Answer:
Step-by-step explanation:
First let us write the given polynomial as in descending powers of x with 0 coefficients for missing items
F(x) = x^3-3x^2+0x+0
We have to divide this by x-2
Leading terms in the dividend and divisor are
x^3 and x
Hence quotient I term would be x^3/x=x^2
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
Multiply x-2 by x square and write below the term and subtract
We get
x-2) x^3-3x^2+0x+0(x^2
x^3-2x^2
---------------
-x^2+0x
Again take the leading terms and find quotient is –x
x-2) x^3-3x^2+0x+0(x^2-x
x^3-2x^2
---------------
-x^2+0x
-x^2-2x
Subtract to get 2x +0 as remainder.
x-2) x^3-3x^2+0x+0(x^2-x-2
x^3-2x^2
---------------
-x^2+0x
-x^2+2x
-------------
-2x-0
-2x+4
------------------
-4
Thus remainder is -4 and quotient is x^2-x-2
The page numbers must have a difference of one, so:
n * (n+1) = 182
n^2 + n -182 =0
Solving by the quadratic formula we see n = 13 and n+1 = 14
Answer:
I think its (6,-3)
Step-by-step explanation:
