Question

Can someone tell me if this is done correct so far?

Answer 1

Factor this polynomial:
-16t^2 + 28t + 8

Factor pairs of -16: -1 and 16, 1 and -16, -2 and 8, 2 and -8, -4 and 4.
Factor pairs of 8: 1 and 8, 2 and 4.

Possible combinations:
(-16t + 1) (t + 8)
(-16t + 8) (t + 1)
(-16t + 2) (t + 4)
(-16t + 4) (t + 2)

Answer 2

I haven't checked which of the combinations is correct yet, but I just want to make sure I did it correctly so far.

Answer 3

I think I kind of understand what you are doing... like factoring a polynomial \(x^2 + bx + c\) and you write it as \((x - ...)(x-...)\)
But you have a factor of -16 in front of \(x^2\). What you can do is:
multiply the -16 coefficient, and the constant 8 to get -128

Now, you need two numbers \(m, n\) when multiplied together, give -128
And then you need these same two numbers \(m, n\), when added together, give 28 (the middle coefficient).

Since -128 is a large number.. it's a bit trickier to find, but two numbers that when multiplied give -128 are "32" and "-4".

If you add these two numbers, you get 28 (i.e 32 + (-4) = 28)

Now you write your polynomial as:
\(-16t^2 + 32 t - 4t + 8\) (essentially the 28t becomes a sum of 32t and -4t)

Now you can pair-wise factor the first two terms, and the last two terms:
\(-16t(t-2)-4(t-2)\)

Now you have a common factor of \((t-2)\), so factor again:
\((-16t-4)(t-2)\)

Now you can do a small additional factoring out a -4 from the first parentheses to get
\(-4(4t+1)(t-2)\)

Answer 4

I didn't understand what you wrote. Did I do it correctly or not?

Answer 5

Well if you do it your way, personally I find it a bit longer trying to think of the different factor possibilities.. the method above seems long to read, but with practice I find it much more efficient. Maybe it's more clear on this site: http://www.mathsisfun.com/algebra/factoring-quadratics.html and look for the example "6x^2 + 5x - 6"

With your method though, you are missing some options as you can have negative factors for 8: namely one possible combination (that works) with your method would be to consider
\((-16t-4)(t-2)\)

Answer 6

Oh Boy, now I'm getting confused.

Answer 7

How did you get (-16t - 4) (t - 2) ?

Answer 8

well for factors of 8, you can have (1, 8), (-1, -8), (2, 4), (-2, -4)

Answer 9

ok

Answer 10

we did this problem earlier

Answer 11

what was the problem?