Can someone please explain factoring quadratic equations?
I'm having trouble understanding the process could someone please explain step by step?
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Could you give us an example of the sort of question that you're struggling with? Is it why we do it, how to do it? What sort of thing?
Reply 2
19
Original post by EeveeSAMA
I'm having trouble understanding the process could someone please explain step by step?
Factorising* first of all
I'm not sure what your level is but I'll explain it as simply as I can. A quadratic equation is a curve that can be mapped on a graph. Factorising a quadratic equation would give you the roots (where the equation meets the x axis). For example: let's take the quadratic equation x2 + 5x + 6=0 From this equation, you are trying to find the double bracket form (the factorised form). For this, it would be (x+3)(x+2)=0 You'd find that expanding (x+3)(x+2) would give x2 + 5x + 6
The solution (root/x intercept) of the quadratic equation is where x+3=0 and x+2=0, so x=-3 and x=-2.
Did this help or you need a more specific explanation?
Reply 3
12
Original post by EeveeSAMA
I'm having trouble understanding the process could someone please explain step by step?
Easiest thing to do is what numbers multiples to give the last number (without x) and what numbers add to give the number with the x! If you send an example I could probably explain in depth!
Reply 4
Original post by lildishh
Easiest thing to do is what numbers multiples to give the last number (without x) and what numbers add to give the number with the x! If you send an example I could probably explain in depth!
Doesn't work in all cases ... depends on coeff of x^2.
Reply 5
Original post by Muttley79
Doesn't work in all cases ... depends on coeff of x^2.
Here is how to extend @lildishh's method to work for any case:
Let's use 3x² – 11x + 6 as an example.
It is in the form ax² + bx + c
1) Find two numbers that multiply to make 18 (a*c). 1,18 2,9
2) I want the numbers to add to make b. -2 and -9 are the numbers that I want, as they add to make -11, and multiply to make 18.
3) This is the weird bit. Put the numbers you got (in this case -2 and -9) over a (in this case 3.) So -2/3, and -9/3
4) You need to simplify where possible. So -2/3 stays as it is, -9/3 becomes -3/1.
5) The denominator becomes the coefficient of x, while the numerator is the number you add/subtract.
6) So it becomes (3x-2)(x-3)
This method always works (except maybe if you have huge numbers as coefficients), and it carried me through my GCSE's, onto A-level maths. 😀
One more thing is that if there is a common factor between a b and c, you need to factorise it out and multiply by it at the end. (But I think that you need to do that for all methods???)
(edited 8 months ago)
Reply 6
21
Original post by CuriousGeorge
Let's use 3x² – 11x + 6 as an example.
It is in the form ax² + bx + c
1) Find two numbers that multiply to make 18 (a*c). 1,18 2,9
2) I want the numbers to add to make b. -2 and -9 are the numbers that I want, as they add to make -11, and multiply to make 18.
3) This is the weird bit. Put the numbers you got (in this case -2 and -9) over a (in this case 3.) So -2/3, and -9/3
4) You need to simplify where possible. So -2/3 stays as it is, -9/3 becomes -3/1.
5) The denominator becomes the coefficient of x, while the numerator is the number you add/subtract.
6) So it becomes (3x-2)(x-3)
This method always works (except maybe if you have huge numbers as coefficients), and it carried me through my GCSE's, onto A-level maths. 😀
One more thing is that if there is a common factor between a b and c, you need to factorise it out and multiply by it at the end. (But I think that you need to do that for all methods???)
It is in the form ax² + bx + c
1) Find two numbers that multiply to make 18 (a*c). 1,18 2,9
2) I want the numbers to add to make b. -2 and -9 are the numbers that I want, as they add to make -11, and multiply to make 18.
3) This is the weird bit. Put the numbers you got (in this case -2 and -9) over a (in this case 3.) So -2/3, and -9/3
4) You need to simplify where possible. So -2/3 stays as it is, -9/3 becomes -3/1.
5) The denominator becomes the coefficient of x, while the numerator is the number you add/subtract.
6) So it becomes (3x-2)(x-3)
This method always works (except maybe if you have huge numbers as coefficients), and it carried me through my GCSE's, onto A-level maths. 😀
One more thing is that if there is a common factor between a b and c, you need to factorise it out and multiply by it at the end. (But I think that you need to do that for all methods???)
just for info, do you understand why the "weird bit" works?
Reply 7
Original post by mqb2766
just for info, do you understand why the "weird bit" works?
