1. Special ProductsThe following special products come from multiplying out the brackets. You'll need these often, so it's worth knowing them well.

a(x + y) = ax + ay

(x + y)(x − y) = x2 − y2 (Difference of 2 squares)

(x + y)2 = x2 + 2xy + y2 (Square of a sum)

(x − y)2 = x2 − 2xy + y2 (Square of a difference)

Examples using the special products

Example 1: Multiply out 2x(a − 3)

Answer

This one uses the first product above. We just multiply the term outside the bracket (the "2x") with the term terms inside the brackets (the "a" and the "−3").

2x(a − 3) = 2ax − 6x

Example 2: Multiply (7s + 2t)(7s − 2t)

Answer

We recognize this one involves the Difference of 2 squares:

(7s + 2t)(7s − 2t)

= (7s)2− (2t)2

= 49s2− 4t2

Example 3: Multiply (12 + 5ab)(12 − 5ab)

Answer

(12 + 5ab)(12 − 5ab)

= (12)2 − (5ab)2

= 144 − 25a2b

This one also had an answer that was the difference of 2 squares.

Example 4: Expand (5a + 2b)2

Answer

This one is the square of a sum of 2 terms.

(5a + 2b)2

= (5a)2 + 2(5a)(2b) + (2b)2

= 25a2 + 20ab + 4b2

Example 5: Expand (q − 6)2

Answer

(q − 6)2

= (q)2 − 2(q)(6) + (6)2

= q2 − 12q + 36

This example involved the square of a difference of 2 terms.

Example 6: Expand (8x − y)(3x + 4y)

Answer

This question is not in any of the formats we have above. So we just need to multiply out the brackets, term-by-term.

It's important to recognize when we have a Special Product and when our question is something else.

(8x − y)(3x + 4y)

= 8x(3x + 4y) − y(3x + 4y)

= 8x(3x) + 8x(4y) − y(3x) − y(4y)

= 24x2 + 32xy −3xy − 4y2

= 24x2 + 29xy − 4y2

Example 7: Expand (x + 2 + 3y)2

Answer

To expand this, we put it in the form (a + b)2 and expand it using the 3rd rule above, which says:

(a + b)2 = a2 + 2ab + b2

I put

a = x + 2

b = 3y

This gives me:

(x + 2 + 3y)2

= ([x + 2] + 3y)2 [This is the (a + b)2 step.]

= [x + 2]2 + 2[x + 2](3y) + (3y)2 [Here I apply (a + b)2 = a2 + 2ab + b2]

= [x2 + 4x + 4] + (2x + 4)(3y) + 9y2 [In this row I just expand out the brackets.]

= x2 + 4x + 4 + 6xy + 12y + 9y2 [This is a "tidy up" step.]

I could have chosen the following and obtained the same answer:

a = x

b = 2 + 3y

Try it!

Special Products involving Cubes

The following products are just the result of multiplying out the brackets.

(x + y)3 = x3 + 3x2y + 3xy2 + y3 (Cube of a sum)

(x − y)3 = x3 − 3x2y + 3xy2 − y3 (Cube of a difference)

(x + y)(x2 − xy + y2) = x3 + y3 (Sum of 2 cubes)

(x − y)(x2 + xy + y2) = x3 − y3 (Difference of 2 cubes)

These are also worth knowing well enough so you recognize the form, and the differences between each of them. (Why? Because it's easier than multiplying out the brackets and it helps us solve more complex algebra problems later.)

