Question

use long division to rewrite this expression in an equivalent form as $q(x) + \frac{r(x)}{b(x)}$, where $q(x)$ is the quotient, $r(x)$ is the remainder, and $b(x)$ is the divisor.\
$x^3 - 2x^2 - 4 \div (x - 3)$\
drag the expressions to the correct locations on the image. not all expressions will be used. for help, see this example.\
expressions: -13, 5, $x^2 - 5x + 15$, $x^2 + 5x - 15$, $x^2 + x + 3$, $x - 3$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(x^3 - 2x^2 - 4\) (which is \(x^3\)) by the leading term of the divisor \(x - 3\) (which is \(x\)). So, \(\frac{x^3}{x}=x^2\). This is the first term of the quotient \(q(x)\).
Multiply the divisor \(x - 3\) by \(x^2\): \(x^2(x - 3)=x^3 - 3x^2\).
Subtract this from the dividend: \((x^3 - 2x^2 - 4)-(x^3 - 3x^2)=x^3 - 2x^2 - 4 - x^3 + 3x^2=x^2 - 4\).

Step2: Divide the new leading term

Now, take the new dividend \(x^2 - 4\). Divide its leading term \(x^2\) by the leading term of the divisor \(x\), giving \(\frac{x^2}{x}=x\). This is the next term of the quotient.
Multiply the divisor \(x - 3\) by \(x\): \(x(x - 3)=x^2 - 3x\).
Subtract this from the new dividend: \((x^2 - 4)-(x^2 - 3x)=x^2 - 4 - x^2 + 3x = 3x - 4\).

Step3: Divide the new leading term again

Take the new dividend \(3x - 4\). Divide its leading term \(3x\) by the leading term of the divisor \(x\), giving \(\frac{3x}{x}=3\). This is the next term of the quotient.
Multiply the divisor \(x - 3\) by \(3\): \(3(x - 3)=3x - 9\).
Subtract this from the new dividend: \((3x - 4)-(3x - 9)=3x - 4 - 3x + 9 = 5\). Wait, no, wait, let's check again. Wait, in the previous step, after subtracting \(x^2 - 3x\) from \(x^2 - 4\), we get \(3x - 4\). Then dividing \(3x\) by \(x\) gives 3. Multiply \(x - 3\) by 3: \(3x - 9\). Subtract: \((3x - 4)-(3x - 9)=3x - 4 - 3x + 9 = 5\)? Wait, but the options have -13. Wait, maybe I made a mistake. Let's redo the long division.

Wait, the dividend is \(x^3 - 2x^2 + 0x - 4\) (we add the missing \(x\) term). Divisor is \(x - 3\).

1. Divide \(x^3\) by \(x\) to get \(x^2\). Multiply \(x - 3\) by \(x^2\): \(x^3 - 3x^2\). Subtract from dividend: \((x^3 - 2x^2 + 0x - 4)-(x^3 - 3x^2)=x^2 + 0x - 4\).

2. Divide \(x^2\) by \(x\) to get \(x\). Multiply \(x - 3\) by \(x\): \(x^2 - 3x\). Subtract: \((x^2 + 0x - 4)-(x^2 - 3x)=3x - 4\).

3. Divide \(3x\) by \(x\) to get \(3\). Multiply \(x - 3\) by \(3\): \(3x - 9\). Subtract: \((3x - 4)-(3x - 9)=5\)? But the options have -13. Wait, maybe the dividend is \(x^3 - 2x^2 - 4\), but maybe I missed a sign. Wait, no, let's check the options. Wait, the options for the quotient: \(x^2 + x + 3\)? Wait, let's try with \(x^2 + x + 3\) as quotient. Multiply \((x - 3)(x^2 + x + 3)=x^3 + x^2 + 3x - 3x^2 - 3x - 9=x^3 - 2x^2 - 9\). Then the dividend is \(x^3 - 2x^2 - 4\), so the remainder would be \((x^3 - 2x^2 - 4)-(x^3 - 2x^2 - 9)=5\). But the options have -13. Wait, maybe the dividend is \(x^3 - 2x^2 - 4\) and divisor \(x - 3\), but let's check another way. Wait, maybe the quotient is \(x^2 + x + 3\), divisor \(x - 3\), remainder? Wait, no, let's use the Remainder Theorem. The remainder when dividing by \(x - a\) is \(f(a)\). So \(f(3)=3^3 - 2*3^2 - 4=27 - 18 - 4=5\). But the options have -13. Wait, maybe the dividend is \(x^3 - 2x^2 - 4\) and divisor \(x + 3\)? No, the divisor is \(x - 3\). Wait, the options: quotient options are \(x^2 - 5x + 15\), \(x^2 + 5x - 15\), \(x^2 + x + 3\). Let's try \(x^2 + x + 3\) as quotient. Then \((x - 3)(x^2 + x + 3)=x^3 + x^2 + 3x - 3x^2 - 3x - 9=x^3 - 2x^2 - 9\). Then dividend is \(x^3 - 2x^2 - 4\), so remainder is \((x^3 - 2x^2 - 4)-(x^3 - 2x^2 - 9)=5\). So the quotient \(q(x)=x^2 + x + 3\), remainder \(r(x)=5\), divisor \(b(x)=x - 3\). Wait, but the options have -13. Wait, maybe I made a mistake in the dividend. Wait, the original problem is \(x^3 - 2x^2 - 4 \div (x - 3)\). Let's do long division again carefully:

Dividend: \(x^3 - 2x^2 + 0x - 4\)

Divisor: \(x - 3\)

Step 1: \(x^3 \div x = x^2\). Multiply divisor by \(x^2\): \(x^3 - 3x^2\). Subtract from dividend: \((x^3 - 2x^2 + 0x - 4) - (x^3 - 3x^2) = x^2 + 0x - 4\)

Step 2: \(x^2 \div x = x\). Multiply divisor by \(x\): \(x^2 - 3x\). Subtract: \((x^2 + 0x - 4) - (x^2 - 3x) = 3x - 4\)

Step 3: \(3x \div x = 3\). Multiply divisor by 3: \(3x - 9\). Subtract: \((3x - 4) - (3x - 9) = 5\)

So the quotient \(q(x)=x^2 + x + 3\), remainder \(r(x)=5\), divisor \(b(x)=x - 3\). So the expression is \(x^2 + x + 3 + \frac{5}{x - 3}\).

Wait, the options: the first box (q(x)) should be \(x^2 + x + 3\), the numerator (r(x)) is 5, and the denominator (b(x)) is \(x - 3\).

Step4: Identify the parts

So \(q(x)=x^2 + x + 3\), \(r(x)=5\), \(b(x)=x - 3\). So the first box (left of +) is \(x^2 + x + 3\), the numerator is 5, and the denominator is \(x - 3\).

Answer:

The left box (q(x)): \(x^2 + x + 3\)
The numerator (r(x)): \(5\)
The denominator (b(x)): \(x - 3\)

(So dragging \(x^2 + x + 3\) to the left box, \(5\) to the numerator, and \(x - 3\) to the denominator.)