Question

the factorization of a trinomial is modeled with algebra tiles. which trinomial is factored? $x^2 + 3x - 6$ $x^2 + 5x - 6$ $x^2 + 3x - 2$ $x^2 + x - 6$

Explanation:

Step1: Analyze the algebra tiles

The length of the rectangle (from the tiles) can be determined by the tiles: one side has \(x\) (from the \(+x\) tile) and \(3\) (from the three \(+\) tiles), so length is \(x + 3\). The other side has \(x\) (from the \(+x\) tile) and \(- 2\) (from the two \(-\) tiles at the top and the two \(-x\) tiles? Wait, no, let's count the tiles properly. Wait, the area of the rectangle (which represents the trinomial) is the product of the length and width. Let's look at the tiles: the orange tiles (positive) and blue tiles (negative). The \(x^2\) tile, then the \(x\) tiles: let's count the number of \(x\) terms and constant terms.

Wait, the length of the rectangle (horizontal) has a \(+x\) tile and two \(-\) tiles? No, maybe better to factor the trinomial by looking at the dimensions. From the tiles, the length is \(x + 3\) (since there are three \(+\) tiles, so constant term \(+3\)?) Wait, no, the vertical side: one \(+x\) tile and three \(+\) tiles? Wait, no, let's do the multiplication. Let's see the factors: one factor is \(x + 3\) (since vertical has \(x\) and three \(+\)s) and the other factor is \(x - 2\) (horizontal has \(x\) and two \(-\)s? Wait, the top has two \(-\) tiles, and the blue tiles have \(-x\) and \(-x\), and then the constant \(-\) tiles. Wait, let's multiply \((x + 3)(x - 2)\).

Step2: Multiply the binomials

Using the distributive property (FOIL method): \((x + 3)(x - 2)=x^2-2x + 3x-6=x^2 + x-6\). Let's check the options. The last option is \(x^2 + x - 6\), which matches the product of \((x + 3)(x - 2)\) from the algebra tiles (since the length and width are \(x + 3\) and \(x - 2\), multiplying them gives the trinomial).

Answer:

\(x^2 + x - 6\) (the last option, which is \(x^2 + x - 6\))