So in math today my teacher introduced us to quadratic expressions where there is no common factor. For example 8x² + 10x - 7
So here is my method that I would like to share (My teacher tried to teach me differently and I didn't understand so I tested things out and found that this worked!
1. Multiple term 1 by term 3
e.g. 8 x 7 = 56
2. Find the factors of this number
e.g. Factors of 56:
- 56 x 1
- 28 x 2
- 14 x 4
3. Find which factors add or subtract to equal term 2
e.g. Which factor equals 10? In this case 14 - 4 = 10
4a. Now you must divide ONE of these factors by term 1, and get a whole number.
e.g. 14 divided by 8 or 4 divided by 8. Seeing none of these numbers would produce a whole number you must do an extra step
4b. If you do not get a whole number, you must split term 1 by its factors, and then find factors that divide into your factors from step 3.
The factors of 8 (Term one) are
- 1 x 8
- 2 x 4
Seeing neither 14 or 4 can be divided by 8 and produce a whole number, you must use 2 x 4.
This will give you 14 ÷ 2 and 4 ÷ 4. This leaves you with 7 and 1.
5. You now have the necessary numbers for the brackets. Use the factors found in 4b
(4x )(2x )
and then use the numbers that were created by dividing in 4b
(4x 7)(2x 1)
6. Now decide what signs will be in the bracket. To do this, guess and check to see which negative and positive/negative numbers add or subtract to equal the middle term
e.g. Which of the combinations will equal 10?
-14 + 4 = -10
-14 - 4 = -18
14 + 4 = 18
14 - 4 = 10
The correct combination is therefore +14 and - 4
Which means that the final factorised product is (4x + 7)(2x - 1)
So here is my method that I would like to share (My teacher tried to teach me differently and I didn't understand so I tested things out and found that this worked!
1. Multiple term 1 by term 3
e.g. 8 x 7 = 56
2. Find the factors of this number
e.g. Factors of 56:
- 56 x 1
- 28 x 2
- 14 x 4
3. Find which factors add or subtract to equal term 2
e.g. Which factor equals 10? In this case 14 - 4 = 10
4a. Now you must divide ONE of these factors by term 1, and get a whole number.
e.g. 14 divided by 8 or 4 divided by 8. Seeing none of these numbers would produce a whole number you must do an extra step
4b. If you do not get a whole number, you must split term 1 by its factors, and then find factors that divide into your factors from step 3.
The factors of 8 (Term one) are
- 1 x 8
- 2 x 4
Seeing neither 14 or 4 can be divided by 8 and produce a whole number, you must use 2 x 4.
This will give you 14 ÷ 2 and 4 ÷ 4. This leaves you with 7 and 1.
5. You now have the necessary numbers for the brackets. Use the factors found in 4b
(4x )(2x )
and then use the numbers that were created by dividing in 4b
(4x 7)(2x 1)
6. Now decide what signs will be in the bracket. To do this, guess and check to see which negative and positive/negative numbers add or subtract to equal the middle term
e.g. Which of the combinations will equal 10?
-14 + 4 = -10
-14 - 4 = -18
14 + 4 = 18
14 - 4 = 10
The correct combination is therefore +14 and - 4
Which means that the final factorised product is (4x + 7)(2x - 1)
-
I nearly lost the wil to live halfway through your method which is a pity because at one point you were very close to the answer.
I have reproduced part of your answer below:
For 8x² + 10x - 7
1. Multiple term 1 by term 3
e.g. 8 x 7 = 56
2. Find the factors of this number
e.g. Factors of 56:
- 56 x 1
- 28 x 2
- 14 x 4
3. Find which factors add or subtract to equal term 2
e.g. Which factor equals 10? In this case 14 - 4 = 10
Now replace the middle term by 14x and -4x.
8x² + 10x - 7 = 8x² + 14x - -4x - 7
Now factorise in groups of 2
8x² + 10x - 7 = 2x(4x + 7) - (4x + 7)
Because you selected the correct two numbers the brackets itself is a common factor.
8x² + 10x - 7 = (4x + 7)(2x - 1)
The bad news is that you have not discovered a new method. What I have done above is a well-known method used by proper maths teachers
I have reproduced part of your answer below:
For 8x² + 10x - 7
1. Multiple term 1 by term 3
e.g. 8 x 7 = 56
2. Find the factors of this number
e.g. Factors of 56:
- 56 x 1
- 28 x 2
- 14 x 4
3. Find which factors add or subtract to equal term 2
e.g. Which factor equals 10? In this case 14 - 4 = 10
Now replace the middle term by 14x and -4x.
8x² + 10x - 7 = 8x² + 14x - -4x - 7
Now factorise in groups of 2
8x² + 10x - 7 = 2x(4x + 7) - (4x + 7)
Because you selected the correct two numbers the brackets itself is a common factor.
8x² + 10x - 7 = (4x + 7)(2x - 1)
The bad news is that you have not discovered a new method. What I have done above is a well-known method used by proper maths teachers