Example: Expand (2s + 3)3

Answer

This involves the Cube of a Sum:

(2s + 3)3

= (2s)3 + 3(2s)2(3) + 3(2s)(3)2 + (3)3

= 8s3 + 36s2 + 54s + 27

Exercises:

Expand:

(1) (s + 2t)(s − 2t)

Answer

Using the Difference of 2 Squares formula

(x + y)(x − y) = x2 − y2,

we have:

(s + 2t)(s − 2t)

= (s)2− (2t)2

= s2 − 4t2

(2) (i1 + 3)2

Answer

Using the Square of a Sum formula

(x + y)2 = x2 + 2xy + y2,

we have:

(i1 + 3)2

= (i1)2 + 2(i1)(3) + (3)2

= i12 + 6i1 + 9

(3) (3x + 10y)2

Answer

Using the Square of a Sum formula

(x + y)2 = x2 + 2xy + y2,

we have:

(3x + 10y)2

= (3x)2+ (2)(3x)(10y) + (10y)2

= 9x2+ 60xy + 100y2

(4) (3p − 4q)2

Answer

Using the Square of a Difference formula

(x − y)2 = x2 − 2xy + y2,

we have:

(3p − 4q)2

= (3p)2− (2)(3p)(4q) + (4q)2

= 9p2− 24pq + 16q2

2. Common Factor and Difference of SquaresFactoring means writing an expression as the product of its simplest factors.

Example 1: Factoring a number

14 = 7 × 2

[7 and 2 are the simplest factors of 14].

Example 2: Factoring an algebraic expression

3x + 15 = 3(x + 5)

This means that the factors of 3x + 15 are

3 and

(x + 5)

To be able to factor successfully, we need to recognise the formulas from Section 1. So it's a good idea to learn those formulas well!

Factoring Difference of Two Squares

To factor the difference of 2 squares, we just apply the formula given in Section 1 - Special Products in reverse. That is:

x2 − y2 = (x + y)(x − y)

Example 3: Factoring difference of 2 squares

Factor 36s2 − 121t2

Answer

We first recognize it is a difference of 2 squares, then we use the formula given above.

36s2 − 121t2

= (6s)2 − (11t)2

= (6s − 11t)(6s + 11t)

Exercises

Factor the following:

(1) 18p3 − 3p2

Answer

We note that 3p2 divides evenly into the 2 terms in the question. So the factorization is given by:

18p3 − 3p2 = 3p2(6p − 1)

(2) 5a + 10ax − 5ay + 20az

Answer

5a + 10ax − 5ay + 20az = 5a(1 + 2x − y + 4z)

(3) 36a2b 2 − 169c2

Answer

This is a difference of 2 squares.

36a2b2 − 169c2

= (6ab)2 − (13c)2

= (6ab + 13c)(6ab − 13c)

(4) (a − b)2 − 1

Answer

Once again, we recognize this as a difference of 2 squares.

(a − b)2 − 1 = (a − b)2 − 12

Put X = a − b and Y = 1

So (a − b)2 − 1

= X2 − Y2

= (X + Y)(X − Y)

= (a − b + 1)(a − b − 1)

(5) y4 − 81

Answer

y4 − 81

= (y2)2 − (9)2

= (y2 + 9)(y2 − 9)

= (y2 + 9)(y + 3)(y − 3)

(6) r2 − s2 + 2st − t2

Answer

We recognize that this involves 2 differences of two squares. We group it as follows:

r2 − s2 + 2st − t2

= r2 − (s2 − 2st + t2)

We recognize that s2 − 2st + t2 is a square, and equals (s − t)2. So we can factor our expression as follows:

r2 − s2 + 2st − t2 = r2 − (s2 − 2st + t2)

= r2 − (s − t)2 [This is also a difference of 2 squares.]

= [r − (s − t)][r + (s − t)]

= (r − s + t)(r + s − t)

3. Factoring TrinomialsA trinomial is a 3 term polynomial. For example, 5x2 − 2x + 3 is a trinomial.

In many applications in mathematics, we need to solve an equation involving a trinomial. Factoring is an important part of this process. [See the related section: Solving Quadratic Equations.]

Example 1:

Factor x2 − 5x − 6

Answer

Here, we are looking for an answer in the form:

x2 − 5x − 6 = (x...)(x...)

Note that the term we have at the beginning of the question is x2, so we put x in each bracket. This gives us (x)(x) = x2, which is what we need for that first term.

