Question
Factorise fully
1. ( x + y ) all squared - 1
2. 4p squared + 7pq - 2q squared
3. X cubed - 1 over x cubed
4. 1 over 27 , x cubed + 216
1. ( x + y ) all squared - 1
2. 4p squared + 7pq - 2q squared
3. X cubed - 1 over x cubed
4. 1 over 27 , x cubed + 216
Answers
GPT-4o mini
Let's factor each expression fully:
1. **\( (x + y)^2 - 1 \)**
This is a difference of squares, which can be factored using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\[
(x + y)^2 - 1^2 = ((x + y) - 1)((x + y) + 1) = (x + y - 1)(x + y + 1)
\]
**Final answer: \( (x + y - 1)(x + y + 1) \)**
2. **\( 4p^2 + 7pq - 2q^2 \)**
We can factor this trinomial by looking for two numbers that multiply to \( (4)(-2) = -8 \) and add to \( 7 \). The two numbers are \( 8 \) and \( -1 \).
Rewrite the middle term:
\[
4p^2 + 8pq - pq - 2q^2
\]
Now, group terms:
\[
(4p^2 + 8pq) + (-pq - 2q^2) = 4p(p + 2q) - q(p + 2q)
\]
Factor out the common term \( (p + 2q) \):
\[
(4p - q)(p + 2q)
\]
**Final answer: \( (4p - q)(p + 2q) \)**
3. **\( \frac{x^3 - 1}{x^3} \)**
First, factor \( x^3 - 1 \) using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) where \( a = x \) and \( b = 1 \):
\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]
Therefore, the expression becomes:
\[
\frac{(x - 1)(x^2 + x + 1)}{x^3}
\]
This can't be further simplified in terms of factored form since it's already in simplest form regarding polynomials.
**Final answer: \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)**
4. **\( \frac{1}{27}(x^3 + 216) \)**
Notice that \( 216 = 6^3 \), so we use the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) where \( a = x \) and \( b = 6 \):
\[
x^3 + 216 = x^3 + 6^3 = (x + 6)(x^2 - 6x + 36)
\]
So, we have:
\[
\frac{1}{27}(x + 6)(x^2 - 6x + 36)
\]
**Final answer: \( \frac{(x + 6)(x^2 - 6x + 36)}{27} \)**
Putting all the final answers together:
1. \( (x + y - 1)(x + y + 1) \)
2. \( (4p - q)(p + 2q) \)
3. \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)
4. \( \frac{(x + 6)(x^2 - 6x + 36)}{27} \)
1. **\( (x + y)^2 - 1 \)**
This is a difference of squares, which can be factored using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\[
(x + y)^2 - 1^2 = ((x + y) - 1)((x + y) + 1) = (x + y - 1)(x + y + 1)
\]
**Final answer: \( (x + y - 1)(x + y + 1) \)**
2. **\( 4p^2 + 7pq - 2q^2 \)**
We can factor this trinomial by looking for two numbers that multiply to \( (4)(-2) = -8 \) and add to \( 7 \). The two numbers are \( 8 \) and \( -1 \).
Rewrite the middle term:
\[
4p^2 + 8pq - pq - 2q^2
\]
Now, group terms:
\[
(4p^2 + 8pq) + (-pq - 2q^2) = 4p(p + 2q) - q(p + 2q)
\]
Factor out the common term \( (p + 2q) \):
\[
(4p - q)(p + 2q)
\]
**Final answer: \( (4p - q)(p + 2q) \)**
3. **\( \frac{x^3 - 1}{x^3} \)**
First, factor \( x^3 - 1 \) using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) where \( a = x \) and \( b = 1 \):
\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]
Therefore, the expression becomes:
\[
\frac{(x - 1)(x^2 + x + 1)}{x^3}
\]
This can't be further simplified in terms of factored form since it's already in simplest form regarding polynomials.
**Final answer: \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)**
4. **\( \frac{1}{27}(x^3 + 216) \)**
Notice that \( 216 = 6^3 \), so we use the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \) where \( a = x \) and \( b = 6 \):
\[
x^3 + 216 = x^3 + 6^3 = (x + 6)(x^2 - 6x + 36)
\]
So, we have:
\[
\frac{1}{27}(x + 6)(x^2 - 6x + 36)
\]
**Final answer: \( \frac{(x + 6)(x^2 - 6x + 36)}{27} \)**
Putting all the final answers together:
1. \( (x + y - 1)(x + y + 1) \)
2. \( (4p - q)(p + 2q) \)
3. \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)
4. \( \frac{(x + 6)(x^2 - 6x + 36)}{27} \)