In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?

In the standard form of quadratic equations, there are three parts to it: ax^2 + bx + c where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant. The -4 at the end of the equation is the constant. This hopefully answers your last question. Now, your first question. So the problem is that you need to find two numbers (a and b) such that the sum of a and b equals 11 and the product equals -12. Correct? Correct indeed. I use a pretty straightforward mental method but I'll introduce my teacher's method of factors first. What you need to do is find all the factors of -12 that are integers. Let's start with 1. 1 * -12 2 * -6 3 * -4 4 * -3 6 * -2 12 * -1 Here we see 6 factor pairs or 12 factors of -12. Let's see which one adds up to 11. It seems like `12 + (-1) = 11`. So we'll split 11 to 12 and -1. My other method is straight out recognising the middle terms. This works well with small numbers. I can clearly see that 12 is close to 11 and all I need is a change of 1. So that leaves out 12 * -1 and -12 * 1. I can see that -12 * 1 makes -11 which is not what I want so I go with 12 * -1. If you misunderstand something I said, just post a comment.

Sometimes I don't understand some of the problems. :(

Just always remember to simplify the expression before u do anything and then make one side equal to zero my subtracting or adding etc. Its easy once u get the hang of it.

Think about what is confusing you. 1) Is it the factoring? If it is, you need to review the lessons and practice problems for factoring. There are multiple factoring techniques. You need to know each of them and when to apply them. 2) If you are ok with the factoring, then what pay close attention to the other topics on the page. They cover the other steps you need to do in quite some detail.

In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?

Yes, the -12 comes from multiplying -4 with the 3 from 3x^2.

How do you factor it when the leading coefficient is more than 1? For example, something like 3xsquared+ x-2.

You factor the trinomial by grouping. See the lessons at this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2 Hope this helps.

My equation is x^2-2x+1=0 ( find x) How do i factor out the 1? Could i make it 1^2?

You need 2 factors of 1 that add to -2. Your choices are to use: 1*1 or (-1)(-1). Which pair adds to -2? Hope this helps.

In the last two problems' explanations there seems to be a mistake. Why when they say "divide by 3" in the second step does the equation look like "x^2+11x+10 = 0" instead of 3(x^2+11x+10) = 0" The way they wrote it, the right hand side of the equation would equal 0/3, wouldn't it?

They can divide the entire equation by 3. That's ok. Remember: 0/3 = 0, which is why you see just o.

I don't get 3x^2-9x-20=x^2+5x+16. What do we do in the first place?

Before trying to factor, you need to put the equation in the standard form: Ax^2+Bx+C=0. To do this, use opposite operations to move each term on the right side to the left side. Then factor. Hope this helps.

Hi! This is Rowel! I have a question about x^2+5x=0. Won't (x+5)(x+0) work also, if we made the original equation into x^2+5x+0=0? Thank you for looking at my question.

x+0 = x. There is no value added by changing x into x+0. Keep it simple. The factors are x(x+5).

I know this is not related to the above questions, but i could not find where to ask normal day to day questions. If someone could please help me answer this would be great.: Factorise the expression as far as possible to prime factors: (x-1)^2-9(y-1)^2 ... ^2(power of 2)

Expand the squares, then distribute the 9. Combine like terms then factor. Post comment if you still feel stuck.

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Algebra 1

Course: Algebra 1 > Unit 14

Lesson 5: Solving quadratics by factoring

Solving quadratics by factoring

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Learn how to solve quadratic equations like (x-1)(x+3)=0 and how to use factorization to solve other forms of equations.

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Amber Lynn
Posted 8 years ago. Direct link to Amber Lynn's post “In the above equation 3x^...”
In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
Button navigates to signup pageComment on Amber Lynn's post “In the above equation 3x^...”
(47 votes)
Answer
- Johnathan
  Posted 8 years ago. Direct link to Johnathan's post “In the standard form of q...”
  In the standard form of quadratic equations, there are three parts to it: ax^2 + bx + c where a is the coefficient of the quadratic term, b is the coefficient of the linear term, and c is the constant. The -4 at the end of the equation is the constant. This hopefully answers your last question. Now, your first question.
  
