Question

assignment
and a negative constant term.
complete the statements to explain how to factor ( x^2 + x - 12 ) using algebra tiles.
represent the trinomial using 1 positive ( x^2 )-tile,
, and .
form a rectangle with the tiles, adding zero pairs as needed. how many zero pairs are needed?

drag tiles onto the board to represent each factor. what is the factorization? ( x^2 + x - 12 =

no factor tiles are in place.
product rectangle: ( x )
not in product rectangle: ( -x )

Explanation:

Step1: Analyze the trinomial

The trinomial is \(x^{2}+x - 12\). The general form of a quadratic trinomial \(ax^{2}+bx + c\) (here \(a = 1\), \(b=1\), \(c=- 12\)). For the \(x\)-terms, we have \(+x\) (since the coefficient of \(x\) is \(1\)), and for the constant term, we have \(- 12\) (which can be represented as 12 negative unit tiles or a combination, but in terms of the tiles, the linear term is \(+x\) (so 1 positive \(x\)-tile) and the constant term is \(-12\) (12 negative unit tiles, but when forming the rectangle, we need to find factors of \(-12\) that add up to \(1\)). The factors of \(-12\) that add up to \(1\) are \(4\) and \(- 3\) (since \(4\times(-3)=-12\) and \(4+( - 3)=1\)). But first, representing the trinomial: we have \(1\) positive \(x^{2}\)-tile, \(1\) positive \(x\)-tile (since the coefficient of \(x\) is \(1\)), and \(12\) negative unit tiles (or \(- 12\) constant term).

Step2: Determine the zero pairs

To form a rectangle, we need to find two binomials \((x + m)(x + n)\) such that \(mn=-12\) and \(m + n = 1\). As above, \(m = 4\), \(n=-3\). But when using algebra tiles, if we have the \(x\)-term as \(+x\) and we need to adjust with zero pairs. Wait, the trinomial is \(x^{2}+x - 12\). The \(x^{2}\)-tile, \(x\)-tile (positive), and \(-12\) (negative unit tiles). Now, to form the rectangle, we need to find the length and width. The length and width will be \((x + 4)\) and \((x-3)\) because \((x + 4)(x - 3)=x^{2}+x-12\). Now, when we arrange the tiles, if we have the \(x^{2}\)-tile, \(x\)-tile, and \(-12\) (unit tiles), we need to see the zero pairs. Wait, the linear term is \(+x\), so we have \(1\) positive \(x\)-tile, and for the constant term \(-12\). Now, the factors of \(-12\) that add to \(1\) are \(4\) and \(-3\). So when forming the rectangle, we need to have \(x^{2}\)-tile, \(+x\)-tile (1), and \(-12\) (constant term, which is 12 negative unit tiles or a combination). But in terms of the first drop - down: the linear term is \(+x\) (so 1 positive \(x\)-tile), and the constant term is \(-12\) (so 12 negative unit tiles, but the question is about the tiles for the trinomial. So the first blank: \(1\) positive \(x\)-tile (since the coefficient of \(x\) is \(1\)), the second blank: \(12\) negative unit tiles (or \(-12\) constant term, represented as negative unit tiles). Then, for the zero pairs: when we factor \((x + 4)(x - 3)\), the \(x\)-terms: if we have \(x\) from one factor and \(-3\) and \(4\) from the other. Wait, maybe a better way: the trinomial \(x^{2}+x - 12=(x + 4)(x - 3)\). The area of the rectangle is \(x^{2}+x - 12\). The length and width are \(x + 4\) and \(x - 3\). When we use algebra tiles, we have \(x^{2}\)-tile, \(x\)-tiles (4 positive \(x\)-tiles and 3 negative \(x\)-tiles? No, wait, no. Wait, the linear term is \(+x\), so we have 1 positive \(x\)-tile. To get the correct factors, we need to find that \(4\times(-3)=-12\) and \(4-3 = 1\). So we need to add 3 zero pairs of \(x\)-tiles? Wait, no. Let's think again. The trinomial is \(x^{2}+x - 12\). The \(x^{2}\)-tile, \(x\)-tile (1 positive), and \(-12\) (constant). We need to find two numbers \(a\) and \(b\) such that \(a\times b=-12\) and \(a + b = 1\). So \(a = 4\), \(b=-3\). So the rectangle will have length \(x + 4\) and width \(x - 3\). So when we arrange the tiles, we have \(x^{2}\)-tile, 4 positive \(x\)-tiles, 3 negative \(x\)-tiles (but we only have 1 positive \(x\)-tile initially). So we need to add 3 positive \(x\)-tiles and 3 negative \(x\)-tiles (zero pairs) to make the total number of \(x\)-tiles equal to \(4-3 = 1\)? Wait, maybe I'm overcomplicating. The first blank: the linear term is \(+x\), so we need 1 positive \(x\)-tile. The second blank: the constant term is \(-12\), so 12 negative unit tiles (or represented as \(-12\) constant). Then, to form the rectangle, we need to find that \((x + 4)(x - 3)=x^{2}+x - 12\). The zero pairs: when we have the \(x\)-terms, if we have \(x\) from one factor and \(-3x\) and \(4x\) from the other? Wait, no. The number of zero pairs: since we have \(x\) (1 positive \(x\)-tile) and we need to get to \(x=4x-3x\), so we need to add 3 zero pairs of \(x\)-tiles (3 positive and 3 negative \(x\)-tiles) to make the total \(x\)-tiles equal to \(4x-3x=x\). So the number of zero pairs needed is 3? Wait, no, maybe the first part: represent the trinomial using 1 positive \(x^{2}\)-tile, 1 positive \(x\)-tile (since the coefficient of \(x\) is 1), and 12 negative unit tiles (or \(-12\) constant). Then, to factor, we find two numbers that multiply to \(-12\) and add to 1, which are 4 and - 3. So the factorization is \((x + 4)(x - 3)\).

Step3: Fill in the blanks

First blank: 1 positive \(x\)-tile (since the coefficient of \(x\) is 1, so we have \(+x\) which is 1 positive \(x\)-tile). Second blank: 12 negative unit tiles (or \(-12\) constant term, represented as negative unit tiles). For the zero pairs: we need to add 3 zero pairs? Wait, no, let's check the factorization. \((x + 4)(x - 3)=x^{2}+4x-3x - 12=x^{2}+x - 12\). So when we have the \(x^{2}\)-tile, \(x\)-tile (1), and \(-12\), we need to have \(4x-3x=x\). So we have 1 \(x\)-tile, but we need to get \(4x-3x\), so we need to add 3 \(x\)-tiles (positive) and 3 \(x\)-tiles (negative) as zero pairs. So the number of zero pairs needed is 3? Wait, maybe the first two blanks: first is "1 positive \(x\)-tile", second is "12 negative unit tiles". Then the zero pairs: 3. And the factorization is \((x + 4)(x - 3)\).

Answer:

First blank: 1 positive \(x\)-tile; Second blank: 12 negative unit tiles; Zero pairs needed: 3; Factorization: \((x + 4)(x - 3)\)

(Note: The exact wording for the blanks may depend on the options in the dropdowns, but based on the trinomial \(x^{2}+x - 12\), the linear term is \(+x\) (1 positive \(x\)-tile), the constant term is \(-12\) (12 negative unit tiles), zero pairs needed: 3 (since \(4x-3x=x\), so we need 3 positive \(x\)-tiles and 3 negative \(x\)-tiles as zero pairs), and factorization is \((x + 4)(x - 3)\))