1. Special Products The 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) = x 2 − y 2 (Difference of 2 squares) (x + y) 2 = x 2 + 2xy + y 2 (Square of a sum) (x − y) 2 = x 2 − 2xy + y 2 (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 = 49s 2 − 4t 2 Example 3: Multiply (12 + 5ab)(12 − 5ab) Answer
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1. Special ProductsThe following special products come from multiplying out the brackets. You'll need these often, so it's worth knowing them well.
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:
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.)
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.]
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.
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:
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.
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.
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.
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.
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.