To find the result of dividing [tex]\(2x^3 - 6x^2 - 19x - 5\)[/tex] by [tex]\(x - 5\)[/tex] using the long division method, follow these steps:
1. Set up the division: Write [tex]\(2x^3 - 6x^2 - 19x - 5\)[/tex] (the dividend) inside the long division symbol and [tex]\(x - 5\)[/tex] (the divisor) outside.
2. Divide the leading terms:
- Divide the leading term of the dividend ([tex]\(2x^3\)[/tex]) by the leading term of the divisor ([tex]\(x\)[/tex]) to get the first term of the quotient.
- [tex]\(\frac{2x^3}{x} = 2x^2\)[/tex].
3. Multiply and subtract:
- Multiply [tex]\(2x^2\)[/tex] (the first term of the quotient) by [tex]\(x - 5\)[/tex]:
[tex]\[
2x^2 \cdot (x - 5) = 2x^3 - 10x^2.
\][/tex]
- Subtract this product from the original dividend:
[tex]\[
(2x^3 - 6x^2) - (2x^3 - 10x^2) = 4x^2.
\][/tex]
4. Bring down the next term: The new dividend is [tex]\(4x^2 - 19x - 5\)[/tex].
5. Repeat the process:
- Divide the leading term of the new dividend ([tex]\(4x^2\)[/tex]) by the leading term of the divisor ([tex]\(x\)[/tex]) to get the next term of the quotient:
[tex]\[
\frac{4x^2}{x} = 4x.
\][/tex]
- Multiply [tex]\(4x\)[/tex] by [tex]\(x - 5\)[/tex]:
[tex]\[
4x \cdot (x - 5) = 4x^2 - 20x.
\][/tex]
- Subtract this product from the new dividend:
[tex]\[
(4x^2 - 19x) - (4x^2 - 20x) = x.
\][/tex]
6. Bring down the next term: The new dividend is [tex]\(x - 5\)[/tex].
7. Repeat the process again:
- Divide the leading term of the new dividend ([tex]\(x\)[/tex]) by the leading term of the divisor ([tex]\(x\)[/tex]) to get the next term of the quotient:
[tex]\[
\frac{x}{x} = 1.
\][/tex]
- Multiply [tex]\(1\)[/tex] by [tex]\(x - 5\)[/tex]:
[tex]\[
1 \cdot (x - 5) = x - 5.
\][/tex]
- Subtract this product from the new dividend:
[tex]\[
(x - 5) - (x - 5) = 0.
\][/tex]
8. Final quotient and remainder:
- The quotient obtained is [tex]\(2x^2 + 4x + 1\)[/tex].
- The remainder is [tex]\(0\)[/tex].
Therefore, the result of [tex]\( \frac{2x^3 - 6x^2 - 19x - 5}{x - 5} \)[/tex] is:
[tex]\[
2x^2 + 4x + 1 \quad \text{with a remainder of} \quad 0.
\][/tex]