Now we need 2 numbers that multiply to give -6 and add to give -5. The possibilities are:

multiply to give add to give

2 and -3 2 × -3 = -6 2 + -3 = -1 Nope

-2 and 3 -2 × 3 = -6 -2 + 3 = 1 Nope

6 and -1 6 × -1 = -6 6 + -1 = 5 Nope

-6 and 1 -6 × 1 = -6 -6 + 1 = -5 OK

So we have:

x2 − 5x − 6 = (x − 6)(x + 1)

This method of factoring can be tedious because we may need to try several combinations before we hit on the correct numbers (and letters).

After some practice, though, you can spot the most likely combination of letters and numbers. I've shown all the possibilities in the table to give you an idea of what can be involved.

NOTE: Always check your answer by multiplying it out!

Example 2:

Factor 2n2 − 13n − 7

Answer

Once again, we will do it the long way so you can see what is involved.

In this case, we need 2n2 so we start with:

2n2 − 13n − 7 = (2n)(n)

Now we need 2 numbers that multiply to give -7 and the sum of the inner and outer products must be -13. Do not forget the 2 in the first bracket.

Possibility 1: -7 and 1

Multiply to give -7

2n2 − 13n − 7 = (2n − 7)(n + 1)

Outer product plus inner product:

(2n)(1) + (-7)(n) = 2n - 7n = -5n

This is not the correct answer.

Possibility 2: 7 and -1

This would give us

Outer product plus inner product:

(2n)(-1) + (7)(n) = -2n + 7n = 5n

Possibility 3: -1 and 7

2n2 − 13n − 7 = (2n − 1)(n + 7)

Outer product plus inner product:

(2n)(7) + (-1)(n) = 14n - n = 13n

This is still not the correct answer.

Possibility 4: 1 and -7

2n2 − 13n − 7 = (2n + 1)(n − 7)

Outer product plus inner product:

(2n)(-7) + (1)(n) = -14n + n = -13n

This is the required answer, finally.

So

2n2 − 13n − 7 = (2n + 1)(n − 7)

I repeat, it won't always involve several steps like this. The more you do them, the easier and quicker they become (like most new skills).

Of course, after some practice, you will get a better sense of the numbers that will most likely work. It is unlikely that you will have to churn through all the possibilities before you find the right combination, like I have done above.

Now I'll show you a better method, one that reduces a lot of the guesswork.

Factoring by Grouping

This method requires the least amount of guessing and is recommended.

Example 3:

Factor 6x2 + x − 12

Answer

If we were doing it the long way, we would need to consider all the factors of 6:

1 & 6; -1 & -6; 2 & 3; -2 & -3

and also many factors of -12:

1 & -12; 2 & -6; 3 & -4 and all the negatives of these.

We could spend a long time finding the correct combination of factors if we use the long method.

Factoring by Grouping Method

Using grouping method, first we find 2 things, the:

(a) Product of the outer 2 terms of the trinomial (b) Inner number of the trinomial

Then, the only "guess and check" we need to do is to look for 2 numbers whose:

Product is the result of (a) above Sum is the inner term (from (b) above)

So in our 6x2 + x − 12 example, we are looking for 2 terms whose:

(a) Product is 6x2 × -12 = -72x2 (multiply outer numbers)

and whose

(b) Sum is x (inner number).

We try some terms and easily get 9x and -8x.

These are correct since:

(9x)(-8x) = -72x 9x + (-8x) = x

Now write the original expression replacing x with (9x − 8x), as follows:

6x2 + x − 12 = 6x2 + (9x − 8x) − 12

We now re-group the right-hand side:

6x2 + (9x − 8x) − 12 = (6x2 + 9x) − (8x + 12)

Now factor each of the bracketed terms:

(6x2 + 9x) − (8x + 12) = 3x(2x + 3) − 4(2x + 3)

On the right-hand side, we notice that each term in brackets is the same, so we can combine them as follows:

3x(2x + 3) − 4(2x + 3) = (3x − 4)(2x + 3)

[What just happened?

