To prove that [tex]\((x+1)\)[/tex] is a factor of the cubic polynomial [tex]\(x^3 + 5x^2 - x - 5\)[/tex] using polynomial division, we'll perform the polynomial long division step-by-step.
### Polynomial Long Division:
1. Divide the first term:
- Divide the leading term of the dividend [tex]\(x^3\)[/tex] by the leading term of the divisor [tex]\(x\)[/tex]:
[tex]\[
\frac{x^3}{x} = x^2
\][/tex]
2. Multiply and subtract:
- Multiply [tex]\(x^2\)[/tex] by the entire divisor [tex]\((x+1)\)[/tex]:
[tex]\[
x^2 \cdot (x + 1) = x^3 + x^2
\][/tex]
- Subtract this from the original polynomial:
[tex]\[
(x^3 + 5x^2 - x - 5) - (x^3 + x^2) = 4x^2 - x - 5
\][/tex]
3. Repeat the process for the new polynomial [tex]\(4x^2 - x - 5\)[/tex]:
- Divide the leading term:
[tex]\[
\frac{4x^2}{x} = 4x
\][/tex]
- Multiply and subtract:
[tex]\[
4x \cdot (x + 1) = 4x^2 + 4x
\][/tex]
- Subtract:
[tex]\[
(4x^2 - x - 5) - (4x^2 + 4x) = -5x - 5
\][/tex]
4. Repeat the process for [tex]\(-5x - 5\)[/tex]:
- Divide the leading term:
[tex]\[
\frac{-5x}{x} = -5
\][/tex]
- Multiply and subtract:
[tex]\[
-5 \cdot (x + 1) = -5x - 5
\][/tex]
- Subtract:
[tex]\[
(-5x - 5) - (-5x - 5) = 0
\][/tex]
The remainder is [tex]\(0\)[/tex], which means [tex]\((x + 1)\)[/tex] is indeed a factor of the polynomial [tex]\(x^3 + 5x^2 - x - 5\)[/tex].
### Factoring the Cubic Polynomial:
From the division process, we have:
[tex]\[
x^3 + 5x^2 - x - 5 = (x + 1)(x^2 + 4x - 5)
\][/tex]
### Completely Factor the Quadratic:
Next, we factor the quadratic [tex]\(x^2 + 4x - 5\)[/tex].
To factor [tex]\(x^2 + 4x - 5\)[/tex], we look for two numbers that multiply to [tex]\(-5\)[/tex] and add up to [tex]\(4\)[/tex]. Those numbers are [tex]\(5\)[/tex] and [tex]\(-1\)[/tex].
Thus:
[tex]\[
x^2 + 4x - 5 = (x + 5)(x - 1)
\][/tex]
### Final Factorization:
Combining all the factors, we get:
[tex]\[
x^3 + 5x^2 - x - 5 = (x + 1)(x + 5)(x - 1)
\][/tex]
Thus, the complete factorization of the cubic polynomial is:
[tex]\[
x^3 + 5x^2 - x - 5 = (x + 1) \times (x + 5) \times (x - 1)
\][/tex]
