Question

drag the tiles to the correct boxes to complete the pairs. not all tiles will be used. rewrite the rational expression \\(\frac{2x^3 - 7x^2 + 6x - 9}{x - 3}\\) in the form \\(q(x) + \frac{r(x)}{b(x)}\\). then match q(x), r(x), and b(x) to the correct expressions. tiles: \\(2x^2 - 7x + 6\\), \\(2x^2 - x + 3\\), \\(2x^3 - 7x^2 + 6x - 9\\), \\(x - 3\\), \\(-9\\), \\(0\\). boxes for q(x), r(x), b(x) with reset and next buttons.

Explanation:

Step1: Perform Polynomial Long Division

Divide \(2x^3 - 7x^2 + 6x - 9\) by \(x - 3\).
- Divide the leading term \(2x^3\) by \(x\) to get \(2x^2\) (this is the first term of \(q(x)\)).
- Multiply \(x - 3\) by \(2x^2\) to get \(2x^3 - 6x^2\).
- Subtract this from the original polynomial: \((2x^3 - 7x^2 + 6x - 9) - (2x^3 - 6x^2) = -x^2 + 6x - 9\).

Step2: Continue the Division

- Divide the leading term \(-x^2\) by \(x\) to get \(-x\) (next term of \(q(x)\)).
- Multiply \(x - 3\) by \(-x\) to get \(-x^2 + 3x\).
- Subtract: \((-x^2 + 6x - 9) - (-x^2 + 3x) = 3x - 9\).

Step3: Continue the Division

- Divide the leading term \(3x\) by \(x\) to get \(3\) (next term of \(q(x)\)).
- Multiply \(x - 3\) by \(3\) to get \(3x - 9\).
- Subtract: \((3x - 9) - (3x - 9) = 0\)? Wait, no, wait. Wait, original polynomial is \(2x^3 -7x^2 +6x -9\). Wait, let's redo the division properly.

Wait, let's use synthetic division with root \(x = 3\) (since divisor is \(x - 3\)).

Coefficients: \(2\), \(-7\), \(6\), \(-9\)

Bring down the \(2\).

Multiply \(2\) by \(3\) to get \(6\). Add to \(-7\): \(-7 + 6 = -1\).

Multiply \(-1\) by \(3\) to get \(-3\). Add to \(6\): \(6 + (-3) = 3\).

Multiply \(3\) by \(3\) to get \(9\). Add to \(-9\): \(-9 + 9 = 0\). Wait, that's different. Wait, so the quotient \(q(x)\) is \(2x^2 - x + 3\) (since the coefficients from synthetic division are \(2\), \(-1\), \(3\) and the remainder is \(0\)? But wait, the original polynomial is \(2x^3 -7x^2 +6x -9\). Wait, no, maybe I made a mistake in the problem statement. Wait, the numerator is \(2x^3 -7x^2 +6x -9\)? Wait, the user wrote "2x³ -7x² +6x -9" in the tiles. Wait, but when we do synthetic division with \(x = 3\):

\(3 | 2\ \ -7\ \ 6\ \ -9\)

\(6\ \ -3\ \ 9\)

\(2\ \ -1\ \ 3\ \ 0\)

So the quotient \(q(x)\) is \(2x^2 - x + 3\) and the remainder \(r(x)\) is \(0\)? But wait, the tiles have \(-9\) and \(0\). Wait, maybe I misread the numerator. Wait, the original rational expression is \(\frac{2x^3 -7x^2 +6x - 9}{x - 3}\)? Wait, no, the user's problem says "2x³ -7x² +6x - 9" in the numerator? Wait, no, the first tile is \(2x^2 -7x +6\), second is \(2x^2 -x +3\), third is \(2x^3 -7x^2 +6x -9\), fourth is \(x -3\), then \(-9\), \(0\).

Wait, maybe the numerator is \(2x^3 -7x^2 +6x - 9\)? Wait, no, let's check the division again. Wait, if we divide \(2x^3 -7x^2 +6x -9\) by \(x - 3\), using synthetic division:

Coefficients: 2 (x³), -7 (x²), 6 (x), -9 (constant)

Bring down 2.

