Question

fill in the missing values below one at a time to find the quotient when 9x³ − 12x² + 15x − 20 is divided by 3x − 4.
image of a grid with 3x and -4 on the left, 9x³ in the top-left grid cell, and empty cells otherwise
answer
attempt 1 out of 2
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Explanation:

Step1: Find the first term of the quotient

Divide the leading term of the dividend \(9x^3\) by the leading term of the divisor \(3x\): \(\frac{9x^3}{3x} = 3x^2\). This is the first term of the quotient (top - left box).

Step2: Multiply \(3x\) by \(3x^2\)

We already know this is \(9x^3\) (given in the table). Now, multiply \(3x\) by the next term of the quotient. Let the second term of the quotient be \(ax + b\) (we will find \(a\) and \(b\) step - by - step). First, multiply \(3x\) by the second term of the quotient. Let's find the second term of the quotient by dividing the next term of the dividend. The next term of the dividend (after \(9x^3\)) is \(- 12x^2\). Divide \(-12x^2\) by \(3x\): \(\frac{-12x^2}{3x}=-4x\). So the second term of the quotient is \(-4x\). Now, multiply \(3x\) by \(-4x\): \(3x\times(-4x)=-12x^2\) (this goes in the middle box of the \(3x\) row).

Step3: Multiply \(3x\) by the third term of the quotient

The third term of the quotient: divide the next term of the dividend \(15x\) by \(3x\): \(\frac{15x}{3x} = 5\). So the third term of the quotient is \(5\). Multiply \(3x\) by \(5\): \(3x\times5 = 15x\) (this goes in the right - most box of the \(3x\) row).

Step4: Multiply \(-4\) by the first term of the quotient (\(3x^2\))

\(-4\times3x^2=-12x^2\) (this goes in the left - most box of the \(-4\) row).

Step5: Multiply \(-4\) by the second term of the quotient (\(-4x\))

\(-4\times(-4x)=16x\) (this goes in the middle box of the \(-4\) row).

Step6: Multiply \(-4\) by the third term of the quotient (\(5\))

\(-4\times5=-20\) (this goes in the right - most box of the \(-4\) row).

Step7: Verify the multiplication

Now, let's check the product of \((3x - 4)(3x^2-4x + 5)\)
\[ \begin{align*} &(3x-4)(3x^2-4x + 5)\\ =&3x\times(3x^2-4x + 5)-4\times(3x^2-4x + 5)\\ =&9x^3-12x^2 + 15x-12x^2+16x - 20\\ =&9x^3-24x^2+31x - 20 \end{align*} \]
Wait, we made a mistake. Let's start over with the correct long - division approach.

Correct long - division:
We are dividing \(9x^3-12x^2 + 15x-20\) by \(3x - 4\)

1. Divide \(9x^3\) by \(3x\) to get \(3x^2\). Multiply \(3x - 4\) by \(3x^2\): \(3x^2\times(3x - 4)=9x^3-12x^2\)
Subtract this from the dividend: \((9x^3-12x^2 + 15x-20)-(9x^3-12x^2)=15x-20\)

2. Now, divide \(15x\) by \(3x\) to get \(5\). Multiply \(3x - 4\) by \(5\): \(5\times(3x - 4)=15x-20\)
Subtract this from \(15x - 20\): \((15x-20)-(15x - 20)=0\)

So the quotient is \(3x^2 + 5\)

Now, filling the table:

- Top row (quotient): \(3x^2\), \(0x\) (wait, no, the quotient is \(3x^2+0x + 5\)), \(5\)
- \(3x\) row:
- First box: \(9x^3\) (given)
- Second box: \(3x\times0x = 0\) (but this is not correct. Wait, our initial long - division shows that after dividing \(9x^3\) by \(3x\) to get \(3x^2\), and subtracting \(3x^2(3x - 4)\) from the dividend, we get \(15x-20\). Then we divide \(15x\) by \(3x\) to get \(5\). So the quotient is \(3x^2+5\) (the middle term of the quotient is \(0\) because there is no \(x\) term in the quotient after the first and third terms).

So:

- Top row (quotient): \(3x^2\), \(0\) (for the \(x\) term), \(5\)
- \(3x\) row:
- First box: \(9x^3\) (given)
- Second box: \(3x\times0 = 0\) (but this is not in the dividend. Wait, the dividend is \(9x^3-12x^2+15x - 20\). Let's use the area model correctly.

The area model for \((3x - 4)(3x^2+5)=9x^3+15x-12x^2 - 20=9x^3-12x^2+15x - 20\)

So:

- Top row (quotient): \(3x^2\), \(0\) (x - term), \(5\)
- \(3x\) row:
- First box: \(9x^3\) ( \(3x\times3x^2\))
- Second box: \(0\) ( \(3x\times0\))
- Third box: \(15x\) ( \(3x\times5\))
- \(-4\) row:
- First box: \(-12x^2\) ( \(-4\times3x^2\))
- Second box: \(0\) ( \(-4\times0\))
- Third box: \(-20\) ( \(-4\times5\))

Answer:

The quotient is \(3x^2 + 5\)

Filling the table:

• Top row (from left to right): \(3x^2\), \(0\) (x - term), \(5\)
• \(3x\) row (from left to right): \(9x^3\), \(0\) (x - term product), \(15x\)
• \(-4\) row (from left to right): \(-12x^2\), \(0\) (x - term product), \(-20\)

But if we consider the non - zero terms:

The quotient is \(3x^2+5\) (the middle term of the quotient is \(0\) as there is no \(x\) term in the quotient). So the final quotient when \(9x^3-12x^2 + 15x-20\) is divided by \(3x - 4\) is \(3x^2+5\)