How to factor a strange trinominal. I know how to factor normal trinomials, however I was stumped when I saw this on my homework, could anyone help me through this?
The trinomial is as such:
$-m^2 + 8m + 18$.
 A: You can complete the square.  $-m^2+8m+18=-(m^2-8m-18)=-((m-4)^2-34)=-(m-4-\sqrt{34})(m-4+\sqrt{34})$
A: This one does not factor nicely, as its discriminant, $136$, is not a perfect square.
A: We can immediately determine of the expression given has rational roots evaluating its *discriminant": if it's evaluates to a number that is not a perfect square, it's roots are not rational:


*

*the discriminant of the quadratic polynomial $\,ax^2+bx+c\,$ is
$$\Delta = \,b^2-4ac.$$ Here that gives us $\Delta\,= \,8^2 - (-72) = 136,\;$ and as Cameron points out, $136$ is not a perfect square, so its roots are therefore irrational.


The best you can do, if you want strict to strictly factor your expression, is $$-(m- 4 -\sqrt{34})(m - 4 + \sqrt{34}),$$
How'd we find the values so we have factors of the form $(m - a_1)(m - a_2)$?
Rewriting the expression as,
$$-(m^2 - 8m - 18)=0 \iff m^2 - 8m - 18 =0\tag{1}$$
we can compute the roots using the tried-and-true quadratic equation to determine the roots (solutions) to the equation $(1)$: those roots are the values of $\,a_1\;\text{and}\;a_2$, respectively.
A: Without having knowledge in quadratic equation, you can approch the factorization in the following way. I have made the steps quite elaborate for you to comprehend and apply it to any other similar form of factorization.
If you need to factorize, first equate the equation to $0$
$$-m^2 + 8m + 18=0$$
$$-m^2 + 8m + 18=0$$
Dividing both sides by $-1$
$$\Rightarrow m^2 - 8m - 18=0$$
Adding $18$ to both sides
$$\Rightarrow m^2 - 8m=18$$
Trying to express the L.H.S as a perfect square
$$\Rightarrow m^2 - 2\cdot 4m +4^2 - 4^2=18$$
Adding $16$ to both sides
$$\Rightarrow m^2 - 2\cdot 4\cdot m +4^2=18+16$$
$$\Rightarrow (m-4)^2=34$$
Subtracting $34$ from both sides
$$\Rightarrow (m-4)^2-34=0$$
$$\Rightarrow (m-4)^2-\sqrt {34}^2=0$$
$$\Rightarrow (m-4-\sqrt {34})(m-4+\sqrt {34})=0$$
Dividing both sides by $-1$ again
$$\Rightarrow -(m-4-\sqrt {34})(m-4+\sqrt {34})=0$$
Hence factorization of 
$-m^2 + 8m + 18 = -(m-4-\sqrt {34})(m-4+\sqrt {34})$
A: Let me offer a completely different approach. By your comments you have not yet covered the quadratic equation. Therefore, it makes sense to assume that you should not be getting trinomials with non-integer roots in your homework. Making a list of all trinomials with integer roots which have smallish integer coefficients, one trinomial stands out as looking almost exactly like yours:
$$-m^2 + 3m + 18$$
You should be able to factor this one without the quadratic equation. Is it possible you misread a $3$ as an $8$?
