In the last practice problem they said x^2+3=4 when solved for is x=+1 or -1 shouldn't it be plus or minus square root of 1

but they are the same because the √1 = 1

how would i factor: 10x(x+1)=8x+12

You need to simplify before trying to factor. You must have all terms on one side. Basically, your equation should be in standard form: Ax^2 + Bx + C = 0 Then, you try to factor. So, start by distributing the 10 to eliminate the parentheses. Then Subtract both the 8x and 12. Once that is done, see if you can factor the new polynomial. If not, comment back and I'll help.

How come on the second problem where it askes for the 2 solutions for x you could get it wrong when you have the correct answer because I got it wrong the first time that I typed my answers in but afer I swaped them places I got it wrong?

I think the website has the problems set up so they expect the small value first and larger value second. This simplifies the programming.

In the example 2: 3x^2 + 33x + 30 = 0 the first step for solving the equation gives me 3(x^2 + 11x + 3)=0 but this is inconsistent because I could perfectly divide both sides into (x^2 + 11x + 3) without violating any math rules and the result will give me 3 = 0 which is not true. Are math rules inconsistent or incoherent when dividing 0?

Joseph, First, you may want to check your arithmetic on your first step: 30/3 = 10. Second, the exercises ask us to solve for `x`. Your second step eliminates `x` from the equation. You cannot solve for a variable that is no longer there. Typically, if you end up with something like `3 = 0`, it indicates that either you have made an error or there is no solution to the problem.

In problem 3 in the practice, once we get to: p^2 = 4p Couldn’t we have divided both sides of the equation by p? And use the exponents property on the left hand side for (p^2)/p leaving us with p^(2-1)=4 p=4 Once you replace p with [(x^2)+3] it also gives us +-1 as a result but without having to deal with the square root of negative 3 (i.e.: imaginary nb) hassle.

Quadratic equations create 2 solutions. Sometimes they are the same solution and the equation degrades to a single solution. By dividing by "p", you destroy / lose the 2nd solution. You can't know that the 2nd solution will be a complex number at this point in solving the equation. And, as you get into higher level math, there are applications where you will want the complex solutions.

I don't understand how you decide the order of the positive and negative factors? -6 x 5 =-30 (in the last question) as well as 6 x -5. I am clearly missing some really basic principle! Please help me!

when multiplying, the order in which you multiply your factors (-5 and 6) doesn't matter.

Main content

Algebra 1

Course: Algebra 1 > Unit 14

Lesson 5: Solving quadratics by factoring

Solving quadratics by factoring review

CCSS.Math:

Factoring quadratics makes it easier to find their solutions. This article reviews factoring techniques and gives you a chance to try some practice problems.

Example 1

Find the solutions of the equation.

2, x, squared, minus, 3, x, minus, 20, equals, x, squared, plus, 34

\begin{aligned}2x^2-3x-20&=x^2+34\\\\ 2x^2-3x-20-x^2-34&=0\\\\ x^2-3x-54&=0\\\\ (x+6)(x-9)&=0\end{aligned}

\begin{aligned}&\swarrow&\searrow\\\\ x+6&=0&x-9&=0\\\\ x&=-6&x&=9\end{aligned}

In conclusion, the solutions are x, equals, minus, 6 and x, equals, 9.

Want to see see another example? Check out this video.

Example 2

Find the solutions of the equation.

3, x, squared, plus, 33, x, plus, 30, equals, 0

\begin{aligned}3x^2+33x+30&=0\\\\ x^2+11x+10&=0\\\\ (x+1)(x+10)&=0 \end{aligned}

\begin{aligned}&\swarrow&\searrow\\\\ x+1&=0&x+10&=0\\\\ x&=-1&x&=-10\end{aligned}

In conclusion, the solutions are x, equals, minus, 1 and x, equals, minus, 10.

Want to see see another example? Check out this video.

Example 3

Find the solutions of the equation.

