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 $3x + 2$.
box with $3x^2$, $3x$, $+2$, $9x^3$
answer attempt 1 out of 2
you must answer all questions above in order to submit.

Explanation:

Step1: Find the first missing term in the top row (next to \(3x^2\))

We know that the divisor is \(3x + 2\), and the first term of the quotient (top row first term) is \(3x^2\). To find the next term in the quotient (top row second term), we multiply \(3x^2\) by \(2\)? No, wait, actually, we look at the middle row (the \(3x\) row) and the first column is \(3x\), first row first term is \(3x^2\), so \(3x\times3x^2 = 9x^3\) (which matches the given \(9x^3\)). Now, the second term in the top row: let's denote the quotient as \(3x^2+ax + b\). So when we multiply \(3x\) (from the divisor's first term) by \(ax\) (second term of quotient), we get \(3x\times ax=3a x^2\). But the original polynomial's second term is \(-6x^2\). Wait, actually, the box method: the top row is the quotient terms. The first term of the quotient is \(3x^2\) (since \(3x\times3x^2 = 9x^3\), which is the first term of the dividend). Now, the next term in the quotient: let's find the coefficient for the \(x\) term. We have the middle row (multiplied by \(3x\)) and the bottom row (multiplied by \(2\)). Let's first find the second term in the top row. The first term of the divisor is \(3x\), and we have the dividend's \(x^2\) term: \(-6x^2\). Wait, the middle row (multiplied by \(3x\)) has the \(x^3\) term \(9x^3\), then the next term in the middle row (under the second column of top row) should be \(3x\times\) (second term of quotient). Let's denote the quotient as \(3x^2 + mx + n\). Then, when we multiply \(3x + 2\) by \(3x^2 + mx + n\), we get:

\((3x)(3x^2)+(3x)(mx)+(3x)(n)+2(3x^2)+2(mx)+2(n)\)

\(=9x^3 + 3m x^2 + 3n x + 6x^2 + 2m x + 2n\)

Combine like terms:

\(9x^3+(3m + 6)x^2+(3n + 2m)x + 2n\)

This should equal \(9x^3 - 6x^2 - 29x - 14\)

So set coefficients equal:

1. For \(x^2\): \(3m + 6=-6\) → \(3m=-12\) → \(m = -4\)
2. For \(x\): \(3n + 2m=-29\). We know \(m=-4\), so \(3n + 2(-4)=-29\) → \(3n - 8=-29\) → \(3n=-21\) → \(n=-7\)
3. For constant term: \(2n=-14\) → \(n=-7\) (which matches)

So the quotient is \(3x^2 - 4x - 7\). Now, let's fill the box:

Top row: first term \(3x^2\), second term \(-4x\), third term \(-7\)

Middle row (multiplied by \(3x\)):

- First column: \(3x\times3x^2 = 9x^3\) (given)
- Second column: \(3x\times(-4x)=-12x^2\)
- Third column: \(3x\times(-7)=-21x\)

Bottom row (multiplied by \(+2\)):

- First column: \(2\times3x^2 = 6x^2\)
- Second column: \(2\times(-4x)=-8x\)
- Third column: \(2\times(-7)=-14\)

Now, let's check the \(x^2\) terms: the middle row's second column is \(-12x^2\), and the bottom row's first column is \(6x^2\). Then, \(-12x^2 + 6x^2=-6x^2\) (which matches the dividend's \(x^2\) term). For the \(x\) terms: middle row third column \(-21x\) and bottom row second column \(-8x\), so \(-21x -8x=-29x\) (matches). Constant term: bottom row third column \(-14\) (matches).

So let's fill the missing values:

Top row (second term): \(-4x\) (since quotient's second term is \(-4x\))

Top row (third term): \(-7\) (quotient's third term)

Middle row (second column): \(3x\times(-4x)=-12x^2\)

Middle row (third column): \(3x\times(-7)=-21x\)

Bottom row (first column): \(2\times3x^2 = 6x^2\)

Bottom row (second column): \(2\times(-4x)=-8x\)

Bottom row (third column): \(2\times(-7)=-14\)

Let's start with the top row second term: when we multiply \(3x\) (middle row first term) by the second term of the quotient, we get the middle row second column. Wait, the middle row is \(3x\) times the quotient terms. So middle row second column is \(3x\times\) (second term of quotient). We found the second term of quotient is \(-4x\), so \(3x\times(-4x)=-12x^2\). So the second term in the top row is \(-4x\) (because top row is quotient, so \(3x^2\) (first term), \(-4x\) (second term), \(-7\) (third term)).

So first, the top row second box: \(-4x\)

Top row third box: \(-7\)

Middle row second box: \(3x\times(-4x)=-12x^2\)

Middle row third box: \(3x\times(-7)=-21x\)

Bottom row first box: \(2\times3x^2 = 6x^2\)

Bottom row second box: \(2\times(-4x)=-8x\)

Bottom row third box: \(2\times(-7)=-14\)

Let's verify step by step.

First, the quotient is \(3x^2 -4x -7\), so the top row is \(3x^2\), \(-4x\), \(-7\).

