We need to determine the missing values [tex]\(A\)[/tex] and [tex]\(B\)[/tex] in Isla's division table while dividing [tex]\(3x^3 + x^2 - 12x - 4\)[/tex] by [tex]\(x + 2\)[/tex].
Here's a structured approach to understanding all the steps involved and finding the missing values:
1. Divide the first term:
- The first term of the dividend [tex]\(3x^3\)[/tex] is divided by the first term of the divisor [tex]\(x\)[/tex]:
[tex]\[
\frac{3x^3}{x} = 3x^2
\][/tex]
- The first term of the quotient is [tex]\(3x^2\)[/tex].
2. Multiply the entire divisor by the first term of the quotient:
- Multiply [tex]\(3x^2\)[/tex] by the divisor [tex]\(x + 2\)[/tex]:
[tex]\[
3x^2 \cdot (x + 2) = 3x^3 + 6x^2
\][/tex]
3. Subtract the result from the dividend:
- Subtract [tex]\(3x^3 + 6x^2\)[/tex] from [tex]\(3x^3 + x^2 - 12x - 4\)[/tex]:
[tex]\[
(3x^3 + x^2 - 12x - 4) - (3x^3 + 6x^2) = x^2 - 6x - 4
\][/tex]
4. Divide the next term:
- Divide [tex]\(x^2\)[/tex] by [tex]\(x\)[/tex]:
[tex]\[
\frac{x^2}{x} = x
\][/tex]
- Subtract the term [tex]\(6x^2\)[/tex] from both sides to continue the division:
[tex]\[
(x^2 - 6x - 4) - x = -5x^2 - 12x - 4
\][/tex]
5. Repeat the process for subsequent terms:
- The details for the further steps will also solve and correct below:
- After dividing [tex]\(x^2-6x\)[/tex] by [tex]\(x\)[/tex], we get the term [tex]\(-5x\)[/tex].
- Multiply [tex]\(-5x\)[/tex] by the divisor [tex]\(x + 2\)[/tex]:
[tex]\[
-5x(x + 2) = -5x^2 - 10x
\][/tex]
- Subtract the result from the previous remainder:
[tex]\[
(x^2 - 6x) - (-5x^2 - 10x) = x^2 + 5x^2 + 4x = 6x^2 + 4x
\][/tex]
6. Finally, the last quotient term is found by subtracting the final multiplication:
- Divide [tex]\(6x^2 + 4x - 4\)[/tex] by [tex]\(x + 2\)[/tex].
[tex]\[
-2(x+2) = -2x - 4 = B
\][/tex]
The [tex]\(A\)[/tex] division value:
[tex]\[
4x \cdot 2 = 2^2 = -10x
\][/tex]
Therefore, the correct values are
[tex]\[
A = -10x, \quad B = -4
\][/tex]
So, the missing values in the table are:
[tex]\[
\boxed{A = -10x; B = -4}
\][/tex]
