Question

polynomial division box method (guided)
score: 0/2 penalty: 1 off
question
fill in the missing values below one at a time to find the quotient when $9x^3 - 6x^2 - 29x - 14$ is divided by $3x + 2$.
image of a box for polynomial division with some filled values: top row first cell $3x^2$, left column first cell $3x$, second cell $+2$; middle row first cell $9x^3$, bottom row first cell $6x^2$; other cells are empty with one square highlighted
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answer attempt 1 out of 2
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Explanation:

Step1: Multiply \(3x\) and the middle term of the quotient (top middle)

We know that the first term of the quotient is \(3x^2\) (top left). Now, to find the middle term of the top row (quotient), we consider the \(x^2\) term in the dividend. The product of \(3x\) and the middle term of the quotient should combine with other terms to give the \(x^2\) term in the dividend. Wait, actually, the box method for polynomial division: the top row is the quotient, left column is the divisor \(3x + 2\). So each cell is the product of the top row term and left column term.

First, let's find the middle term of the top row (quotient). Let the middle term of the top row be \(ax\) and the last term be \(b\). Then:

- Product of \(3x\) (left column, first row) and \(3x^2\) (top row, first column) is \(9x^3\) (which matches the dividend's \(x^3\) term).
- Product of \(+2\) (left column, second row) and \(3x^2\) (top row, first column) is \(6x^2\) (which matches the dividend's \(x^2\) term when combined? Wait, the dividend's \(x^2\) term is \(-6x^2\). So we need to find the middle term of the top row such that when we multiply \(3x\) (left column, first row) by that middle term and add the product of \(+2\) (left column, second row) and that middle term, we get the \(x^2\) term of the dividend. Wait, no, the box is structured as:

Divisor: \(3x + 2\) (left column: \(3x\) and \(+2\))

Quotient: \(3x^2 + ax + b\) (top row: \(3x^2\), \(ax\), \(b\))

Then the cells are:

- \(3x \times 3x^2 = 9x^3\) (top row first, left column first)
- \(3x \times ax = 3ax^2\) (top row second, left column first)
- \(3x \times b = 3bx\) (top row third, left column first)
- \(+2 \times 3x^2 = 6x^2\) (top row first, left column second)
- \(+2 \times ax = 2ax\) (top row second, left column second)
- \(+2 \times b = 2b\) (top row third, left column second)

Now, sum the \(x^2\) terms: \(3ax^2 + 6x^2 = (3a + 6)x^2\). This should equal the dividend's \(x^2\) term, which is \(-6x^2\). So:

\(3a + 6 = -6\)

\(3a = -12\)

\(a = -4\)

So the middle term of the top row is \(-4x\).

Now, let's find the cell where left column is \(3x\) and top row is \(-4x\): that's \(3x \times (-4x) = -12x^2\). Wait, but the first cell in the middle column (left column \(3x\), top row middle) should be \(3x \times (-4x) = -12x^2\)? Wait, no, the dividend's \(x^2\) term is \(-6x^2\), and we have \(6x^2\) (from \(+2 \times 3x^2\)) and \(3x \times (-4x) = -12x^2\), so \(6x^2 - 12x^2 = -6x^2\), which matches. Good.

Now, moving to the \(x\) term. The sum of the \(x\) terms: \(3bx + 2ax\) should equal the dividend's \(x\) term, which is \(-29x\). We know \(a = -4\), so \(2ax = 2*(-4)x = -8x\). So:

\(3bx - 8x = -29x\)

\(3b - 8 = -29\)

\(3b = -21\)

\(b = -7\)

Now, let's check the constant term: \(2b = 2*(-7) = -14\), which matches the dividend's constant term \(-14\). Perfect.

So now, let's fill the cells:

- Top row: \(3x^2\), \(-4x\), \(-7\)
- Left column: \(3x\), \(+2\)

Now, the cell where left column is \(3x\) and top row is \(-4x\) is \(3x \times (-4x) = -12x^2\). Wait, but the problem shows a box with:

Top row: \(3x^2\), [blank], [blank]

Left column: \(3x\), \(+2\)

Cells:

- \(3x\) row, \(3x^2\) column: \(9x^3\) (filled)
- \(3x\) row, middle column: [blank] (let's call this Cell 1)
- \(3x\) row, last column: [blank] (Cell 2)
- \(+2\) row, \(3x^2\) column: \(6x^2\) (filled)
- \(+2\) row, middle column: [blank] (Cell 3)
- \(+2\) row, last column: [blank] (Cell 4)

From our earlier calculation, Cell 1 (3x row, middle column) is \(3x \times (-4x) = -12x^2\)? Wait, no, wait: top row middle term is \(-4x\), so Cell 1 (3x * (-4x)) is \(-12x^2\)? Wait, but the dividend's \(x^2\) term is \(-6x^2\), which is \(6x^2\) (from +2*3x^2) plus Cell 1 (3x*(-4x)): \(6x^2 -12x^2 = -6x^2\), correct.

