Mathematics: Polynomials

2. If a: b = 4:5, find (5a - 2b) : (2a + 5b) .

Ans:
a:b=4:5
a:b=4k:5k , where k is a constant ,

so , a=4k and b=5k,

(5a - 2b) : (2a + 5b)
=5(4k)-2(5k):2(4k)+5(5k)
=(20-10)k:(8+25)k
=10:33//


3. It is given that y partly varies directly as x and partly variesinversely as x. When x = 2, y = 9 ; when x = 3 , y = 11, find

let y=k1x+k2/x

when x=2,y=9
=>9=2k1+k2/2
=>4k1+k2-18=0--(1)

when x = 3 , y = 11;
=>11=3k1+k2/3
=>33=9k1+k2--(2)

(1)+(2):
4k1+k2-18+33=9k1+k2
5=5k1
k1=3,
k2=6

so y=3x+6/x//

(b) the value(s) of x when y = 19

19=3x+6/x
19x=3x^2+6
3x^2-19x+6
(3x-1)(x-6)=0
x=1/3 or x=6//

Q.1
Assume F(x)=ax^3 + bx^2 -5x -2.
By Remainder theorem,
F(-1) = -5, that is -a +b +5 -2 = -5, -a +b = -8............(1)
F(-1/2) = 0, that is -a/8 +b/4 +5/2 - 2 = 0, -a +2b + 20 - 16 = 0
that is -a + 2b = -4.................(2)
(2)-(1) we get b = 4, so a=12.
(b) F(x+1) = a(x+1)^3 + b(x+1)^2 -5(x+1) -2.
Put x = -1 F(x+1) = -2. Therefore, remainder =-2.

Question 1 has some mistake about a and b.