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2.6 E) Factorising Quadratics (A ≠ 1) – Part 2
2.6 E) Factorising Quadratics (A ≠ 1) – Part 2
The content in this section builds on the content that was discussed in the previous section. Make sure that you have gone through the content in the previous section before working through this section (click here to be taken through to the previous section).
Example 1
Factorise the quadratic expression below:
Factorise the quadratic expression below:
The first step to factorise the quadratic expression is to multiply a (the coefficient of x2) by c (the constant term; the number at the end). a in the expression is 6 and c is -3. When we multiply these two together we get -18.
We are now looking for two numbers that add together to give -7 (the value of b) and multiply together to give -18 (a x c). Here are the pairs of numbers that multiply to give -18:
The only pair out of the above pairs that add to give -7 are -9 and 2. The whole point of finding this pair of numbers is so that we can replace -7x in the expression with -9x + 2x (which still gives -7x). It does not matter which way we sub -9x + 2x into the expression; we could have subbed it in as 2x – 9x (which I will do later on). The expression becomes:
The next step is to factorise the first two terms and the last two terms separately. When we are factorising the first two terms and the last two terms, we should get the same factors inside the bracket. There are no common factors in the last two terms, so we just factorise out 1.
As we can see from the expression, both of the brackets are 2x – 3.
One of the brackets in the final factorised expression will be the common factor (2x – 3) and the other bracket will be the terms that are outside each of the brackets.
One of the brackets in the final factorised expression will be the common factor (2x – 3) and the other bracket will be the terms that are outside each of the brackets.
We can multiply the brackets out to check that we get the expression that we have factorised; I am not going to do this, but feel free to expand and check.
Earlier I said that it does not matter the order that you sub the x values into the expression. This is because you will obtain the same factorised answer; just with the brackets in a different order. Earlier we replaced -7x with -9x + 2x. Let’s now replace -7x with 2x – 9x.
Earlier I said that it does not matter the order that you sub the x values into the expression. This is because you will obtain the same factorised answer; just with the brackets in a different order. Earlier we replaced -7x with -9x + 2x. Let’s now replace -7x with 2x – 9x.
Like before, we factorise the first two terms and the final two terms separately. The brackets that we get should be the same. The first bracket is (3x + 1) and in order for the second bracket to be (3x + 1), we need to factorise the last two terms by -3.
One of the brackets in the final factorised expression will be the common factor of (3x + 1). The other bracket is the terms that is outside each of the brackets.
This is exactly the same brackets as when we subbed in -7x as -9x + 2x. Therefore, you can see that it does not matter the order in which you sub the x values into the expression.
Example 2
Factorise the expression below:
Factorise the expression below:
The value for a is 10, b is -11 and c is -6. The first step to factorise this expression is to multiply a by c, which is -60 (10 x -6). We are now looking for a pair of numbers that multiplies together to give -60 (a x c) and adds together to give -11 (the value of b).
Here are the pairs of numbers that multiply to give -60:
The only pair that adds to give -11 is -15 and 4. Therefore, we replace -11x in the expression with -15x + 4x (it does not matter the way that we sub -15x + 4x into the expression; we could have subbed it in as 4x – 15x).
- 60 and -1
- -60 and 1
- 30 and -2
- -30 and 2
- 20 and -3
- -20 and 3
- 15 and -4
- -15 an 4
- 12 and -5
- -12 and 5
- 10 and -6
- -10 and 6
The only pair that adds to give -11 is -15 and 4. Therefore, we replace -11x in the expression with -15x + 4x (it does not matter the way that we sub -15x + 4x into the expression; we could have subbed it in as 4x – 15x).
The next step is to factorise the first two terms and final two terms separately.
The first bracket in the final factorise expression will be the common factor of (2x – 3). The second bracket is the terms that are outside each of the brackets.
Final Note
This technique is quite tricky, so it may be worth going through the examples in this section and the section before again before giving the quiz a go.
This technique is quite tricky, so it may be worth going through the examples in this section and the section before again before giving the quiz a go.