Multiply 2 by 3: 6. Add to -7: -1.

Multiply -1 by 3: -3. Add to 6: 3.

Multiply 3 by 3: 9. Add to -9: 0. So the quotient is \(2x^2 - x + 3\) and the remainder is 0. But wait, the tiles have \(-9\) and \(0\). Wait, maybe the numerator is \(2x^3 -7x^2 +6x - 9\)? Wait, no, maybe I made a mistake. Wait, let's do polynomial long division:

Divide \(2x^3 -7x^2 +6x -9\) by \(x - 3\).

First term: \(2x^3 / x = 2x^2\). Multiply \(x - 3\) by \(2x^2\): \(2x^3 -6x^2\). Subtract from dividend: \((2x^3 -7x^2 +6x -9) - (2x^3 -6x^2) = -x^2 +6x -9\).

Next term: \(-x^2 / x = -x\). Multiply \(x - 3\) by \(-x\): \(-x^2 +3x\). Subtract: \((-x^2 +6x -9) - (-x^2 +3x) = 3x -9\).

Next term: \(3x / x = 3\). Multiply \(x - 3\) by \(3\): \(3x -9\). Subtract: \((3x -9) - (3x -9) = 0\). So the quotient \(q(x)\) is \(2x^2 -x +3\), the remainder \(r(x)\) is \(0\), and the divisor \(b(x)\) is \(x -3\). Wait, but the numerator is \(2x^3 -7x^2 +6x -9\), which is the original dividend, so \(a(x) = 2x^3 -7x^2 +6x -9\), \(b(x) = x -3\), \(q(x) = 2x^2 -x +3\), \(r(x) = 0\). But wait, the tiles have \(-9\) and \(0\). Wait, maybe the numerator was supposed to be \(2x^3 -7x^2 +6x - 9\)? Wait, no, maybe I misread the numerator. Wait, the problem says "2x³ -7x² +6x - 9" in the tile. So:

- \(q(x)\) is the quotient: \(2x^2 -x +3\)
- \(r(x)\) is the remainder: \(0\)
- \(b(x)\) is the divisor: \(x -3\)

Wait, but let's check again. Wait, if we have \(\frac{a(x)}{b(x)} = q(x) + \frac{r(x)}{b(x)}\), so \(a(x) = q(x)b(x) + r(x)\). Let's verify:

\(q(x) = 2x^2 -x +3\), \(b(x) = x -3\), \(r(x) = 0\).

Then \(q(x)b(x) + r(x) = (2x^2 -x +3)(x -3) + 0\)

Multiply out: \(2x^3 -6x^2 -x^2 +3x +3x -9 = 2x^3 -7x^2 +6x -9\), which matches the numerator \(a(x) = 2x^3 -7x^2 +6x -9\). So that's correct.

Wait, but the tiles have \(2x^2 -7x +6\) as a tile. Maybe I made a mistake in the division. Wait, let's try dividing \(2x^3 -7x^2 +6x -9\) by \(x - 3\) again. Wait, maybe the numerator was \(2x^3 -7x^2 +6x - 9\), but maybe the user made a typo? No, the tile has \(2x^3 -7x^2 +6x -9\). So:

- \(q(x)\): \(2x^2 -x +3\) (from the tiles, this is one of the options)
- \(r(x)\): \(0\) (from the tiles, this is an option)
- \(b(x)\): \(x -3\) (from the tiles, this is an option)

Wait, but let's check the other tile \(2x^2 -7x +6\). If we divide \(2x^3 -7x^2 +6x -9\) by \(x - 3\), is that the quotient? Let's multiply \(2x^2 -7x +6\) by \(x -3\):

\(2x^3 -6x^2 -7x^2 +21x +6x -18 = 2x^3 -13x^2 +27x -18\), which is not the numerator. So that's wrong. So the correct quotient is \(2x^2 -x +3\), remainder \(0\), divisor \(x -3\).

Answer:

• \(q(x)\): \(2x^2 - x + 3\)
• \(r(x)\): \(0\)
• \(b(x)\): \(x - 3\)