3, x, squared, minus, 9, x, minus, 20, equals, x, squared, plus, 5, x, plus, 16

\begin{aligned}3x^2-9x-20&=x^2+5x+16\\\\ 3x^2-9x-20-x^2-5x-16&=0\\\\ 2x^2-14x-36&=0\\\\ x^2-7x-18&=0\\\\ (x+2)(x-9)&=0 \end{aligned}

\begin{aligned}&\swarrow&\searrow\\\\ x+2&=0&x-9&=0\\\\ x&=-2&x&=9\end{aligned}

In conclusion, the solutions are x, equals, minus, 2 and x, equals, 9.

Want to see see another example? Check out this video.

Practice

Problem 1

Current

Solve for x.

x, squared, plus, 14, x, plus, 49, equals, 0

x, equals

Want more practice? Check out these exercises:

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Tanishka
Posted 6 years ago. Direct link to Tanishka's post “In the last practice prob...”
In the last practice problem they said x^2+3=4 when solved for is x=+1 or -1 shouldn't it be plus or minus square root of 1
Button navigates to signup pageButton navigates to signup page
(10 votes)
Answer
- David Severin
  Posted 6 years ago. Direct link to David Severin's post “but they are the same bec...”
  but they are the same because the √1 = 1
  Comment on David Severin's post “but they are the same bec...”
  (12 votes)
Aileen
Posted 5 years ago. Direct link to Aileen's post “how would i factor: 10x(...”
how would i factor: 10x(x+1)=8x+12
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- Kim Seidel
  Posted 5 years ago. Direct link to Kim Seidel's post “You need to simplify befo...”
  You need to simplify before trying to factor. You must have all terms on one side.
  Basically, your equation should be in standard form: Ax^2 + Bx + C = 0
  Then, you try to factor.
  So, start by distributing the 10 to eliminate the parentheses.
  Then Subtract both the 8x and 12.
  Once that is done, see if you can factor the new polynomial.
  If not, comment back and I'll help.
  Button navigates to signup page
  (5 votes)
Kiley
Posted 3 years ago. Direct link to Kiley's post “How come on the second pr...”
How come on the second problem where it askes for the 2 solutions for x you could get it wrong when you have the correct answer because I got it wrong the first time that I typed my answers in but afer I swaped them places I got it wrong?
Button navigates to signup pageComment on Kiley's post “How come on the second pr...”
(5 votes)
Answer
- Kim Seidel
  Posted 3 years ago. Direct link to Kim Seidel's post “I think the website has t...”
  I think the website has the problems set up so they expect the small value first and larger value second. This simplifies the programming.
  Button navigates to signup page
  (2 votes)
Hope
Posted 3 years ago. Direct link to Hope's post “How many others are in pr...”
How many others are in pre-cal 11, and want to become a math teacher going into final exams with only 60%. Seriously tho, that's what i have! 🤔
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(5 votes)
Answer
Joseph Arcila
Posted 4 years ago. Direct link to Joseph Arcila's post “In the example 2: 3x^2 + ...”
In the example 2: 3x^2 + 33x + 30 = 0
the first step for solving the equation gives me
3(x^2 + 11x + 3)=0 but this is inconsistent because I could perfectly divide both sides into (x^2 + 11x + 3) without violating any math rules and the result will give me 3 = 0 which is not true. Are math rules inconsistent or incoherent when dividing 0?
Button navigates to signup pageComment on Joseph Arcila's post “In the example 2: 3x^2 + ...”
(3 votes)
Answer
- AD Baker
  Posted 4 years ago. Direct link to AD Baker's post “Joseph, First, you may ...”
  Joseph,
  
  First, you may want to check your arithmetic on your first step: 30/3 = 10.
  
  Second, the exercises ask us to solve for x. Your second step eliminates x from the equation. You cannot solve for a variable that is no longer there.
  
