factoring using zero principal. I can't figure this problem out can anyone help5k^2 -39k - 8 = 0?

Answer 1

5k^2 - 39k -8 = 0
(5k +1)(k-8) = 0
k = -1/5 or k = 8

Hope that helps :)

Answer 2

to factor whenever a > 1 is a simple process the should eliminate all of the guessing

multiply a( c) and list all of the factors for this product

for your problem a = 5 and c = - 8 so the product is -40
the factors for - 40 are:
1 and 40
2 and 20
4 and 10
5 and 8
of course one of these is a negative number to get a negative 40
which of these factors will add to give the middle term or "b" in the equation?
1 and -40 = -39
Once you have found that part of the factoring process, the next step is to write the equation in factored form.
Take the two factors that you found, 1 and -40 and make a fraction using them as the numerator and "a" as the denominator.

1/5 and -40/5 simplify (reduce) if possible

1/5 and -8 / 1
The denominators are the x value in your factored problem and the numerators are the constants.
( 5x +1 )( x - 8 )
This will work every time.

Answer 3

First: multiply the 1st and 3rd coefficient to get "- 40." Find two numbers that give you "- 40" when multiplied and "-39" (2nd/middle coefficient) when added/subtracted. The numbers are = 40 and 1 >>>

Sec: rewrite the equation with the new middle coefficients....

5k^2 - 40k + k - 8 = 0

*When you have 4 terms - group "like" terms & factor....

(5k^2 + k) - (40k - 8) = 0

k(5k + 1) - 8(5k + 1) = 0

(5k + 1)(k - 8) = 0

Third: solve the two k-variables > set both parenthesis to equal "0"....

1. 5k + 1 = 0
5k + 1 - 1 = 0 - 1
5k = -1
5k/5 = -1/5
k = -1/5

2. k - 8 = 0
k - 8 + 8 = 0 + 8
k = 8

Solutions: -1/5 and 8

Answer 4

Apply the formula for the 2nd degree equatioThe roots are 8 and
-1/5

So factoring = 5(x-8)*(x+1/5)

Answer 5

(5k +1 )*(k -8 )=5k^2-39k-8