Factoring Polynomials

To factor the trinomial, let's start by determining if the terms have any common factor. To do this, we'll write each term as a product of its factors. ${\color{#0000FF}{x}} \cdot x \cdot x + 2 \cdot {\color{#0000FF}{x}} \cdot x -3 \cdot {\color{#0000FF}{x}}$ It can be seen that each term has an $x.$ Thus, we can factor the GCF of $x$ out of the entire expression. $x^3+2x^2-3x=x\left(x^2+2x-3\right)$ The parentheses now contain a quadratic trinomial, which can be factored. Notice that $3$ and $\text{-} 1$ multiply to equal $\text{-} 3$ and add to equal $2.$ The expression $x^2+2x-3$ can be written as the product $(x+3)(x-1).$ Thus, $x^3+2x^2-3x$ can be factored into $x(x+3)(x-1).$

To begin, we must determine which type of factoring we can use to solve the given polynomial equation. Notice that there is no common factor amongst all four terms. $(2 \cdot x \cdot x \cdot x) + (3 \cdot x \cdot x) - (2 \cdot 3 \cdot 3 \cdot x) - (3 \cdot 3 \cdot 3)$ The first three terms all have an $x$ in common, while the last three terms have a $3$ in common. Thus, we must factor by grouping the first two terms and the last two terms. In the first pair, the GCF is $x^2,$ and in the second it is $\text{-} 9.$ Let's factor out the GCFs separately. Now, both terms have the factor $2x+3,$ which can be factored out. $x^2(2x+3)-9(2x+3)=\left(x^2-9\right)(2x+3)=0$ Using the Zero Product Property, we can separate this equation into two new equations, that can be solved individually. Thus, the original equation has three solutions: $x=\text{-}3,$ $x=\text{-}1.5,$ and $x=3.$