If we have 3xA − 4A, we can factor A out of each term, and write (3x − 4)A. This is how grouping method works. You always end up with brackets that have the same terms inside, and these can be factored out.]

So our answer is:

6x2 + x − 12 = (3x − 4)(2x + 3)

Always check your answer by multiplying it out!

NOTE: Of course, we may need to re-arrange our trinomial to get it into the correct form for grouping method to work. Normally this means we write our polynomial terms in decreasing powers of x.

Example 4:

Let's return to Example 2 from above and do it again, but this time use grouping method.

Factor: 2n2 − 13n − 7

Answer

This time, the product of the outer terms is 2n2 × -7 = -14n2.

The inner term is -13n.

So we are looking for 2 terms whose product is -14n2 and whose sum is -13n.

Those 2 terms are -14n and n.

(This step is nearly always easier to do with grouping method, compared to what we were doing at the top of the page.)

So we write:

2n2 − 13n − 7 = 2n2 − 14n + n − 7

= (2n2 − 14n) + (n − 7)

= 2n(n − 7) + (1)(n − 7)

= (2n + 1)(n − 7)

Exercises

Factorise each of the following:

(1) 3n2 − 20n + 20 [Care with this one!!]

Answer

3n2− 20n + 20

Cannot be further factored!!

When factoring, it's important to know when we cannot go any further. This example looks like we may be able to factor it, but we cannot.

(2) 3x2 + xy − 14y2

Answer

Using grouping method, we have:

Product of the outer 2 terms is (3x2)(−14y2) = −42x2y2

Inner term: xy

So we are looking for 2 terms whose product is −42x2y2 and whose sum is xy.

A little bit of thinking gives us: 7xy and -6xy.

So we can factor our trinomial as follows:

3x2 + xy − 14y2 = 3x2 + 7xy − 6xy − 14y2

= (3x2 + 7xy) − (6xy + 14y2)

= x(3x + 7y) − 2y(3x + 7y)

= (x − 2y)(3x + 7y)

(3) 4r2 + 11rs − 3s2

Answer

Using grouping method, we have:

Product of the outer 2 terms is: (4r2)(−3s2) = −12r2s2

Inner term: 11rs

So we are looking for 2 terms whose product is −12r2s2and whose sum is 11rs.

Those 2 terms will be 12rs and -rs.

So we can factor our trinomial as follows:

4r2 + 11rs − 3s2 = 4r2 + 12rs − rs − 3s2

= (4r2 + 12rs) − (rs + 3s2)

= 4r(r + 3s) − s(r + 3s)

= (4r − s)(r + 3s)

(4) 6x4 − 13x3 + 5x2

Answer

The first step with this example is to factor out x2 from each term:

6x4 − 13x3 + 5x2 = x2(6x2 − 13x+ 5)

Using grouping method on the (6x2 − 13x + 5) part, we have:

Product of the outer 2 terms is: (6x2)(5) = 30x2

Inner term: -13x

So we are looking for 2 terms whose product is 30x2 and whose sum is -13x.

Those 2 terms are -10x and -3x.

So we can factor our trinomial as follows:

6x4 − 13x3 + 5x2  = x2(6x2 − 13x+ 5)

= x2[6x2 − 10x − 3x + 5]

= x2[(6x2 − 10x) − (3x − 5)]

= x2[2x(3x − 5) − (1)(3x − 5)]

= x2[(2x − 1)(3x − 5)]

= x2(2x − 1)(3x − 5)

4. The Sum and Difference of CubesWe came across these before:

[Sum of two cubes]: x3 + y3 = (x + y)(x2 − xy + y2)

[Difference of 2 cubes]: x3 − y3 = (x − y)(x2 + xy + y2)

We can use these to factor expressions involving cubes, as in the following examples.

