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# Products and factors: Fractions

## Simplification of Fractions

In some cases of simplifying an algebraic expression, the expression will be a fraction. For example,

x 2 + 3 x x + 3 x 2 + 3 x x + 3
(1)

has a quadratic in the numerator and a binomial in the denominator. You can apply the different factorisation methods to simplify the expression.

x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x provided x - 3 x 2 + 3 x x + 3 = x ( x + 3 ) x + 3 = x provided x - 3
(2)

If xx were 3 then the denominator, x-3x-3, would be 0 and the fraction undefined.

### Exercise 1: Simplification of Fractions

Simplify: 2x-b+x-abax2-abx2x-b+x-abax2-abx

#### Solution

1. Step 1. Factorise numerator and denominator :

Use grouping for numerator and common factor for denominator in this example.

= ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b ) = ( a x - a b ) + ( x - b ) a x 2 - a b x = a ( x - b ) + ( x - b ) a x ( x - b ) = ( x - b ) ( a + 1 ) a x ( x - b )
(3)
2. Step 2. Cancel out same factors :

= a + 1 a x = a + 1 a x
(4)

### Exercise 2: Simplification of Fractions

Simplify:x2-x-2x2-4÷x2+xx2+2xx2-x-2x2-4÷x2+xx2+2x

#### Solution

1. Step 1. Factorise numerators and denominators :
= ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 ) = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) ÷ x ( x + 1 ) x ( x + 2 )
(5)
2. Step 2. Multiply by factorised reciprocal :
= ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 ) = ( x + 1 ) ( x - 2 ) ( x + 2 ) ( x - 2 ) × x ( x + 2 ) x ( x + 1 )
(6)
3. Step 3. Cancel out same factors :

= 1 = 1
(7)

### Simplification of Fractions

1. Simplify:
 (a) 3a153a15  (b) 2a+1042a+104  (c) 5a+20a+45a+20a+4  (d) a2-4aa-4a2-4aa-4  (e) 3a2-9a2a-63a2-9a2a-6  (f) 9a+279a+189a+279a+18  (g) 6ab+2a2b6ab+2a2b  (h) 16x2y-8xy12x-616x2y-8xy12x-6  (i) 4xyp-8xp12xy4xyp-8xp12xy  (j) 3a+914÷7a+21a+33a+914÷7a+21a+3  (k) a2-5a2a+10÷3a+154aa2-5a2a+10÷3a+154a  (l) 3xp+4p8p÷12p23x+43xp+4p8p÷12p23x+4  (m) 162xp+4x÷6x2+8x12162xp+4x÷6x2+8x12  (n) 24a-812÷9a-3624a-812÷9a-36  (o) a2+2a5÷2a+420a2+2a5÷2a+420  (p) p2+pq7p÷8p+8q21qp2+pq7p÷8p+8q21q  (q) 5ab-15b4a-12÷6b2a+b5ab-15b4a-12÷6b2a+b  (r) f2a-fa2f-af2a-fa2f-a
2. Simplify: x2-13×1x-1-12x2-13×1x-1-12

Using the concepts learnt in simplification of fractions, we can now add and subtract simple fractions. To add or subtract fractions we note that we can only add or subtract fractions that have the same denominator. So we must first make all the denominators the same and then perform the addition or subtraction. This is called finding the lowest common denominator or multiple.

For example, if you wanted to add: 1212 and 3535 we would note that the lowest common denominator is 10. So we must multiply the first fraction by 5 and the second fraction by 2 to get both of these with the same denominator. Doing so gives: 510510 and 610610. Now we can add the fractions. Doing so, we get 11101110.

### Exercise 3

Simplify the following expression: x-2x2-4+x2x-2-x3+x-4x2-4x-2x2-4+x2x-2-x3+x-4x2-4

#### Solution

1. Step 1. Factorise numerators and denominators :
x-2 (x+2)(x-2) + x2 x-2 - x3+x-4 (x+2)(x-2) x-2 (x+2)(x-2) + x2 x-2 - x3+x-4 (x+2)(x-2)
(8)
2. Step 2. Make the denominators the same :

We make all the denominators the same so that we can add or subtract the fractions. The lowest common denominator is (x-2)(x+2)(x-2)(x+2).

