SOLUTION: OK, last weary question for the night: have I gotten this one right?? 4x^2 - 21x + 20 (4x^2 - 25x) + (4x = 20) x(4^2 - 25) + (4(x + 5) = (4^2 -25)(x+4)?????????????? thanks in

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Question 384695: OK, last weary question for the night: have I gotten this one right??
4x^2 - 21x + 20
(4x^2 - 25x) + (4x = 20)
x(4^2 - 25) + (4(x + 5) = (4^2 -25)(x+4)??????????????
thanks in advance - and where've y'all been all my life????????
Found 2 solutions by jim_thompson5910, jsmallt9:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!

Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .


Now multiply the first coefficient by the last term to get .


Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?


To find these two numbers, we need to list all of the factors of (the previous product).


Factors of :
1,2,4,5,8,10,16,20,40,80
-1,-2,-4,-5,-8,-10,-16,-20,-40,-80


Note: list the negative of each factor. This will allow us to find all possible combinations.


These factors pair up and multiply to .
1*80 = 80
2*40 = 80
4*20 = 80
5*16 = 80
8*10 = 80
(-1)*(-80) = 80
(-2)*(-40) = 80
(-4)*(-20) = 80
(-5)*(-16) = 80
(-8)*(-10) = 80

Now let's add up each pair of factors to see if one pair adds to the middle coefficient :


First NumberSecond NumberSum
1801+80=81
2402+40=42
4204+20=24
5165+16=21
8108+10=18
-1-80-1+(-80)=-81
-2-40-2+(-40)=-42
-4-20-4+(-20)=-24
-5-16-5+(-16)=-21
-8-10-8+(-10)=-18



From the table, we can see that the two numbers and add to (the middle coefficient).


So the two numbers and both multiply to and add to


Now replace the middle term with . Remember, and add to . So this shows us that .


Replace the second term with .


Group the terms into two pairs.


Factor out the GCF from the first group.


Factor out from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.


Combine like terms. Or factor out the common term


===============================================================


Answer:


So factors to .


In other words, .


Note: you can check the answer by expanding to get or by graphing the original expression and the answer (the two graphs should be identical).



If you need more help, email me at jim_thompson5910@hotmail.com

Also, feel free to check out my tutoring website

Jim
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!

By the way you have attempted this problem it is apparent that you are trying to split the middle term in such a way that factoring by grouping will work. This is a new approach to factoring expressions like this and I have yet to see that it makes the factoring easier.

Another way boils down to trial and error. We try different possibilities until we find the right one. Depending on how many possibilities there are this can be quick or time-consuming. (At the end I will show you yet another way to factor expressions like this.)

With an understanding of how multiplication and addition work we can often rule out some of the possibilities. With your expression, for example, the 4 and 20 are positive so the factors of 4 and 20 must be both positive or both negative. And with the middle term being negative, we now know that both factors of one of the numbers must be negative. So the factors must look like:
(ax - b)(cx - d)
Additionally, using logic I explain before, we cannot use pairs of even factors and still get an odd number like -21 in the middle. So this rules out the possible
(2x - b)(2x -d)
and
(ax - 2)(bx - 10)
possibilities. Since there is only one other pair of factors for 4 we now know that the factors will look like:
(4x - b)(x - d)
and that b and d must be 1 and 20 or 4 and 5.

We have now narrowed down the possibilities from 24 to 4. It should not take long to find the only one that works:
(4x - 5)(x - 4)

The logic which explains why a pair of even factors of 4 or 20 could not produce an odd middle coefficient:

Another way to factor quadratic trinomials like this is to use the Quadratic Formula in an unusual way:

which simplifies as follows:



IMPORTANT: If the expression inside the square root is not a perfect square like 121 is, then stop here because the expression will not factor!



In long form this is:
or
Simplifying each equation we get:
or
or
Using the Quadratic Formula for factoring we want to keep the answers as fractions. Now we take each equation above and use it to write a factor. From we get the factor:
(1x - 4)
Note that we use a "minus" between the terms. And note where the numerator and the denominator of the fraction went. From the second equation we get the factor:
(4x - 5)
These are the same two factor we found above. NOTE: If the formula gives you a negative answer like then the factor we would get would be: (9x - (-5)) or (9x + 5).

This method requires
The advantages of this approach are:
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