28x^2-5x-3?

Question

Can you help me factor it? I'm having a hard time doing this.

Answer 1

(4x+1)(7x-3)
it's right, but I don't really know how to explain to you, it's almost like guess and check-but im really use to factoring, i've went to a tutoring and practiced it a lot, so it's just sort of natural for me, so yeah, sorry for no explanation...

Answer 2

There is a way called the AC method that works every time and takes all the guesswork out of these types of problems.

First, multiply the coefficient of the x^2 term (28) and the constant at the end (-3). This gives you -84.

Now you need to come up with 2 numbers that multiply to give you -84, but also add to give you -5. How about +7 and -12?

Now you use these 2 numbers to rewrite the middle term (be sure to include the variable) like this:
28x^2 + 7x - 12x -3

Notice the 2 new middle terms add to -5x which is what you had before.

Now you factor by grouping (hopefully you know this technique already.) Group the first 2 and the last 2 terms and factor out a common term.

28x^2 and 7x have 7x in common, so factor that out, leaving (4x + 1.)
-12x and -3 have -3 in common, so factor that out, leaving (4x + 1.) Now you have:

7x(4x + 1) - 3(4x + 1)

See how the 2 terms in parentheses match? That's exactly what you want to happen! So (4x + 1) is a common term, so you can factor it out too to get:

(4x + 1)(7x - 3)

It seems like a lot of work at first, but I promise it works every time and it sure beats the "trial and error" method mentioned earlier.

Good luck!

Answer 3

(7x -3)(4x+1)

x=3/7 or x=-1

Answer 4

your equation is in the form a*x^2 + b*x + c = 0

roots of this equation are

{x= [-b + sqrt(b^2-4ac)] /2a} and {x=[-b - sqrt(b^2-4ac)] /2a }

square root only applies to the numerator (b^2-4ac)