Please answer the following
#8
Answers
p(y) = y^3 + 2 y^2 + 3 y + 2 = q(y) * f(y) + q(y) = q(y) [ 1 + f(y) ]
Obviously the degree of q(y) is less than that of f(y). Given that it is more than or equal to 1. If f(y) has a degree 3, then quotient will be of degree 0 and remainder will be of degree 2.
If f(y) is of degree 1, then the remainder will be of degree 0 and quotient will be of degree 2. Thus only possibility is that f(y) has degree 2 and q(y) has degree 1.
(f(y))^2 has a degree 4 and (q(y))^2 has a degree 2. Thus (f(y))^2 - (q(y))^2 has a degree of 4. So the option (3) is correct.
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Let a, b, and c be integers.
Let f(y) = a y^2 + b y + (c-1) Let q(y) = d x + e
y^3 + 2 y^2 + 3 y + 2
= [ a y^2 + b y + c ] * (d y + e) = a d y^3 + (ae + b d) y^2 + (c d + b e) y + c e
Comparing coefficients, d = 1/a e = 2/c
2a/c + b/a = 2 => 2 a^2 + b c = 2 a c ---(1)
c/a + 2b/c = 3 => 2 a b + c^2 = 3 a c --- (2)
from: (1): b = 2 a(c-a)/ c and from (2) b = c (3a -c)/2a ---- (3)
So, 2 a (c - a) /c = c (3a-c)/2a
=> 4 a^2 (c - a) = c^2 (3 a - c) ----- (4)
=> c^3 - 4 a^3 + 4 a^2 c - 3 a c^2 = 0 divide by c^3 and let a/c = x.
=> 1 - 4 x^3 + 4 x^2 - 3 x = 0
=> 4 x^3 - 4 x^2 + 3 x - 1 = 0
for x = 1/2 the equation is satisfied.
So c = 2 a and using (3), b = a
so then d = 1/a and e = 1/a
So f(y) = a y^2 + a y + (2 a - 1)
and so q(y) = (y + 1)/a
Now calculate (f(y))^2 - (q(y))^2 ....
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Finding out the values or a. From (4) we have:
4 a^2 (c - a) + c^2 (c - a) - 2 a c^2 = 0
(c^2 + 4 a^2) (c – a) = 2 c^2 * a
Since a, b and c are integers, the possibilities are: that c > a and
1) c – a = a and c^2 + 4 a^2 = 2 c^2 => c = 2 a
2) c – a = 2 and c^2 + 4 a^2 = a c^2 => 4 a^2 = (a-1)(a+2)^2
a^3 - a^2 – 4 = 0 => a = 2
so : f(y) = 2 y^2 + 2 y + 3 and q(y) = (y + 1)/2
Answer:
Answer will be (3).
p(y) = y^3 + 2y^2 + 3y + 2 q(y) = q(y) [1+ f(y)]
Y
=q(y) *f(y) +
Obviously the degree of q(y) is less than that of f(y). Given that it is more than or equal to 1. If f(y) has a degree 3, then quotient will be of degree 0 and remainder will be of degree 2.
If f(y) is of degree 1, then the remainder will be of degree 0 and quotient will be of degree 2. Thus only possibility is that f(y) has degree 2 and q(y) has degree 1.
(f(y))^2 has a degree 4 and (q(y))^2 has a degree 2. Thus (f(y))^2-(q(y))^2 has a degree of 4. So the option (3) is correct.
Let a, b, and c be integers.
Let f(y) = a y^2 +by+ (c-1)
=dx+e
y^3 + 2y^2 + 3y + 2
[ay^2+by+c] (dy+e) = a dy^3 + (ae + b d) y^2 + (cd+be)y+ce
Comparing coefficients, d = 1/a
Y
e = 2/c
2a/c+b/a=2>2 a^2 + bc=2ac ---(1) c/a + 2b/c = 3 => 2ab+c^2 = 3 a c ---(2)
from: (1): b=2 a(c-a)/ c (2) b = c (3a-c)/2a --- (3)
So, 2 a (c-a)/c = c (3a-c)/2a => 4 a^2 (c-a) = c^2 (3 a-c)
(4) => c^3-4 a^3 + 4 a^2 c-3 a c^2 = 0 divide by c^3 and let a/c = x.
=> 1-4 x^3 + 4x^2-3x = 0
=> 4x^3 - 4x^2+3x-1=0 for x = 1/2 the equation is satisfied.
So c = 2 a and using (3), b = a so then d=1/a and e=1/a
So f(y) = a y^2 + ay+ (2a-1) and so q(y) = (y + 1)/a
Now calculate (f(y))^2 - (q(y))^2
Finding out the values or a. From (4) we have:
4 a^2 (c-a) + c^2 (c-a) -2 a c^2 = 0
(c^2 + 4 a^2) (c-a) = 2 c^2" a
Since a, b and c are integers, the possibilities are: that c> a and
1) c-a = a and c^2 + 4 a^2 = 2 c^2
2) c-a=2 and c^2 + 4 a^2 = a c^2 => 4 a^2 = (a-1)(a+2)^2
a^3 a^2-4 = 0
a = 2
so: f(y) = 2 y^2+2y+3 q(y) = (y +1)/2