Question

a polynomial is factored using algebra tiles. what are the factors of the polynomial? (x - 1) and (x + 3); (x + 1) and (x - 3); (x - 2) and (x + 3); (x + 2) and (x - 3)

Explanation:

Step1: Analyze the algebra tiles

The algebra tiles represent the polynomial. The orange tiles: \(+x^2\) and \(+x\). The blue tiles: three \(-x\) and three \(-1\) (since each small blue square is \(-1\)). So combining like terms: \(x^2 + x - 3x - 3 = x^2 - 2x - 3\).

Step2: Factor the polynomial

Factor \(x^2 - 2x - 3\). We need two numbers that multiply to \(-3\) and add to \(-2\). Those numbers are \(-3\) and \(+1\). So, \(x^2 - 2x - 3=(x + 1)(x - 3)\)? Wait, no, wait. Wait, let's re - check the tile count. Wait, the \(x\) terms: \(+x-3x=-2x\), and the constant term is \(-3\). Wait, but let's do the factoring correctly. Wait, maybe I made a mistake in the tile interpretation. Wait, the top row: \(x^2\) and \(x\) (so the length of the top side for the \(x^2\) and \(x\) is \(x + 1\)? Wait, no, let's look at the columns and rows. The first column (vertical) has \(x^2\), \(-x\), \(-x\), \(-x\)? No, wait the grid: the first row (horizontal) has a white square (maybe the constant term for the first factor) and then the \(x^2\) and \(x\) in the next columns. Wait, maybe the length of the horizontal side (the first factor) is \(x + 1\) and the vertical side (the second factor) is \(x - 3\)? No, wait let's re - express the polynomial from the tiles.

Wait, the \(x^2\) tile, one \(+x\) tile, three \(-x\) tiles, and three \(-1\) tiles. So \(x^2+x - 3x-3=x^2 - 2x - 3\). Now, factor \(x^2 - 2x - 3\). We need two binomials \((x + a)(x + b)\) where \(a\times b=-3\) and \(a + b=-2\). The pairs of factors of \(-3\) are \((-3,1)\) and \((3, - 1)\). If \(a = 1\) and \(b=-3\), then \((x + 1)(x - 3)=x^2-3x+x - 3=x^2 - 2x - 3\). Wait, yes! So \((x + 1)(x - 3)=x^2-2x - 3\). But wait, the options: one of the options is \((x + 1)\) and \((x - 3)\)? Wait no, the options are:

Option 1: \((x - 1)\) and \((x + 3)\)

Option 2: \((x + 1)\) and \((x - 3)\)

Option 3: \((x - 2)\) and \((x + 3)\)

Option 4: \((x + 2)\) and \((x - 3)\)

Wait, but when we factor \(x^2 - 2x - 3\), we get \((x + 1)(x - 3)\), which is option 2? Wait, no, wait let's check the multiplication: \((x + 1)(x - 3)=x^2-3x+x - 3=x^2 - 2x - 3\), which matches the polynomial from the tiles. But wait, maybe I misread the tiles. Wait, another way: the number of \(x\) tiles: \(+x-3x=-2x\), constant term \(-3\). So the polynomial is \(x^2-2x - 3\), which factors to \((x + 1)(x - 3)\), which is option B: \((x + 1)\) and \((x - 3)\). Wait, but let's check the options again. Wait, the options are:

1. \((x - 1)\) and \((x + 3)\): \((x - 1)(x + 3)=x^2+3x - x - 3=x^2+2x - 3\) (not our polynomial)

1. \((x + 1)\) and \((x - 3)\): \((x + 1)(x - 3)=x^2-3x+x - 3=x^2 - 2x - 3\) (matches)

1. \((x - 2)\) and \((x + 3)\): \((x - 2)(x + 3)=x^2+3x - 2x - 6=x^2+x - 6\) (no)

1. \((x + 2)\) and \((x - 3)\): \((x + 2)(x - 3)=x^2-3x+2x - 6=x^2 - x - 6\) (no)

So the correct factors are \((x + 1)\) and \((x - 3)\), which is option 2. Wait, but wait, maybe I made a mistake in the tile count. Wait, the \(x\) terms: \(+x\) (1 \(x\)) and three \(-x\) (so total \(x-3x=-2x\)), and the constant terms: three \(-1\) (so \(-3\)). So the polynomial is \(x^2 - 2x - 3\), which factors to \((x + 1)(x - 3)\). So the factors are \((x + 1)\) and \((x - 3)\), which is option B.

Answer:

\((x + 1)\) and \((x - 3)\) (the second option: \((x + 1)\) and \((x - 3)\))