Then, middle row (multiplied by \(3x\)):

- \(3x \times 3x^2 = 9x^3\) (given)
- \(3x \times (-4x) = -12x^2\) (so middle row second box: \(-12x^2\))
- \(3x \times (-7) = -21x\) (middle row third box: \(-21x\))

Bottom row (multiplied by \(+2\)):

- \(2 \times 3x^2 = 6x^2\) (bottom row first box: \(6x^2\))
- \(2 \times (-4x) = -8x\) (bottom row second box: \(-8x\))
- \(2 \times (-7) = -14\) (bottom row third box: \(-14\))

Now, let's check the addition of columns:

First column: \(9x^3 + 6x^2\)? Wait, no, the box is structured as:

The top row is the quotient terms (each term is a part of the quotient). The leftmost column is the divisor terms (\(3x\) and \(+2\)). So each cell in the middle and bottom rows is the product of the leftmost column term and the top row term.

So:

- Top row, first cell: \(3x^2\) (quotient term 1)
- Top row, second cell: \(ax\) (quotient term 2)
- Top row, third cell: \(b\) (quotient term 3)

- Middle row (left term \(3x\)):
- First cell: \(3x \times 3x^2 = 9x^3\)
- Second cell: \(3x \times ax = 3a x^2\)
- Third cell: \(3x \times b = 3b x\)

- Bottom row (left term \(+2\)):
- First cell: \(2 \times 3x^2 = 6x^2\)
- Second cell: \(2 \times ax = 2a x\)
- Third cell: \(2 \times b = 2b\)

Now, the sum of the middle and bottom rows for each column should give the dividend terms (or parts of it). Wait, actually, the dividend is the sum of all middle and bottom row cells? No, the dividend is \((3x + 2)(3x^2 + ax + b)=9x^3 + 3a x^2 + 3b x + 6x^2 + 2a x + 2b=9x^3 + (3a + 6)x^2 + (3b + 2a)x + 2b\), which should equal \(9x^3 -6x^2 -29x -14\). So:

1. Coefficient of \(x^2\): \(3a + 6 = -6\) → \(3a = -12\) → \(a = -4\)
2. Coefficient of \(x\): \(3b + 2a = -29\). Substitute \(a = -4\): \(3b + 2(-4) = -29\) → \(3b - 8 = -29\) → \(3b = -21\) → \(b = -7\)
3. Constant term: \(2b = -14\) → \(b = -7\) (consistent)

So quotient is \(3x^2 -4x -7\), so:

- Top row second cell: \(ax = -4x\)
- Top row third cell: \(b = -7\)

Middle row second cell: \(3x \times (-4x) = -12x^2\)
Middle row third cell: \(3x \times (-7) = -21x\)

Bottom row first cell: \(2 \times 3x^2 = 6x^2\)
Bottom row second cell: \(2 \times (-4x) = -8x\)
Bottom row third cell: \(2 \times (-7) = -14\)

So filling these in:

Top row (second): \(-4x\)

Top row (third): \(-7\)

Middle row (second): \(-12x^2\)

Middle row (third): \(-21x\)

Bottom row (first): \(6x^2\)

Bottom row (second): \(-8x\)

Bottom row (third): \(-14\)

Let's confirm with the box:

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

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

Middle row (under \(3x\)): \(9x^3\), \(-12x^2\), \(-21x\)

Bottom row (under \(+2\)): \(6x^2\), \(-8x\), \(-14\)

Now, if we add the middle row and bottom row for each column (but actually, the middle row is multiplied by \(3x\) and bottom by \(2\), so the total is \((3x + 2)(3x^2 -4x -7)\), which we already verified equals the dividend.

So the missing values are:

Top row (second): \(-4x\)

Top row (third): \(-7\)

Middle row (second): \(-12x^2\)

Middle row (third): \(-21x\)

Bottom row (first): \(6x^2\)

Bottom row (second): \(-8x\)

Bottom row (third): \(-14\)

Let's start with the top row second term: since the first term of the quotient is \(3x^2\) (top row first), the second term is \(ax = -4x\) (as we found \(a = -4\)). So top row second box: \(-4x\)

Top row third box: \(b = -7\), so \(-7\)

Middle row second box: \(3x \times (-4x) = -12x^2\)

Middle row third box: \(3x \times (-7) = -21x\)

Bottom row first box: \(2 \times 3x^2 = 6x^2\)

Bottom row second box: \(2 \times (-4x) = -8x\)

Bottom row third box: \(2 \times (-7) = -14\)

So these are the missing values. Let's check each step:

1. Top row second term: quotient's second term is \(-4x\) (since \(3x^2 -4x -7\) is the quotient)
2. Top row third term: quotient's third term is \(-7\)
3. Middle row second term: \(3x \times (-4x) = -12x^2\)
4. Middle row third term: \(3x \times (-7) = -21x\)
5. Bottom row first term: \(2 \times 3x^2 = 6x^2\)
6. Bottom row second term: \(2 \times (-4x) = -8x\)
7. Bottom row third term: \(2 \times (-7) = -14\)

All these values satisfy the polynomial multiplication, so they are correct.

Answer:

The missing values are filled as follows:

• Top row (second cell): \(\boldsymbol{-4x}\)
• Top row (third cell): \(\boldsymbol{-7}\)
• Middle row (second cell): \(\boldsymbol{-12x^2}\)
• Middle row (third cell): \(\boldsymbol{-21x}\)
• Bottom row (first cell): \(\boldsymbol{6x^2}\)
• Bottom row (second cell): \(\boldsymbol{-8x}\)
• Bottom row (third cell): \(\boldsymbol{-14}\)

(If the problem expects filling one at a time, follow the order: top row second, top row third, middle row second, middle row third, bottom row first, bottom row second, bottom row third with these values.)