Answer:
P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)
Step-by-step explanation:
Since P(1) = 0 and P(- 2) = 0, then
(x - 1) and (x + 2) are factors of P(x)
(x - 1)(x + 2) = x² + x - 2 ← is also a factor of P(x)
dividing 3
- 5x³ - 17x² + 13x + 6 by x² + x - 2 gives
P(x) = (x - 1)(x + 2)(3x² - 8x - 3) = (x - 1)(x + 2)(x - 3)(3x + 1)
First, distribute the 100 to the (x-1) portion:
500 = 100x - 100
then, add 100 to both sides of your equation to get “x” alone on the right side:
500 + 100 = 100x - 100 + 100
600 = 100x
next, divide 100x by 100 and 600 by 100 to isolate a singular “x”:
600/100 = 100x/100
6 = x
therefore, x=6 is your solution
Answer:
C. 1920
Step-by-step explanation:
25*8*10
1920
Answer:
Proofs are in the explantion.
Step-by-step explanation:
We are given the following:
1)
for integer
.
1)
for integer
.
a)
Proof:
We want to show
.
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(a+c)-(b+d)=rn. If we do that we would have shown that
.
kn+mn = (a-b)+(c-d)
(k+m)n = a-b+ c-d
(k+m)n = (a+c)+(-b-d)
(k+m)n = (a+c)-(b+d)
k+m is is just an integer
So we found integer r such that (a+c)-(b+d)=rn.
Therefore,
.
//
b) Proof:
We want to show
.
So we have the two equations:
a-b=kn and c-d=mn and we want to show for some integer r that we have
(ac)-(bd)=tn. If we do that we would have shown that
.
If a-b=kn, then a=b+kn.
If c-d=mn, then c=d+mn.
ac-bd = (b+kn)(d+mn)-bd
= bd+bmn+dkn+kmn^2-bd
= bmn+dkn+kmn^2
= n(bm+dk+kmn)
So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.
Therefore,
.
//