Example

Factor 64x3 + 125

Answer:

We use the Sum of 2 Cubes formula above.

64x3 + 125

= (4x)3 + (5)3

= (4x + 5)[(4x)2 − (4x)(5) + (5)2]

= (4x + 5)(16x2 − 20x + 25)

Exercises

Factor:

(1) x3 + 27

Answer

Using the Sum of 2 Cubes formula, we obtain:

x3 + 27

= (x)3 + (3)3

= (x + 3)[(x)2 − (x)(3) + (3)2]

= (x + 3)(x2 − 3x + 9)

(2) 3m3 − 81

Answer

Using the Difference of 2 Cubes formula, we obtain:

3m3 − 81

= 3(m3 − 27)

= 3(m3 − (3)3)

= 3(m − 3)[(m)2 + (m)(3) + (3)2]

= 3(m − 3)(m2 + 3m + 9)

5. Equivalent Fractions

Recall the following fraction properties:

This is true because we have multiplied both the top (numerator) and the bottom (denominator) by 5. We say 3/4 and 15/20 are equivalent fractions.

This is true because we have divided both the numerator and the denominator by 7. We say 7/21 and 1/3 are equivalent fractions.

We now apply these ideas to fractions involving algebraic expressions.

Example 1

Divide the numerator and the denominator of by 3ab2.

Answer:

NOTE: This answer is not in simplest form. We could divide top and bottom again by a2.

KNOW WHEN TO STOP!

The following expression cannot be simplified further because there is an addition sign in the numerator and a subtraction in the denominator:

We cannot cancel the x and the x2.

However, if the terms in the numerator and denominator are multiplied, then we can do further simplifying like this:

Example 2

Reduce to simplest form:

Answer:

We start by factoring the numerator and then observe we can divide top and bottom by one of the factors:

Example 3

Reduce to simplest form:

Answer

We need to factor both numerator and denominator:

At this point we can only cancel the [leading] 2 and 4:

Now, we recognize that the numerator is the difference of 2 squares:

Now, since the terms in brackets are connected by multiplication, we can cancel the (x + 2) from top and bottom:

Exercises

Simplify:

(1)

Answer

(2)

Answer

We factor the denominator (bottom) of the fraction first:

(3)

Answer

We cannot simplify this any further!

(4)

Answer

We first factor the numerator (top) then cancel.

On the first line of the solution, I am changing the order of the (y − x) into (x − y) by placing a minus in front.

This is because

−(x − y)

= −x − (-y)

= −x + y

= y − x

6. Multiplication and Division of FractionsRecall the following fraction facts:

When multiplying by a fraction, multiply numerators and multiply denominators:

If you can, simplify first :

(I canceled the 13 & 39 to give 1/3 and the 12 with the 24 to give 1/2.)

When dividing by a fraction, invert and multiply:

(I multiplied by the inverse of 2/7, which is 7/2)

When we do the same things with algebraic expressions, remember to SIMPLIFY FIRST, so that the problem is easy to perform

Example 1

Simplify

Answer;

Simplifying first, we cancel the 11 in the first fraction with the 33 on the bottom of the second fraction:

Example 2

Simplfy:

Answer

Simplifying:

The first step in this problem involved expanding out the (ax)2 on the top of the second fraction to give a2x2.

Also, because it is multiplication, we can write it as one fraction (multiply tops, multiply bottoms).

Next, we cancel the 18 on top with the 3 on bottom (giving 6 on top).

Also we can cancel the s on top and bottom.

The a2 on top cancels with the a on bottom to give a on top.

The x2 on top cancels with the x2 on bottom to give 1.

We then multiply out what is left to give the final answer.

Example 3

Simplify:

Answer

First, we invert the st/4 term to give 4/st and then multiply by that inverse:

To get from the first to the second row, I cancelled out the two s's on the top and bottom; and also cancelled the 4 on top with the 2 on bottom to give 2 on top.