  So the problem is that you need to find two numbers (a and b) such that the sum of a and b equals 11 and the product equals -12. Correct? Correct indeed. I use a pretty straightforward mental method but I'll introduce my teacher's method of factors first. What you need to do is find all the factors of -12 that are integers. Let's start with 1.
  1 * -12
  2 * -6
  3 * -4
  4 * -3
  6 * -2
  12 * -1
  Here we see 6 factor pairs or 12 factors of -12. Let's see which one adds up to 11. It seems like 12 + (-1) = 11. So we'll split 11 to 12 and -1.
  My other method is straight out recognising the middle terms. This works well with small numbers. I can clearly see that 12 is close to 11 and all I need is a change of 1. So that leaves out 12 * -1 and -12 * 1. I can see that -12 * 1 makes -11 which is not what I want so I go with 12 * -1.
  
  If you misunderstand something I said, just post a comment.
  Comment on Johnathan's post “In the standard form of q...”
  (30 votes)
Jose Santizo
Posted 8 years ago. Direct link to Jose Santizo's post “Sometimes I don't underst...”
Sometimes I don't understand some of the problems. :(
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(17 votes)
Answer
- ChrisTry162
  Posted 8 years ago. Direct link to ChrisTry162's post “Just always remember to s...”
  Just always remember to simplify the expression before u do anything and then make one side equal to zero my subtracting or adding etc. Its easy once u get the hang of it.
  Comment on ChrisTry162's post “Just always remember to s...”
  (19 votes)
Clark Burch
Posted 5 months ago. Direct link to Clark Burch's post “still confused”
still confused
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(8 votes)
Answer
- Kim Seidel
  Posted 5 months ago. Direct link to Kim Seidel's post “Think about what is confu...”
  Think about what is confusing you.
  1) Is it the factoring? If it is, you need to review the lessons and practice problems for factoring. There are multiple factoring techniques. You need to know each of them and when to apply them.
  
  2) If you are ok with the factoring, then what pay close attention to the other topics on the page. They cover the other steps you need to do in quite some detail.
  Comment on Kim Seidel's post “Think about what is confu...”
  (10 votes)
pascal5
Posted 7 years ago. Direct link to pascal5's post “In the last two problems'...”
In the last two problems' explanations there seems to be a mistake. Why when they say "divide by 3" in the second step does the equation look like
"x^2+11x+10 = 0"
instead of
3(x^2+11x+10) = 0"

The way they wrote it, the right hand side of the equation would equal 0/3, wouldn't it?
Button navigates to signup pageComment on pascal5's post “In the last two problems'...”
(3 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “They can divide the entir...”
  They can divide the entire equation by 3. That's ok. Remember: 0/3 = 0, which is why you see just o.
  Button navigates to signup page
  (12 votes)
Cooky_0901
Posted 5 months ago. Direct link to Cooky_0901's post “I don't get 3x^2-9x-20=x^...”
I don't get 3x^2-9x-20=x^2+5x+16. What do we do in the first place?
Button navigates to signup pageComment on Cooky_0901's post “I don't get 3x^2-9x-20=x^...”
(3 votes)
Answer
- Kim Seidel
  Posted 5 months ago. Direct link to Kim Seidel's post “Before trying to factor, ...”
  Before trying to factor, you need to put the equation in the standard form: Ax^2+Bx+C=0. To do this, use opposite operations to move each term on the right side to the left side. Then factor.
  
