Question
Factorise fully
1. x squared minus x
2. (x + y ) all squared minus 1
3. 4p squared + 7 pq minus 2 q squared
4. 5x cubed + 40
5. X cubed minus 1 over x cubed
6. (1 over 27)x cubed plus 216
1. x squared minus x
2. (x + y ) all squared minus 1
3. 4p squared + 7 pq minus 2 q squared
4. 5x cubed + 40
5. X cubed minus 1 over x cubed
6. (1 over 27)x cubed plus 216
Answers
GPT-4o mini
Let's factorise each expression step by step:
### 1. \( x^2 - x \)
Factor out the common factor \( x \):
\[
x^2 - x = x(x - 1)
\]
### 2. \( (x + y)^2 - 1 \)
This expression can be recognized as a difference of squares:
\[
(a^2 - b^2) = (a - b)(a + b)
\]
Where \( a = (x + y) \) and \( b = 1 \):
\[
= ((x + y) - 1)((x + y) + 1) = (x + y - 1)(x + y + 1)
\]
### 3. \( 4p^2 + 7pq - 2q^2 \)
We will factor this quadratic expression. We can try to find two numbers that multiply to \( 4 \times (-2) = -8 \) and add to \( 7 \). These numbers are \( 8 \) and \( -1 \).
\[
4p^2 + 8pq - pq - 2q^2 = (4p^2 + 8pq) + (-pq - 2q^2)
\]
Factoring by grouping:
\[
= 4p(p + 2q) - q(p + 2q) = (4p - q)(p + 2q)
\]
### 4. \( 5x^3 + 40 \)
Factor out the greatest common factor, which is \( 5 \):
\[
5(x^3 + 8)
\]
Now, \( x^3 + 8 \) can be factored as a sum of cubes:
\[
x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)
\]
So the full factorization is:
\[
5(x + 2)(x^2 - 2x + 4)
\]
### 5. \( \frac{x^3 - 1}{x^3} \)
Factor \( x^3 - 1 \) as a difference of cubes:
\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]
Now, substitute it back:
\[
= \frac{(x - 1)(x^2 + x + 1)}{x^3} = \frac{x - 1}{x} \cdot (x^2 + x + 1)
\]
This can also be expressed as:
\[
\frac{x - 1}{x}(x^2 + x + 1) = (1 - \frac{1}{x})(x^2 + x + 1)
\]
### 6. \( \frac{1}{27}x^3 + 216 \)
First, factor out \( \frac{1}{27} \):
\[
= \frac{1}{27}(x^3 + 27 \cdot 216)
\]
Recognizing that \( 216 = 6^3 \) gives us:
\[
= \frac{1}{27}(x^3 + 6^3)
\]
Which can be factored as:
\[
= \frac{1}{27}((x + 6)(x^2 - 6x + 36))
\]
In summary, the fully factored forms are:
1. \( x(x - 1) \)
2. \( (x + y - 1)(x + y + 1) \)
3. \( (4p - q)(p + 2q) \)
4. \( 5(x + 2)(x^2 - 2x + 4) \)
5. \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)
6. \( \frac{1}{27}(x + 6)(x^2 - 6x + 36) \)
### 1. \( x^2 - x \)
Factor out the common factor \( x \):
\[
x^2 - x = x(x - 1)
\]
### 2. \( (x + y)^2 - 1 \)
This expression can be recognized as a difference of squares:
\[
(a^2 - b^2) = (a - b)(a + b)
\]
Where \( a = (x + y) \) and \( b = 1 \):
\[
= ((x + y) - 1)((x + y) + 1) = (x + y - 1)(x + y + 1)
\]
### 3. \( 4p^2 + 7pq - 2q^2 \)
We will factor this quadratic expression. We can try to find two numbers that multiply to \( 4 \times (-2) = -8 \) and add to \( 7 \). These numbers are \( 8 \) and \( -1 \).
\[
4p^2 + 8pq - pq - 2q^2 = (4p^2 + 8pq) + (-pq - 2q^2)
\]
Factoring by grouping:
\[
= 4p(p + 2q) - q(p + 2q) = (4p - q)(p + 2q)
\]
### 4. \( 5x^3 + 40 \)
Factor out the greatest common factor, which is \( 5 \):
\[
5(x^3 + 8)
\]
Now, \( x^3 + 8 \) can be factored as a sum of cubes:
\[
x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)
\]
So the full factorization is:
\[
5(x + 2)(x^2 - 2x + 4)
\]
### 5. \( \frac{x^3 - 1}{x^3} \)
Factor \( x^3 - 1 \) as a difference of cubes:
\[
x^3 - 1 = (x - 1)(x^2 + x + 1)
\]
Now, substitute it back:
\[
= \frac{(x - 1)(x^2 + x + 1)}{x^3} = \frac{x - 1}{x} \cdot (x^2 + x + 1)
\]
This can also be expressed as:
\[
\frac{x - 1}{x}(x^2 + x + 1) = (1 - \frac{1}{x})(x^2 + x + 1)
\]
### 6. \( \frac{1}{27}x^3 + 216 \)
First, factor out \( \frac{1}{27} \):
\[
= \frac{1}{27}(x^3 + 27 \cdot 216)
\]
Recognizing that \( 216 = 6^3 \) gives us:
\[
= \frac{1}{27}(x^3 + 6^3)
\]
Which can be factored as:
\[
= \frac{1}{27}((x + 6)(x^2 - 6x + 36))
\]
In summary, the fully factored forms are:
1. \( x(x - 1) \)
2. \( (x + y - 1)(x + y + 1) \)
3. \( (4p - q)(p + 2q) \)
4. \( 5(x + 2)(x^2 - 2x + 4) \)
5. \( \frac{(x - 1)(x^2 + x + 1)}{x^3} \)
6. \( \frac{1}{27}(x + 6)(x^2 - 6x + 36) \)