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# Solving quadratics by factoring review

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
Solve for x.
x, squared, plus, 14, x, plus, 49, equals, 0
x, equals

Want more practice? Check out these exercises:

## Want to join the conversation?

• 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 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! 🤔
• 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.
• 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 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.
• 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?
• 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.
(1 vote)
• 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!