Answer :
Certainly! Let's work through the steps to factor the given polynomial [tex]\( P(x) = x^3 + 5x^2 - x - 5 \)[/tex] by grouping.
1. Group the terms:
[tex]\[ P(x) = x^3 + 5x^2 - x - 5 \][/tex]
We will group the polynomial into two sets of terms for easier factoring:
[tex]\[ P(x) = (x^3 + 5x^2) + (-x - 5) \][/tex]
2. Factor each group:
- For the first group [tex]\(x^3 + 5x^2\)[/tex], we can factor out [tex]\(x^2\)[/tex]:
[tex]\[ x^3 + 5x^2 = x^2(x + 5) \][/tex]
- For the second group [tex]\(-x - 5\)[/tex], we can factor out [tex]\(-1\)[/tex]:
[tex]\[ -x - 5 = -1(x + 5) \][/tex]
Thus, the polynomial becomes:
[tex]\[ P(x) = x^2(x + 5) - 1(x + 5) \][/tex]
3. Factor out the common factor [tex]\(x + 5\)[/tex] from the groups:
- Both terms have a common factor of [tex]\(x + 5\)[/tex].
[tex]\[ P(x) = (x^2 - 1)(x + 5) \][/tex]
4. Factor [tex]\(x^2 - 1\)[/tex] as a difference of squares:
- Recall that [tex]\(x^2 - 1\)[/tex] can be factored as [tex]\((x - 1)(x + 1)\)[/tex]:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
5. Write the final factored form of [tex]\(P(x)\)[/tex]:
- Now combine these factors:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
So, the final factored form of the polynomial [tex]\(P(x) = x^3 + 5x^2 - x - 5\)[/tex] is:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
To summarize the steps as requested in the original layout:
- Grouping:
[tex]\[ P(x) = x^2(x + 5) - 1(x + 5) \][/tex]
- Factoring out the common term:
[tex]\[ P(x) = (x^2 - 1)(x + 5) \][/tex]
- Factoring the difference of squares:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
Thus, completing the steps, we get:
[tex]\[ \begin{array}{l} P(x)=x^2(x + 5) - (x + 5) \\ P(x)=\left(x^2 - 1\right)(x + 5) \\ P(x)=(x - 1)(x+1)(x+5) \end{array} \][/tex]
1. Group the terms:
[tex]\[ P(x) = x^3 + 5x^2 - x - 5 \][/tex]
We will group the polynomial into two sets of terms for easier factoring:
[tex]\[ P(x) = (x^3 + 5x^2) + (-x - 5) \][/tex]
2. Factor each group:
- For the first group [tex]\(x^3 + 5x^2\)[/tex], we can factor out [tex]\(x^2\)[/tex]:
[tex]\[ x^3 + 5x^2 = x^2(x + 5) \][/tex]
- For the second group [tex]\(-x - 5\)[/tex], we can factor out [tex]\(-1\)[/tex]:
[tex]\[ -x - 5 = -1(x + 5) \][/tex]
Thus, the polynomial becomes:
[tex]\[ P(x) = x^2(x + 5) - 1(x + 5) \][/tex]
3. Factor out the common factor [tex]\(x + 5\)[/tex] from the groups:
- Both terms have a common factor of [tex]\(x + 5\)[/tex].
[tex]\[ P(x) = (x^2 - 1)(x + 5) \][/tex]
4. Factor [tex]\(x^2 - 1\)[/tex] as a difference of squares:
- Recall that [tex]\(x^2 - 1\)[/tex] can be factored as [tex]\((x - 1)(x + 1)\)[/tex]:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
5. Write the final factored form of [tex]\(P(x)\)[/tex]:
- Now combine these factors:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
So, the final factored form of the polynomial [tex]\(P(x) = x^3 + 5x^2 - x - 5\)[/tex] is:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
To summarize the steps as requested in the original layout:
- Grouping:
[tex]\[ P(x) = x^2(x + 5) - 1(x + 5) \][/tex]
- Factoring out the common term:
[tex]\[ P(x) = (x^2 - 1)(x + 5) \][/tex]
- Factoring the difference of squares:
[tex]\[ P(x) = (x - 1)(x + 1)(x + 5) \][/tex]
Thus, completing the steps, we get:
[tex]\[ \begin{array}{l} P(x)=x^2(x + 5) - (x + 5) \\ P(x)=\left(x^2 - 1\right)(x + 5) \\ P(x)=(x - 1)(x+1)(x+5) \end{array} \][/tex]