I have no idea 😆
Reply 8
21
Original post by CuriousGeorge
I have no idea 😆
It goes back to the babylonians, so for your example
3x² – 11x + 6
we want to transform it into a standard quadratic with unity squared coefficient. So consider the new expression whch is 3 times that so
(3x)² – 11(3x) + 18
and sub z=3x we get the quadratic
z² – 11z + 18
which can be factorized normally as
(z-2)(z-9) = (3x-2)(3x-9)
we have to divide that by 3 (as we multiplied the expression by 3 at the start) and we can pick either factor to divide by 3, but the second is neater as both terms are multiples of 3 so
(3x-2)(x-3)
So its the old maths trick of transforming the problem into one which you can solve.
(edited 8 months ago)
Reply 9
Original post by CuriousGeorge
Let's use 3x² – 11x + 6 as an example.
It is in the form ax² + bx + c
1) Find two numbers that multiply to make 18 (a*c). 1,18 2,9
2) I want the numbers to add to make b. -2 and -9 are the numbers that I want, as they add to make -11, and multiply to make 18.
3) This is the weird bit. Put the numbers you got (in this case -2 and -9) over a (in this case 3.) So -2/3, and -9/3
4) You need to simplify where possible. So -2/3 stays as it is, -9/3 becomes -3/1.
5) The denominator becomes the coefficient of x, while the numerator is the number you add/subtract.
6) So it becomes (3x-2)(x-3)
This method always works (except maybe if you have huge numbers as coefficients), and it carried me through my GCSE's, onto A-level maths. 😀
One more thing is that if there is a common factor between a b and c, you need to factorise it out and multiply by it at the end. (But I think that you need to do that for all methods???)
It is in the form ax² + bx + c
1) Find two numbers that multiply to make 18 (a*c). 1,18 2,9
2) I want the numbers to add to make b. -2 and -9 are the numbers that I want, as they add to make -11, and multiply to make 18.
3) This is the weird bit. Put the numbers you got (in this case -2 and -9) over a (in this case 3.) So -2/3, and -9/3
4) You need to simplify where possible. So -2/3 stays as it is, -9/3 becomes -3/1.
5) The denominator becomes the coefficient of x, while the numerator is the number you add/subtract.
6) So it becomes (3x-2)(x-3)
This method always works (except maybe if you have huge numbers as coefficients), and it carried me through my GCSE's, onto A-level maths. 😀
One more thing is that if there is a common factor between a b and c, you need to factorise it out and multiply by it at the end. (But I think that you need to do that for all methods???)
You are changing your method at point 3 and so invalidate your previous post.
Reply 10
14
Original post by Muttley79
You are changing your method at point 3 and so invalidate your previous post.
Just a little change. What is your solution to the problem where the coefficient of x^2 is higher than 1?
Reply 11
6
Original post by sdfj
Factorising* first of all
I'm not sure what your level is but I'll explain it as simply as I can. A quadratic equation is a curve that can be mapped on a graph. Factorising a quadratic equation would give you the roots (where the equation meets the x axis). For example: let's take the quadratic equation x2 + 5x + 6=0 From this equation, you are trying to find the double bracket form (the factorised form). For this, it would be (x+3)(x+2)=0 You'd find that expanding (x+3)(x+2) would give x2 + 5x + 6
The solution (root/x intercept) of the quadratic equation is where x+3=0 and x+2=0, so x=-3 and x=-2.
Did this help or you need a more specific explanation?
I'm not sure what your level is but I'll explain it as simply as I can. A quadratic equation is a curve that can be mapped on a graph. Factorising a quadratic equation would give you the roots (where the equation meets the x axis). For example: let's take the quadratic equation x2 + 5x + 6=0 From this equation, you are trying to find the double bracket form (the factorised form). For this, it would be (x+3)(x+2)=0 You'd find that expanding (x+3)(x+2) would give x2 + 5x + 6
The solution (root/x intercept) of the quadratic equation is where x+3=0 and x+2=0, so x=-3 and x=-2.
Did this help or you need a more specific explanation?
That's a beautiful explanation! Thank you! I'm in grade 11, mixed, I live in Canada (incase our education systems are different). Could you please help me by walking through this problem?: 4x2 - 16x + 15 (Complex Trinomials (when a = 1))
Reply 12
6
Original post by sdfj
Factorising* first of all
I'm not sure what your level is but I'll explain it as simply as I can. A quadratic equation is a curve that can be mapped on a graph. Factorising a quadratic equation would give you the roots (where the equation meets the x axis). For example: let's take the quadratic equation x2 + 5x + 6=0 From this equation, you are trying to find the double bracket form (the factorised form). For this, it would be (x+3)(x+2)=0 You'd find that expanding (x+3)(x+2) would give x2 + 5x + 6
The solution (root/x intercept) of the quadratic equation is where x+3=0 and x+2=0, so x=-3 and x=-2.
Did this help or you need a more specific explanation?