x-2 (x+2)(x-2) + (x2) (x+2) (x+2)(x-2) - x3+x-4 (x+2)(x-2) x-2 (x+2)(x-2) + (x2) (x+2) (x+2)(x-2) - x3+x-4 (x+2)(x-2)
(9)
3. Step 3. Write all the fractions as one:

Since the fractions all have the same denominator we can write them all as one fraction with the appropriate operator

x-2 + (x2) (x+2) - x3+x-4 (x+2)(x-2) x-2 + (x2) (x+2) - x3+x-4 (x+2)(x-2)
(10)
4. Step 4. Simplify the numerator:
2x2 +2x-6 (x+2)(x-2) 2x2 +2x-6 (x+2)(x-2)
(11)
5. Step 5. Write the final answer:
2(x2 +x-3) (x+2)(x-2) 2(x2 +x-3) (x+2)(x-2)
(12)

## Two interesting mathematical proofs

We can use the concepts learnt in this chapter to demonstrate two interesting mathematical proofs. The first proof states that n2+nn2+n is even for all nZnZ. The second proof states that n3-nn3-n is divisible by 6 for all nZnZ. Before we demonstrate that these two laws are true, we first need to note some other mathematical rules.

If we multiply an even number by an odd number, we get an even number. Similarly if we multiply an odd number by an even number we get an even number. Also, an even number multiplied by an even number is even and an odd number multiplied by an odd number is odd. This result is shown in the following table:

 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 n2+nn2+n is even for all nZnZ. If we factorise this expression we get: n(n+1)n(n+1). If nn is even, than n+1n+1 is odd. If nn is odd, than n+1n+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 n2+nn2+n is always even. Try this for a few values of nn and you should find that this is true.

To demonstrate that n3-nn3-n is divisible by 6 for all nZnZ, we first note that the factors of 6 are 3 and 2. So if we show that n3-nn3-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(n+1)(n-1)n(n+1)(n-1). Now we note that we are multiplying three consecutive numbers together (we are taking nn and then adding 1 or subtracting 1. This gives us the two numbers on either side of nn.) For example, if n=4n=4, then n+1=5n+1=5 and n-1=3n-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 n3-nn3-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 n2+nn2+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 n3-nn3-n is divisible by 6 for all nZnZ.

## 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 ( a x + b ) ( a x - b ) ( a x + b ) ( a x - b )
• 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 ( ( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E ) ( A + B ) ( C + D + E ) = A ( C + D + E ) + B ( C + D + E ) ) helps us to multiply a binomial and a trinomial.
• The sum of cubes is: ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 ( x + y ) ( x 2 - x y + y 2 ) = x 3 + y 3 and the difference of cubes is: x3-y3=(x-y)(x2+xy+y2)x3-y3=(x-y)(x2+xy+y2)
• 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. a2-9a2-9
2. m2-36m2-36
3. 9b2-819b2-81
4. 16b6-25a216b6-25a2
5. m2-(1/9)m2-(1/9)
6. 5-5a2b65-5a2b6
7. 16ba4-81b16ba4-81b
8. a2-10a+25a2-10a+25
9. 16b2+56b+4916b2+56b+49
10. 2a2-12ab+18b22a2-12ab+18b2
11. -4b2-144b8+48b5-4b2-144b8+48b5
2. Factorise completely:
1. (16x4)(16x4)
2. 7x214x+7xy14y7x214x+7xy14y
3. y27y30y27y30
4. 1xx2+x31xx2+x3
5. 3(1p2)+p+13(1p2)+p+1
3. Simplify the following:
1. (a-2)2-a(a+4)(a-2)2-a(a+4)
2. (5a-4b)(25a2+20ab+16b2)(5a-4b)(25a2+20ab+16b2)
3. (2m-3)(4m2+9)(2m+3)(2m-3)(4m2+9)(2m+3)
4. (a+2b-c)(a+2b+c)(a+2b-c)(a+2b+c)
4. Simplify the following:
1. p2-q2p÷p+qp2-pqp2-q2p÷p+qp2-pq
2. 2x+x2-2x32x+x2-2x3
5. Show that (2x-1)2-(x-3)2(2x-1)2-(x-3)2 can be simplified to (x+2)(3x-4)(x+2)(3x-4)

6. What must be added to x2-x+4x2-x+4

to make it equal to (x+2)2(x+2)2

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