  Typically, if you end up with something like 3 = 0, it indicates that either you have made an error or there is no solution to the problem.
  Comment on AD Baker's post “Joseph, First, you may ...”
  (5 votes)
louisaandgreta
Posted a year ago. Direct link to louisaandgreta's post “In problem 3 in the pract...”
In problem 3 in the practice, once we get to:

p^2 = 4p

Couldn’t we have divided both sides of the equation by p?
And use the exponents property on the left hand side for (p^2)/p leaving us with
p^(2-1)=4
p=4

Once you replace p with [(x^2)+3] it also gives us +-1 as a result but without having to deal with the square root of negative 3 (i.e.: imaginary nb) hassle.
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted a year ago. Direct link to Kim Seidel's post “Quadratic equations creat...”
  Quadratic equations create 2 solutions. Sometimes they are the same solution and the equation degrades to a single solution.
  
  By dividing by "p", you destroy / lose the 2nd solution. You can't know that the 2nd solution will be a complex number at this point in solving the equation. And, as you get into higher level math, there are applications where you will want the complex solutions.
  Comment on Kim Seidel's post “Quadratic equations creat...”
  (4 votes)
Tams Fletcher
Posted 2 years ago. Direct link to Tams Fletcher's post “One of the ways I ended u...”
One of the ways I ended up solving this problem is the following:

(x^2 +3)^2 = 4x^2+12
(x^2 +3).(x^2 +3)= 4(x^2+3)
/(x^2 +3) = /(x^2 +3)
x^2 +3 = 4
x^2 +3 -3 = 4 - 3
x^2 = 1
x = 1 or -1

Why does this different method of simplification get rid of the +/-sqrt-3 solution, while the method in the explanation above does not, and doing the long-hand, expansion and combination does not. Did I do something wrong or is this an interesting property of the equations we will learn about later?
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(4 votes)
Answer
- David Severin
  Posted 2 years ago. Direct link to David Severin's post “When you divided by the x...”
  When you divided by the x^2+3, you eliminated these two solutions from your problem. If either of these are used as solutions, you would get 0=0 on the second step, but they are not present on the third step. Think about a simpler problem, x^2=x If you do x^2-x, factor out an x to get x(x-1)=0, you get solutions of 0 and 1, If you divide by x to get x=1, you have misplaced the x=0 that disappeared when you divided by x.
  Comment on David Severin's post “When you divided by the x...”
  (1 vote)
Marion Read
Posted 4 years ago. Direct link to Marion Read's post “I don't understand how yo...”
I don't understand how you decide the order of the positive and negative factors? -6 x 5 =-30 (in the last question) as well as 6 x -5. I am clearly missing some really basic principle! Please help me!
Button navigates to signup pageComment on Marion Read's post “I don't understand how yo...”
(3 votes)
Answer
- rylan.wetsell
  Posted 6 months ago. Direct link to rylan.wetsell's post “when multiplying, the ord...”
  when multiplying, the order in which you multiply your factors (-5 and 6) doesn't matter.
  Button navigates to signup page
  (2 votes)
hiroyasu
Posted 4 years ago. Direct link to hiroyasu's post “Does anyone know why the ...”
Does anyone know why the answer to 2x^2+8x+16? I got x=-2±2i. But I wonder what caused it to become a weird answer. It looked simple to me, but it seems like it was more complicated than I thought.
Button navigates to signup pageComment on hiroyasu's post “Does anyone know why the ...”
(3 votes)
Answer
- David Severin
  Posted 4 years ago. Direct link to David Severin's post “Most of the time, the ans...”
  Most of the time, the answer will be no real solutions, your complex number answer is based on b^2-4ac being negative. No real solutions mean that the parabola does not cross or touch the x axis at all,
  Button navigates to signup page
  (1 vote)
Brian Nyachia
Posted 3 years ago. Direct link to Brian Nyachia's post “Im tring to solve the qua...”
Im tring to solve the quadratic 18x^2-3x-6 but the problem is I can't factor the equation completely since it ends up becoming 6x^2-x-2. How can I solve this?
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(3 votes)
Answer
- David Severin
  Posted 3 years ago. Direct link to David Severin's post “So if you multiply 6*2=12...”
  So if you multiply 6*2=12, then what 2 numbers multiply to be 12 and add to be -1 (3 and -4), break middle term into these, so 6x^2 + 3x - 4x - 2 = 0. factor first two and last two, then combine with common factor and solve.
  Button navigates to signup page
  (1 vote)