Factoring polynomials of the form ax^2 + bx + c

[coughsyrup78]

I'm having trouble factoring this polynomial.

12x² - 25xy + 7y²

My work:

12x² - 25xy + 7y² =
12x² - 28xy + 3xy + 7y² =
(12x² - 28xy) + (3xy + 7y²)=
3x(4x - 7y) - y(3x - 7y) =
I can't continue any further because the (3x - 7y) should be (4x - 7y).
I think I might be grouping incorrectly. I tried rearranging the terms different ways and I still can't get it to work.

Thanks in advance for your help.

 

[MarkFL]

I would look at the product of the first and last coefficients:

\(\displaystyle 12\cdot7=84\)

We want to then find two factors of 84 whose sum is the middle coefficient -25, and these are -4 and -21. Now we want to factor the first coefficient such that one factor goes into 4 and the other into 21, and this is \(\displaystyle 12=4\cdot3\) so we have:

\(\displaystyle 12x^2-25xy+7y^2=(4x-7y)(3x-y)\)

 

[Deleted member 4993]

I'm having trouble factoring this polynomial.

12x² - 25xy + 7y²

My work:

12x² - 25xy + 7y² =.............. Next line should be

12x² - 28xy + 3xy + 7y² =
(12x² - 28xy) + (3xy + 7y²)=

3x(4x - 7y) - y(3x - 7y) = .................. This line is incorrect - however does not lead to correct factorization.
I can't continue any further because the (3x - 7y) should be (4x - 7y).
I think I might be grouping incorrectly. I tried rearranging the terms different ways and I still can't get it to work.

Thanks in advance for your help.

y²[12(x/y)² - 25(x/y) + 7]

Let u = x/y

12u² - 25u + 7 .... using quadratic formula we get

u_1,2 = [25 ± √(625 - 336)]/24

Now continue......................

 

[HallsofIvy]

I would suggest the same thing except from a different point of view. Consider this a quadratic equation in x with the coefficients involving y and \(\displaystyle y^2\): \(\displaystyle (12)x^2- (25y)x+ (7y^2)= 0\) and solve using the quadratic formula. You will find that every term involves a "y" that you can factor out. That is Denis's "Enter y at the end". Once you know that you should be able to determine the factors.

 

[coughsyrup78]

I would look at the product of the first and last coefficients:

\(\displaystyle 12\cdot7=84\)

We want to then find two factors of 84 whose sum is the middle coefficient -25, and these are -4 and -21. Now we want to factor the first coefficient such that one factor goes into 4 and the other into 21, and this is \(\displaystyle 12=4\cdot3\) so we have:

\(\displaystyle 12x^2-25xy+7y^2=(4x-7y)(3x-y)\)

I see what I did wrong now. I have to use -4 and -21.

I tried to use -28 and 3, that adds up to -25, but I didn't notice that it doesn't multiply to positive 84.

Thanks!

 

[Deleted member 4993]

That is why I use quadratic equation and let the chips fall where they may...

 

[coughsyrup78]

That is why I use quadratic equation and let the chips fall where they may...

At this point they are teaching us the grouping method, so we have to do it that way for now.

 

[agentmulder]

FORGET the "y"; enter it at end. 12x^2 - 25x + 7

I like this idea.