factor: n^2-10n+24=0 ??

n² - 10n + 24 = 0
n² - 5n = - 24 + (- 5)²
n² - 5n = - 24 + 25
(n - 5)² = 1
n - 5 = +/- 1

= n - 5 - 1, = n - 6
= n - 5 + 1, = n - 4

Answer: (n - 6)(n - 4)

Proof:
= (n - 6)(n - 4)
= n² - 4n - 6n + 24
= n² - 10n + 24

(n-4)(n-6)=0
n=4 n=6

(n-6)(n-4)=0

n = 4 or n = 6

clear?!!

( n - 6 ) ( n - 4 ) = 0
n = 4 , n = 6

n^2-10n+24=0
just think:
- what two numbers multiplied by each other = 24?
- what two numbers when added together = -10

- 6 and -4 are the two numbers
because
(-6)(-4) = +24, which is the constant of the equation
(-6) + (-4) = -10, which is the coefficient of n term

so the answer is
(n - 6)(n-4) = 0

If you are also asked to solve for n, simply set each factor to zero and solve:

n- 6 = 0 ==> n = 6
n - 4 = 0 ==> n = 4

a trick to factorising quadratics is to think what two numbers add to make -10 and times together to make 24.
if we have an^2 + bn + c = 0
then say we have two roots -x and -y where the quadratic equals 0
x + y = b and x * y = c ( you cant solve these using algebra you need to use the method of trial and error)

so (n - 6)(n - 4) = 0

or just use the quadratic formula

n = -b + - sqrt(b^2 - 4ac)/2a

The left factors to (n-6)(n-4). Now we have:
(n-6)(n-4)=0

Now, (n-6)=0 or (n-4)=0.
n=6 or n=4 (answers)

n^2 - 10n + 24 =0
(n - 6) (n - 4) =0
n = 6 or n = 4

n^2 - 10n + 24
= n^2 - 4n - 6n + 24
= (n^2 - 4n) - (6n - 24)
= n(n - 4) - 6(n - 4)
= (n - 4)(n - 6)