I'm not sure what your level is but I'll explain it as simply as I can. A quadratic equation is a curve that can be mapped on a graph. Factorising a quadratic equation would give you the roots (where the equation meets the x axis). For example: let's take the quadratic equation x2 + 5x + 6=0 From this equation, you are trying to find the double bracket form (the factorised form). For this, it would be (x+3)(x+2)=0 You'd find that expanding (x+3)(x+2) would give x2 + 5x + 6
The solution (root/x intercept) of the quadratic equation is where x+3=0 and x+2=0, so x=-3 and x=-2.
Did this help or you need a more specific explanation?
Reply 13
19
Original post by EeveeSAMA
That's a beautiful explanation! Thank you! I'm in grade 11, mixed, I live in Canada (incase our education systems are different). Could you please help me by walking through this problem?: 4x2 - 16x + 15 (Complex Trinomials (when a = 1))
Hiya, no problem I'll do my best. I think grade 11 is my age in Canada but not sure (UK based here)?
I can solve this problem. My one query is that I don't think a = 1 because the way I learnt it, a is the coefficient (number) in front of the x2. Or do you mean something different by a? I'll show you how I solve this equation as an example (where the coefficient of x is greater than 1).
Reply 14
Original post by EeveeSAMA
That's a beautiful explanation! Thank you! I'm in grade 11, mixed, I live in Canada (incase our education systems are different). Could you please help me by walking through this problem?: 4x2 - 16x + 15 (Complex Trinomials (when a = 1))
This ia strictly not allowed by TSR
Reply 15
Original post by sdfj
Hiya, no problem I'll do my best. I think grade 11 is my age in Canada but not sure (UK based here)?
I can solve this problem. My one query is that I don't think a = 1 because the way I learnt it, a is the coefficient (number) in front of the x2. Or do you mean something different by a? I'll show you how I solve this equation as an example (where the coefficient of x is greater than 1).
I can solve this problem. My one query is that I don't think a = 1 because the way I learnt it, a is the coefficient (number) in front of the x2. Or do you mean something different by a? I'll show you how I solve this equation as an example (where the coefficient of x is greater than 1).
You will be breaking the rules if you post a solution - do not do this
Reply 16
19
Original post by Muttley79
This ia strictly not allowed by TSR
Original post by Muttley79
You will be breaking the rules if you post a solution - do not do this
I am aware of the community guidelines and won't give a full solution or the final answer. I will, however, help this user with the initial steps.
Reply 17
Original post by sdfj
I am aware of the community guidelines and won't give a full solution or the final answer. I will, however, help this user with the initial steps.
Just give hints as the rules require ...
Reply 18
6
Original post by Muttley79
You will be breaking the rules if you post a solution - do not do this
Hey there, all I'm wanting are clear steps to get to a solution. That's all. I'm by no means expecting a full solution. I'm a person who needs clear and concise steps to get to a solution. I'm sure we all here are aware of the rules of the community, I think we need to allow people the chance to respond, maybe they won't give the answer. If they do, remind them respectfully that that is not within the community guidelines. Do that before you call someone out please. I'd appreciate no arguing in this thread as school is stressful enough, I'm just here to get help to understand the process as clearly as possible so I can tackle my tougher questions. Please do not comment on my thread if your here just to say someone is wrong and not provide a rebuttal in a kind way or if you do not give one at all. If you think someone's "steps to the solution" is not quite right, you can for sure share your own thoughts, but again respectfully please. Thank you all for your time. I really appreciate it! PS I know your just trying to help. I also know you have provided steps and I am very grateful, I just wanted to take the time to gently say this
.(edited 8 months ago)
Reply 19
19
Original post by EeveeSAMA
That's a beautiful explanation! Thank you! I'm in grade 11, mixed, I live in Canada (incase our education systems are different). Could you please help me by walking through this problem?: 4x2 - 16x + 15 (Complex Trinomials (when a = 1))
I'm afraid, I cannot provide you with a full answer but I'm happy to help you understand the initial steps and guide you through the question. Taking inspiration from the initial example I gave, you need to find the bracketed form of 4x2 - 16x + 15 = 0
Now I'm going to use the letters a and b to represent the numbers you are trying to find. The unexpanded form of 4x2 - 16x + 15 will either be in the form of (4x + a)(x + b) or (2x + a)(2x + b). This is because you need the number in front of both the x in the brackets to multiply to equal four. That's how you get the 4x2 in the original equation.
For the purposes of trying to help you, you need to use the (2x + a)(2x + b)=0 version to solve this equation. You now need to find the values of a and b which expands to form 4x2 - 16x + 15=0 Using trial and error is fine for now (tip is to try values of a and b that multiply together to get 15 - you should know why), but for me, I can solve that in 15 seconds because of practice - the method feels a lot more intuitive very quickly. The values of a and b may or may not be negative.
You should then be able to find the solution/root/value of x.
Tell me whether you get it, or if not, how far you get. The important thing is to think of a and b as numbers!
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