# 1.6 Fractions  (Page 2/2)

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 Odd number Even number Odd number Odd Even Even number Even Even

If we take three consecutive numbers and multiply them together, the resulting number is always divisible by three. This should be obvious since if we have any three consecutive numbers, one of them will be divisible by 3.

Now we are ready to demonstrate that ${n}^{2}+n$ is even for all $n\in {Z}$ . If we factorise this expression we get: $n\left(n+1\right)$ . If $n$ is even, than $n+1$ is odd. If $n$ is odd, than $n+1$ is even. Since we know that if we multiply an even number with an odd number or an odd number with an even number, we get an even number, we have demonstrated that ${n}^{2}+n$ is always even. Try this for a few values of $n$ and you should find that this is true.

To demonstrate that ${n}^{3}-n$ is divisible by 6 for all $n\in {Z}$ , we first note that the factors of 6 are 3 and 2. So if we show that ${n}^{3}-n$ is divisible by both 3 and 2, then we have shown that it is also divisible by 6! If we factorise this expression we get: $n\left(n+1\right)\left(n-1\right)$ . Now we note that we are multiplying three consecutive numbers together (we are taking $n$ and then adding 1 or subtracting 1. This gives us the two numbers on either side of $n$ .) For example, if $n=4$ , then $n+1=5$ and $n-1=3$ . But we know that when we multiply three consecutive numbers together, the resulting number is always divisible by 3. So we have demonstrated that ${n}^{3}-n$ is always divisible by 3. To demonstrate that it is also divisible by 2, we can also show that it is even. We have shown that ${n}^{2}+n$ is always even. So now we recall what we said about multiplying even and odd numbers. Since one number is always even and the other can be either even or odd, the result of multiplying these numbers together is always even. And so we have demonstrated that ${n}^{3}-n$ is divisible by 6 for all $n\in {Z}$ .

## Summary

• A binomial is a mathematical expression with two terms. The product of two identical binomials is known as the square of the binomial. The difference of two squares is when we multiply $\left(ax+b\right)\left(ax-b\right)$
• Factorising is the opposite of expanding the brackets. You can use common factors or the difference of two squares to help you factorise expressions.
• The distributive law ( $\left(A+B\right)\left(C+D+E\right)=A\left(C+D+E\right)+B\left(C+D+E\right)$ ) helps us to multiply a binomial and a trinomial.
• The sum of cubes is: $\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)={x}^{3}+{y}^{3}$ and the difference of cubes is: ${x}^{3}-{y}^{3}=\left(x-y\right)\left({x}^{2}+xy+{y}^{2}\right)$
• To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.
• We can also factorise a quadratic by grouping. This is where we find a common factor in the quadratic and take it out and then see what is left over.
• We can simplify fractions by using the methods we have learnt to factorise expressions.
• Fractions can be added or subtracted. To do this the denominators of each fraction must be the same.

## End of chapter exercises

1. Factorise:
1. ${a}^{2}-9$
2. ${m}^{2}-36$
3. $9{b}^{2}-81$
4. $16{b}^{6}-25{a}^{2}$
5. ${m}^{2}-\left(1/9\right)$
6. $5-5{a}^{2}{b}^{6}$
7. $16b{a}^{4}-81b$
8. ${a}^{2}-10a+25$
9. $16{b}^{2}+56b+49$
10. $2{a}^{2}-12ab+18{b}^{2}$
11. $-4{b}^{2}-144{b}^{8}+48{b}^{5}$
2. Factorise completely:
1. $\left(16-{x}^{4}\right)$
2. ${7x}^{2}-14x+7xy-14y$
3. ${y}^{2}-7y-30$
4. $1-x-{x}^{2}+{x}^{3}$
5. $-3\left(1-{p}^{2}\right)+p+1$
3. Simplify the following:
1. ${\left(a-2\right)}^{2}-a\left(a+4\right)$
2. $\left(5a-4b\right)\left(25{a}^{2}+20\mathrm{ab}+16{b}^{2}\right)$
3. $\left(2m-3\right)\left(4{m}^{2}+9\right)\left(2m+3\right)$
4. $\left(a+2b-c\right)\left(a+2b+c\right)$
4. Simplify the following:
1. $\frac{{p}^{2}-{q}^{2}}{p}÷\frac{p+q}{{p}^{2}-\mathrm{pq}}$
2. $\frac{2}{x}+\frac{x}{2}-\frac{2x}{3}$
5. Show that ${\left(2x-1\right)}^{2}-{\left(x-3\right)}^{2}$ can be simplified to $\left(x+2\right)\left(3x-4\right)$

6. What must be added to ${x}^{2}-x+4$ to make it equal to ${\left(x+2\right)}^{2}$

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
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20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
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Abhi
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Abhi
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Abhi
Commplementary angles
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Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Cied
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Porter
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Cesar
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Stotaw
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I'm interested in Nanotube
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