polynomial into irreducible factors (factors that cannot be simplified further).
After factoring out the common factor \(a+b\) and simplifying the quadratic expression inside the parentheses, our fully factored polynomial is: \[ (a+b)(x + 4)(x - 3) \]
This factorization shows the polynomial in its simplest product form, making it easier to work with for further algebraic operations or solving. Factoring is essential for simplifying expressions, solving polynomial equations, and revealing the roots or solutions of the polynomial. Understanding how to factorize polynomials efficiently helps handle more complex algebraic problems effectively.