(x - 5i√2)(x +5i√2)
given the roots of a polynomial p(x), say x = a and x = b
then the factors are (x - a)(x - b)
and p(x) is the product of the factors ⇒ p(x) = (x - a)(x - b)
here x² + 50 = 0 ⇒ x² = - 50 → ( set = 0 for roots)
take the square root of both sides
x = ± √-50 = ± √(25 × 2 × -1) = √25 × √2 × √-1 = ± 5i√2
The roots are x = ± 5i√2
thus the factors are ( x - ( - 5i√2)) and (x - (+5i√2))
x² + 50 = (x + 5i√2)(x - 5i√2)
Answer:
C. 12 months
Step-by-step explanation:
145-25= 120
120/10= 12
she pays for 12 months
So find g(0) and g(3)
G(0) = 0^2 -5(0) +3 = 3
And g(3) = 3^2 - 5(3) +3 = 9 - 15 + 3 = 3 so there is no change in g(x) for that interval.
H(0) = 8(0) + 10 = 10
And h(3) = 8(3) + 10 = 34
So the change in h(x) for that interval is 24.
Answer:
not factorable
Step-by-step explanation:
The easiest way to factor is to find two numbers of the "a" value (the coefficient before 
) and the factors of the c value (10) that ADD up to give you -9, in this case it would be:
1 -10 --> 1(-10) gives you -10
1 -1 --> 1(-1) gives you -1
adding these together -10 + (-1) = -11 which does not equal -9.
You could stop here and conclude that this is not factorable by inspection.
Another method: using the quadratic formula to find its roots.
Roots: 
and 
