Strategy to factor equations?

Question

Hi, I was wondering if there was any specific strategy to factor equations, for example, x²-5x+4=0. If anybody could go through step by step and explain to me how I could factor other equations similar to this easily, I would greatly appreciate it.

Answer 1

ax^2 + bx + c = 0

multiply a and c, you need to come up with factors of this multiple that add up to b.

x^2 - 5x + 4, for example 1 x 4 = 4
So we need to come up with factors of 4 that add up to -5
-1 x -4 = 4 and -1 + -4 = -5

So, (x -1)(x - 4) = 0

Answer 2

There is an easy way to check this.

Take your first term in this case 1 and multiply it by the last term 4 and figure any factors of that answer that add together to make the middle term in this case -5.

in this case -4 and -1 add together to make -5 and multiplied they = 4 After that figure which would be the easiest to factor with

IE you can have x^2-4x-x+4=0 but nothing really factors this way so you would put x^2-x-4x+4 = 0

In the second way you can then factor out an x from the first term and a -4 in the second to get

x(x-1) + 4(x-1) leaving you with (x+4)(x-1) = 0 or in this case x=-4 or x=1

This trick will work with any trinomial where the leading term is squared. IN any case where none of the factors from the first term multiplied by the last term will not add up and give you the middle term then you can not factor it.

Also always make sure to look for a greatest common factor first.

Answer 3

What do you know?
look at the end and the middle
what two numers add up to -5 but when you multiply them by each other equal 4?

-1 and -4 right?

(x-1)(x-4)=0

use FOIL to check your answer...

x times x = x^2
x times -4 = -4x
-1 times x = -1x
-1 times -4 = 4
add them all together and you get what you started with.

Next step: substitute values that will make one of the parenthesis equal to 0.
if x = 4 then (4-1)(4-4)=0
(3)(0)=0
same for x = 1
solution set (1,4)

Answer 4

x^2 = 5x + 4 = 0

You expect the factors to look like

(x + a)(x + b)

Which multiplies out to

x^2 + (a + b)x + ab

So look for two numbers whose sum is - 5 and whose product is + 4

-1 and -4 do it

(x - 1)(x - 4) = 0

x = 1, x = 4


For harder factoring problems (coefficient of the x^2 term other than 1), see the lesson at the link below

Answer 5

x^2 - 5x + 4 = 0
(x - 1) (x - 4) = 0

x - 1 = 0, x = 1
x - 4 = 0, x = 4

Answer: x = 1 or x = 4

Answer 6

There are many special techniques which are easy and quick, but here is a method which always works for quadratic expressions. If the expression is ax^2 + bx + c, the factors are
[x - (-b + sqrt(b^2 - 4ac))/(2a)][x - (-b - sqrt(b^2 - 4ac))/(2a)] .

Answer 7

Hi,

On every type of factoring problem ALWAYS look first for a GCF, a greatest common factor. There won't always be one, but if there is, you should divide it out first. A GCF will carry along through the problem and be the first thing in an answer with a GCF.

I' m going to explain factoring trinomials the way that I teach it, which is probably different from what you've done. Try to follow the steps. This method will work on any trinomial.

For x² - 5x + 4 = 0, there is no GCF.
Start both parentheses with the first number and variable.
(1x.......)(1x..........) First sign goes in first parentheses.
(x..-....)(x..........) Product of signs goes in 2nd parentheses.
(x.-....)(x.-.....) <== neg is because neg x pos = negative

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 1 x 4 = 4 So, out to the side list pairs of factors of 4.

4
------
1, 4
2, 2

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

Your signs are the same, so you want to add factors to get 5. Those factors are 1 and 4. The bigger number is pushy - it always goes first. So your factors now are:
(x.-..)(x.-..)
(x.-.4)(x.-.1)

If there were numbers in front of the x terms instead of 1, you would have to try to reduce the factors. In this problem, you are done, so the factors are ( x - 4)(x - 1).

3y² - 13y - 10 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses with the first number and variable.
(3y.......)(3y..........) First sign goes in first parentheses.
(3y..-....)(3y..........) Product of signs goes in 2nd parentheses.
(3y..-....)(3y...+.....) <== plus is because neg x neg = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 3 x 10 = 30 So, out to the side list pairs of factors of 30.

30
------
1, 30
2, 15
3, 10
5, 6

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(3y..-....)(3y...+.....) Your signs are different, so you want to subtract factors to get 13. Those factors are 2 and 15. ( Notice that 3 and 10 would have added to 13, so you HAD to know to subtract this time.) When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(3y..-..15)(3y.+.2.)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 3 but the second parentheses does not reduce.

(3y..-..15)(3y.+.2.)
-------------
.......3 This reduces to your final factors of

(y - 5)(3y + 2)

NEXT PROBLEM !!

2x² +15x + 7 Look for a GCF None this time. If there was one, factor it out. Then temporarily start both parentheses
............................. with the first number and variable.
(2x.......)(2x..........) First sign goes in first parentheses.
(2x..+....)(2x..........) Product of signs goes in 2nd parentheses.
(2x..+....)(2x...+.....) <== plus is because pos x pos = positive

Now multiply your first and third numbers together. Ignore their signs - you've already done them. 2 x 7 = 14 So, out to the side list pairs of factors of 14.

14
------
1, 14
2, 7

Now you want to pick which factors go in your parentheses, using these rules:

If the signs you put on your parentheses are the SAME, find the factors that ADD up to the middle number in the problem.
If the signs you put on your parentheses are the DIFFERENT, find the factors that SUBTRACT to the middle number in the problem. (Note you look at signs in the parentheses, not in the problem.)

(2x..+....)(2x...+.....) Your signs are the same, so you want to add factors to get 15. Those factors are 1 and 14. When you put the numbers in the parentheses, the bigger number is pushy - it always goes first. So your factors now are:
(2x.+.14)(2x.+.1)

Now you have to reduce either or both parentheses by dividing each parentheses' terms by the largest possible divisor. In our problem, the first parentheses is divisible by 2 but the second parentheses does not reduce.

(2x.+.14)(2x.+.1)
-------------
.......2 This reduces to your final factors of

(x + 7)(2x + 1)

This takes care of trinomials!!


Now we are going to look at problems with only 2 terms. Even on these, always check first for a GCF. Then the problems could be the difference of perfect squares, the difference of perfect cubes, or the sum of perfect cubes. First we will look at the difference of perfect squares.

Suppose you had 4x² - 81
This is the difference of perfect squares. 4x² is really (2x) squared. 81 is 9². So both factors start with 2x and both end with 9. One binomial has a "-" and the other has a "+".

The factors are (2x - 9)(2x + 9).

Suppose you had 16a² - 49
This is the difference of perfect squares. 16a² is really (4a) squared. 49 is 7². So both factors start with 4a and both end with 7. One binomial has a "-" and the other has a "+". The order does not matter.

The factors are (4a + 7)(4a - 7).

Suppose you had 18x² - 50
18 and 50 are not perfect squares because no number times itself equals either 18 or 50. The problem is that we did not look first for a GCF! This time both terms are divisible by 2. If we divide that out first, we get:
2(9x² - 25) Now we have a difference of perfect squares!
Keep the GCF of 2 out front. The parentheses has a difference of perfect squares. 9x² is really (3x) squared. 25 is 5². So both factors start with 3x and both end with 5. One binomial has a "-" and the other has a "+".

The factors are 2(3x - 5)(3x + 5).

Suppose you had 16x^4 - 81
This is the difference of perfect squares. 16x^4 is really (4x²) squared. 81 is 9². So both factors start with 4x² and both end with 9. One binomial has a "-" and the other has a "+". The order does not matter.

(4x² + 9)(4x² - 9) This is correct so far, but it is not done this time. Notice that with x² in terms with 4 and 9, that they are perfect squares again, so the factor that is the DIFFERENCE of perfect squares can factor again. (The sum of perfect squares can not factor.) So (4x² - 9) factors again into:

(4x² + 9)(4x² - 9)
(4x² + 9)(2x - 3)(2x + 3) <== This is the answer in factored form.

