1/4x^2-9 the 1/4 is a fraction she is in algebra 1 9th grade

Robert makes a good point to use the difference of squares formula but if you are unable to see it at first, I would suggest that you look at it the long way:

Set it up as a regular polynomial:

(1/4)x^{2} +0x - 9

Notice that since we don't have an x in the original problem, we just insert 0x. This 0x will come into play later on.

It is now in the form:

ax^{2}+bx+c

From there, you multiply your a and c numbers, or in this case 1/4 and -9. You will get -9/4.

Now we figure out what two factors(numbers) we can multiply together to add up to get b, which in this case is 0. Since it is 0, we know that the two numbers will have to cancel each other out, which means this two numbers have to be the same. This means our two numbers are 3/2 and -3/2.

We can now replace the middle term with these two terms:

1/4x^{2} +3/2x -3/2x - 9

We then do factor by grouping and group the first two terms and the last two terms together:

(1/4x^{2} +3/2x) +(-3/2x -9)

Factor out what is common:

1/4x(x+6) - 3/2(x + 6)

Notice that when we factor out a negative, both of the terms in the parenthesis become positive.

Both parenthesis have x+6, so we can bring them together to get:

**(1/4x - 3/2)(x+6)
**

We check our work:

1/4x^{2} - 3/2x +6/4x - 18/2

Simplify and we get:

1/4x^{2} -3/2x +3/2x -9 or **1/4x ^{2} -9**