  Hope this helps.
  Button navigates to signup page
  (11 votes)
gmblim
Posted 6 years ago. Direct link to gmblim's post “In the above equation 3x^...”
In the above equation 3x^2+11x-4 = 0, I understand where we need to find two numbers were a+b need to equal 11 to satisfy the 11x, however, I'm having trouble connecting where -12 came from where it states that we need to find numbers to satisfy (a)(b) = -12. I'm seeing a -4 at the end of the equation. Not sure where -12 came from. Was it from multiplying -4 to the co-effiecient of the 3 in 3x^2?
Button navigates to signup pageComment on gmblim's post “In the above equation 3x^...”
(4 votes)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “Yes, the -12 comes from m...”
  Yes, the -12 comes from multiplying -4 with the 3 from 3x^2.
  Button navigates to signup page
  (6 votes)
Danny
Posted 8 years ago. Direct link to Danny's post “I know this is not relate...”
I know this is not related to the above questions, but i could not find where to ask normal day to day questions. If someone could please help me answer this would be great.:
Factorise the expression as far as possible to prime factors:
(x-1)^2-9(y-1)^2 ...
^2(power of 2)
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Vu
  Posted 8 years ago. Direct link to Vu's post “Expand the squares, then ...”
  Expand the squares, then distribute the 9. Combine like terms then factor. Post comment if you still feel stuck.
  Comment on Vu's post “Expand the squares, then ...”
  (13 votes)
Anushka
Posted 4 years ago. Direct link to Anushka's post “How do you factor it when...”
How do you factor it when the leading coefficient is more than 1? For example, something like 3xsquared+ x-2.
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “You factor the trinomial ...”
  You factor the trinomial by grouping. See the lessons at this link: https://www.khanacademy.org/math/algebra/polynomial-factorization#factoring-quadratics-2
  
  Hope this helps.
  Comment on Kim Seidel's post “You factor the trinomial ...”
  (3 votes)
Laney
Posted 4 years ago. Direct link to Laney's post “My equation is x^2-2x+1=0...”
My equation is x^2-2x+1=0 ( find x) How do i factor out the 1? Could i make it 1^2?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “You need 2 factors of 1 t...”
  You need 2 factors of 1 that add to -2.
  Your choices are to use: 1*1 or (-1)(-1). Which pair adds to -2?
  Hope this helps.
  Button navigates to signup page
  (3 votes)
Rowel the owl! 🦉
Posted 10 days ago. Direct link to Rowel the owl! 🦉's post “Hi! This is Rowel! I hav...”
Hi! This is Rowel!

I have a question about x^2+5x=0.

Won't (x+5)(x+0) work also, if we made the original equation into x^2+5x+0=0?

Thank you for looking at my question.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kim Seidel
  Posted 10 days ago. Direct link to Kim Seidel's post “x+0 = x. There is no val...”
  x+0 = x. There is no value added by changing x into x+0.
  Keep it simple. The factors are x(x+5).
  Comment on Kim Seidel's post “x+0 = x. There is no val...”
  (3 votes)

Algebra 1

Course: Algebra 1 > Unit 14

Solving quadratics by factoring

What you should be familiar with before taking this lesson

What you will learn in this lesson

Solving factored quadratic equations

Reflection question

A note about the zero-product property

Solving by factoring

Solve $x^{2} + 5 x = 0$ ‍.

Solve $x^{2} - 11 x + 28 = 0$ ‍.

Solve $4 x^{2} + 4 x + 1 = 0$ ‍.

Solve $3 x^{2} + 11 x - 4 = 0$ ‍.

Arranging the equation before factoring

One of the sides must be zero.

Removing common factors

Want to join the conversation?

Algebra 1

Course: Algebra 1 > Unit 14

What you should be familiar with before taking this lesson

What you will learn in this lesson

Solving factored quadratic equations

Reflection question

A note about the zero-product property

Solving by factoring

Solve x2+5x=0‍ .

Solve x2−11x+28=0‍ .

Solve 4x2+4x+1=0‍ .

Solve 3x2+11x−4=0‍ .

Arranging the equation before factoring

One of the sides must be zero.

Removing common factors

Want to join the conversation?

Solve $x^{2} + 5 x = 0$ ‍.

Solve $x^{2} - 11 x + 28 = 0$ ‍.

Solve $4 x^{2} + 4 x + 1 = 0$ ‍.

Solve $3 x^{2} + 11 x - 4 = 0$ ‍.