Question

can someone help me
solve by factoring n^2+2n-24=0

Answer 1

n=-6
n=4

Answer 2

if you want how to get @Etownkirk 's solution
here:
we know that the standard equation for the quadratic formula is \(\Large an^2+bn+c=o\)
i used "n" as a variable, so you wont get confused

to find the solutions/zeros/roots of this equation, you have to use the quadratic formula which is: \[n=\frac{ -b \pm \sqrt{b^2-4ac} }{2a }\]
where \(a=1, b=2, c=-24\), plug this values in the formula and you'll get the answer

Answer 3

The question asks that it be solved by factoring, not using the quadratic (or simply asking someone for the answer).

To factor a trinomial like this:
\[n^2+2n-24=0\]Multiply the coefficients of the first and final terms: 1*-24 = -24
Now, your job is to find a pair of factors of -24 that add up to 2, which is the coefficient of the middle term. -4*6 = -24, and -4 + 6 = 2, so -4 and 6 are the factors.

Rewrite the trinomial, splitting the middle term with the factors we found:
\[n^2 + 6n - 4n - 24 = 0\]Now group pairs of terms\[(n^2+6n) - 1(4n+24)\]Notice that I factored out a -1 and changed the sign of the last two terms.
Factor each group separately:\[n(n+6) -4(n+6)\]Now each product term has a common factor, \((n+6)\), so we factor that out as well, giving us\[(n+6)(n-4)\]Let's check our factoring by multiplying the expression

\[(n+6)(n-4) = n(n-4)+6(n-4) = n^2 -4n + 6n - 24 = n^2 +2n - 24\checkmark\]Our factoring is valid.

Now, we can solve our original equation using the 0 product property:
\[(n+6)(n-4) = 0\]For that to be true, one of the following will be true\[(n+6) = 0\]\[(n-4) = 0\]

Therefore, our solutions are \(n = -6,~n=4\)

If we substitute those values into the original equation, we'll see that they both produce a true statement:

\[n^2+2n-24=0\]\[(-6)^2+2(-6)-24 = 0\]\[36-12-24=0\]\[0=0\]
\[(4)^2+2(4)-24 = 0\]\[16+8-24=0\]\[0=0\]

Answer 4

thanks everyone for your help

Answer 5

oh yeah, sorry for not reading it properly :/