數學 Factoring Special Quadratic

(1) I guess the question should be the whole (thing) to the power 4/3 ? Instead of to the power 4 than the whole thing divided by 3. Tell me if I got you wrongly.

(-54x^6 y / 2x^-3 y^4)^(4/3)
= {-27 x^[6-(-3)] y^(1-4)}^(4/3) <-- -54/2 = -27
= (-27 x^9 y^-3)^(4/3)
= 27^(4/3) x^(9*4/3) y^(-3*4/3) <-- the "-" sign is gone because -1^(4/3) = 1
= 3^4 x^12 y^-4 <-- 3^3 = 27 ==> 27^-3 = 3
= 81 x^12 y^-4


(2) The answer is B.

Factorize 5 x^2 - 20
= 5 (x^2 - 4)
= 5 (x^2 - 2^2) <-- a^2 - b^2 = (a+b)(a-b)
= 5 (x+2) (x-2) <-- 3 factors obtained.


(3) The answer is A.

If one of the factor is (x-7), you can dedude the other factor as quadratic equation in the form ax^2 + bx + c = 0 can always be expressed as (px + q)(rx + s) = 0.

Now given p = 1, q = -7.

The first term is 3x^2, r must be 3 in order to get 3x^2 from (x-7)(rx + s).
Now we need to find s.
(x - 7)(3x + s)
= 3x^2 - 21x + sx - 7s
= 3x^2 + (s - 21)x - 7s

The second terms is -16x, thereform s - 21 = -16
s = 5

The last term k = -7s = -7*5 = -35.