Question

Factor the trinomial: -2x^2+17x+1

Answer 1

Any ideas?

Answer 2

\[\sf -2x^2 +17x+1\]\[=\sf x^2 -\frac{17}{2}x -\frac{1}{2}\]\[=\sf \left(x^2 -\frac{17}{2}x\right) -\frac{1}{2}\]Now you're going to treat it like completing the square. DO you know how to do that?

Answer 3

I'm sorry but I don't know how. Could you please show me? Thank you

Answer 4

Sure.

Answer 5

First we analyze the stuff in () We're going to use it to create a quadratic within a quadratic, in a sense.

You know that the equation for a quadratic formula is: \(\sf ax^2 +bx +c=0\)

We're trying to create a quadratic formula within the parenthesis.

Answer 6

In order o do this, we take our "b" value, \(\sf -\frac{17}{2}\) and turn it into a "c" value by the format : \(\sf c= \left(\frac{b}{2}\right)^2\)

Answer 7

So we get \[\sf -\frac{\frac{17}{2}}{2} = \left(-\frac{17}{4}\right)^2=\frac{289}{16}\] this is our "new" c value for our quadratic inside our parenthesis. \[\sf \left(x^2 -\frac{17}{2}x +\frac{289}{16}\right)-\frac{1}{2}-\frac{289}{16} \]Now, you must remember your original function, that is :\(~\sf -x^2 -\frac{17}{2}x-\frac{1}{2}\)

Since we created a new "c" value, we also have to subtract the same c value outside to keep the function in balance.

Answer 8

A quick way to remember how the stuff inside the () simplifies is to remember :\(\sf c=\left(\frac{b}{2}\right)^2\)

in our case, we would have \[\sf \left(x-\frac{17}{4}\right)^2 - \frac{297}{16}\]

Answer 9

I think there is also an alternative method.

Answer 10

Alternative method :
the trinomial is prime and cannot be factored in rational numbers ;)

Answer 11

\[\sf -2x^2+17x+1\]\[=\sf -2\left(x^2-\frac{17}{2}x\right)+1\]\[=\sf -2\left(x^2 -\frac{17}{2}x +\frac{289}{16}\right) +1 +\frac{289}{8}\]\[=\sf -2\left(x-\frac{17}{4}\right)^2 +\frac{297}{8}\]

Answer 12

Thank you so much! :)

Answer 13

No problem :) Hope you understand how to complete the square a little better!