Given the polynomial function
![P(x)=2 x^{2} -4x+30](https://tex.z-dn.net/?f=P%28x%29%3D2%20x%5E%7B2%7D%20-4x%2B30)
If (x-3) is a factor of P(x), then
![P(x)=2 x^{2} -4x+30=(x-3)*Q(x)](https://tex.z-dn.net/?f=P%28x%29%3D2%20x%5E%7B2%7D%20-4x%2B30%3D%28x-3%29%2AQ%28x%29)
, for some polynomial Q of 1st degree,
Then according to the factor theorem P(3)=0, because P(3)=(3-3)Q(x)=0*Q(3)=0.
Check
![P(3)=2 (3)^{2} -4(3)+30=18-12+30=36](https://tex.z-dn.net/?f=P%283%29%3D2%20%283%29%5E%7B2%7D%20-4%283%29%2B30%3D18-12%2B30%3D36)
≠0
we see that P(3) is not 0, so (x-3) is not a factor of P(x).
Answer: no
Answer:
a) 0.3.
b) 1.117287138 to 9 dec. places.
d)3.
Step-by-step explanation:
a and d: the powers 1/3 and 0.333... is the cube root.
c: 0.5 = the square root.
b is found using a calculator
Answer:
Jerry owns 18 comic books
Step-by-step explanation:
24 - 6 = 18