Now we're going to look at the sum and difference of perfect cubes. Again always look first for a GCF. If there is one, factor it out. Perfect cube problems will always factor into a binomial times a trinomial, like this:

x³ + 125 = (x + 5)(x² - 5x + 25)

The general rule for these is:

1) Factor out a GCF if there is one.
2) Make 2 parentheses for a binomial and a trinomial.
3) The sign from the problem goes in the binomial. The opposite sign from what you just used goes in the trinomial, and is always followed by a "+" as the second sign in the trinomial. This is the only step where what you do is different on the sum of cubes from on the difference of cubes.
4) Figure out what you would cube to get the first term of the problem and put that expression as the first term of your binomial. Then figure out what you would cube to get the second term of the problem and put that expression as the second term of your binomial.
5) Your trinomial already has its signs, so don't worry about the signs. To find terms of the trinomial, write the first term of your binomial 3 times followed by the second term of your binomial 3 times out to the side. Multiply the first 2 things together to get the trinomial's first term. Multiply the next 2 things together to get the trinomial's middle term. Multiply the last 2 things together to get the trinomial's third term.

So for, x³ + 125, there is no GCF.
Parentheses would be ( + )( - + )
To get x³, you'd cube x.
To get 125, you'd cube 5.
So your binomial becomes (x + 5)( - + )
Now write x x x 5 5 5
x*x = x²
x*5 = 5x
5*5 = 25
So your trinomial becomes (x + 5)(x² - 5x + 25)

Here's another example:

So for 8x³ + 27y³, there is no GCF
Parentheses would be ( + )( - + )
To get 8x³, you'd cube 2x.
To get 27y³, you'd cube 3y.
So your binomial becomes (2x + 3y)( - + )
Now write 2x 2x 2x 3y 3y 3y
2x*2x = 4x²
2x*3y = 6xy
3y*3y = 9y²
So your trinomial becomes (2x + 3y)(4x² - 6xy + 9y²)



The last thing we will look at is 4 term problems. As always look for a GCF first. Then try to factor a GCF out of just the first 2 terms and then try to factor a GCF out of the last 2 terms. If you can do this, the expression left behind both times must be the exact same expression. The 2 GCFs go together to make one factor and the repeated expression is the second factor. If either expression has an exponent in it, check to see if it can be factored again.

Given the problem: 5x³ - 10x² +3x - 6

There is no GCF for the entire problem.
The first 2 terms has a GCF of 5x². Factor it out.

5x³ - 10x² + 3x - 6
5x²(x - 2) + 3x - 6

Now factor out the GCF of the last 2 terms.

5x²(x - 2) + 3x - 6
5x²(x - 2) + 3(x - 2)

Now put 5x² and + 3 together as one factor. (x - 2) is the other factor.

(5x² + 3)(x - 2) are the factors. Nothing factors again.

Another problem is 4a³ + 28a² -9a - 63

There is no GCF for the entire problem.
The first 2 terms has a GCF of 4a². Factor it out.

4a²(a + 7) -9a - 63

Now factor out the GCF of the last 2 terms.

4a²(a + 7) -9(a + 7) Note the GCF had to be a -9 so that the parentheses would be the same.

Now put 4a² and - 9 together as one factor. (a + 7) is the other factor.

(4a² - 9)(a + 7) are the factors so far, but 4a² - 9 is the difference of perfect squares. It factors again.

(2a - 3)(2a + 3)(a + 7) These are the final factors.

A second type of 4 term problem is something like this:

9x² + 12xy + 4y² - 16z²

In this problem there is no GCF for all 4 terms. The first 2 terms have a GCF and the last 2 terms have a GCF, but the expressions are not the same, so the 2 term - 2 term method doesn't work.

9x² + 12xy + 4y² - 16z²
3x(3x + 4y) + 4(y² - 4z²) Note parentheses are different, so this doesn't work.

Instead of grouping 2 and 2, try grouping 3 and 1.

9x² + 12xy + 4y²....... - 16z²

Try factoring the first 3 terms as a trinomial, hoping to get a binomial squared.

9x² + 12xy + 4y²....... - 16z²
(3x + 2y)² - 16z²

Now factor this as the difference of perfect squares.

Square (3x + 2y) to get (3x + 2y)² and square 4z to get 16z²

That means the factors both start with 3x + 2y and both end with 4z, with a "+" or "-" in front of 4z.

(3x + 2y)² - 16z²
(3x + 2y - 4z)(3x + 2y + 4z) These are the factors!!

That's a good idea how to factor the most common types of problems!!

I hope that helps you!! :-)