Then Cell 3 ( +2 row, middle column) is \(+2 \times (-4x) = -8x\).

Then Cell 2 (3x row, last column) is \(3x \times (-7) = -21x\).

Cell 4 ( +2 row, last column) is \(+2 \times (-7) = -14\).

Now, let's check the sum of the \(x\) terms: Cell 2 ( -21x) + Cell 3 ( -8x) = -29x, which matches the dividend's \(x\) term. Perfect.

So the first missing value (the middle cell in the 3x row) is \(3x \times (-4x) = -12x^2\)? Wait, no, wait: the top row middle term is \(-4x\), so the cell in the 3x row (left column first) and middle top column is \(3x \times (-4x) = -12x^2\). Wait, but let's confirm with the steps.

First, we found the middle term of the quotient (top row middle) is \(-4x\), so the cell under 3x (left column) and next to \(3x^2\) (top row) is \(3x \times (-4x) = -12x^2\). Then the cell under +2 (left column) and next to \(3x^2\) (top row) is \(2 \times (-4x) = -8x\)? Wait, no, the top row middle is \(-4x\), so:

- 3x (left) * (-4x) (top middle) = -12x² (cell in 3x row, middle column)
- +2 (left) * (-4x) (top middle) = -8x (cell in +2 row, middle column)
- 3x (left) * (-7) (top last) = -21x (cell in 3x row, last column)
- +2 (left) * (-7) (top last) = -14 (cell in +2 row, last column)

So the first missing value (the middle cell in the 3x row) is \(-12x^2\)? Wait, but let's check the dividend's terms:

Dividend: \(9x^3 -6x^2 -29x -14\)

Divisor: \(3x + 2\)

Quotient: \(3x^2 -4x -7\) (since \( (3x + 2)(3x^2 -4x -7) = 3x(3x^2 -4x -7) + 2(3x^2 -4x -7) = 9x^3 -12x^2 -21x + 6x^2 -8x -14 = 9x^3 -6x^2 -29x -14\), which matches the dividend. Perfect.

So now, the cell in the 3x row (left column: 3x) and middle top column (quotient's middle term) is \(3x \times (-4x) = -12x^2\). Wait, but the problem's box has the 3x row with \(9x^3\) (first cell), then a blank, then another blank. The +2 row has \(6x^2\), then a blank, then another blank.

So the first missing value (the middle cell in the 3x row) is \(-12x^2\)? Wait, no, wait: the top row is the quotient: \(3x^2\), \(-4x\), \(-7\). Left column: \(3x\), \(+2\). So:

- 3x (left) * \(3x^2\) (top) = \(9x^3\) (filled)
- 3x (left) * \(-4x\) (top middle) = \(-12x^2\) (middle cell in 3x row)
- 3x (left) * \(-7\) (top last) = \(-21x\) (last cell in 3x row)
- +2 (left) * \(3x^2\) (top) = \(6x^2\) (filled)
- +2 (left) * \(-4x\) (top middle) = \(-8x\) (middle cell in +2 row)
- +2 (left) * \(-7\) (top last) = \(-14\) (last cell in +2 row)

So the first missing value (the middle cell in the 3x row) is \(-12x^2\)? Wait, but let's check the problem statement: "Fill in the missing values below one at a time". Let's start with the middle cell in the 3x row.

Wait, maybe the problem is asking for the first missing value, which is the cell in the 3x row and middle top column. So that cell is \(3x \times (-4x) = -12x^2\). Wait, but let's confirm with the quotient. The quotient is \(3x^2 -4x -7\), so the top row is \(3x^2\), \(-4x\), \(-7\). Then the cell under 3x (left) and \(-4x\) (top middle) is \(3x \times (-4x) = -12x^2\).

So the first missing value (the middle cell in the 3x row) is \(-12x^2\).

Step1: Determine the middle term of the quotient

Let the quotient be \(3x^2 + ax + b\). The \(x^2\) term in the dividend is \(-6x^2\). The \(x^2\) terms come from \(2 \times 3x^2 = 6x^2\) and \(3x \times ax = 3ax^2\). So:
\(6x^2 + 3ax^2 = -6x^2\)
\(3a + 6 = -6\)
\(3a = -12\)
\(a = -4\)

Step2: Calculate the middle cell in the \(3x\) row

The middle cell in the \(3x\) row is the product of \(3x\) (left column) and \(-4x\) (middle term of the quotient):
\(3x \times (-4x) = -12x^2\)

Answer:

The first missing value (middle cell in the \(3x\) row) is \(\boxed{-12x^2}\)