Exercises

Simplify:

(1)

Answer

This one involves inverting the 25/13 to give 13/25 and then multiplying by 13/25.

Then we cancel the 5 on top with the 25 on bottom, to give 1/5.

To get the last line is simply:

1 × 13 = 13 on top; and

16 × 5 = 80 on bottom

(2)

Answer

We factor the (9x2 − 16) into (3x + 4)(3x − 4) using the Difference of Squares that we learned before.

Dividing by (4 − 3x) is the same as multiplying by 1/(4 − 3x). (To see why, think about this example: dividing by 2 is the same as muiltiplying by 1/2.)

The second line involves the following useful trick:

(4 − 3x) = −(3x − 4)

(To see why this works, just multiply out the right hand side.)

After cancelling, we are left with a factor of (−1) from the cancelled fraction and this negative is placed out the front for convenience.

(3)

Answer

The first line in this question involves factoring everything on the top and bottom of the fractions.

Now, for the top of the first fraction:

2x2 − 18 = 2(x2 − 9)

We recognise the expression in brackets to be a difference of 2 squares, which we learned about before. We can write:

x2 − 9 = (x + 3)(x − 3)

The bottom of the first fraction also uses difference of 2 squares:

x3 − 25x = x(x2 − 25) = x(x + 5)(x −5)

The top of the second fraction requires taking the common factor of 3 outside:

3x − 15 = 3(x − 5)

The bottom of the second fraction has 2x as a common factor :

2x2 + 6x = 2x(x + 3)

The third line involves cancelling out the following from top and bottom:

2(x + 3)(x − 5)

The final step is to multiply tops and multiply bottoms, since we cannot cancel anything else.

7. Addition and Subtraction of FractionsRecall that to add or subtract fractions, we need to have the same denominator.

Example 1

Here, the lowest common denominator which we can use is 15. So we have:

Subtraction works in the same way, as we see in the next example.

Example 2

Answer

Our lowest common denominator this time is 14. So we have:

When we have algebra using fractions, use the same rules.

Example 3

Answer

The lowest common denominator here will be 6y4.

So we have:

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Example 4

Answer

We factor the first denominator to get an idea of what to do:

We can see now that if we multiply the second fraction by 2x2, we will have a common denominator:

We would normally write this as:

Exercises

Simplify the following.

(1)

Answer

(2)

Answer

The lowest common denominator is 4.

(3)

Answer

We have factored out the 3 at the bottom of the first fraction and the 4 at the bottom of the second fraction.

In the 3rd line, we find the lowest common denominator, 12(2y + 1), and multiply top and bottom of the 2 fractions accordingly.

The last line is a tidy up step.

8. Equations Involving FractionsIn this section, we can find the solution easily by multiplying throughout by the lowest common denominator.

Example 1

An aquarium can be filled by one hose in 7 minutes and a second thinner hose in 10 minutes. How long will it take to fill the tank if both hoses operate together?

Answer

Since each hose makes the filling time less, we have to add the reciprocals together and take the reciprocal of the result.

We need to use:

So we have:

So

So it will take 4.1 minutes to fill the tank with both hoses operating together.

Example 2

For 2 resistors with resistances R1and R2 in parallel, the combined resistance R is given by:

For a particular circuit, the combined resistance R was found to be 4 ohms (Ω), and R1= 10 Ω. Find R2.

Answer

We have:

We need to multiply throughout by the lowest common denominator: 20R2

Example 3

A car averaged 30 km/h going from home to work and 40 km/h on the return journey. If the total time for the two journeys is 50 minutes, how far is it from home to work?

Answer

Let the length of the journey from home to work be x km.

Recall that speed

So time .

We must use the same time units throughout. We will use hours.

Now 50 minutes hours

In the forward journey, the car's time was hours.

For the return journey, the time was hours.

The total time was hours.

So

This gives us

That is

So the distance from home to work is 14.3 km.

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