Question

Sum:
If ax²+bx+6 is divided by (2x+1)
and the remainder is I and
2bx²+6x+a is divided by (3x-1), the
remainder is 2, the find value of a and b.


please reply fast people!​

Answer 1

Dividend = Divisor * Quotient + Remainder

Part I
———
Let Quotient be “cx + d”

ax^2 + bx + 6 = (2x + 1) (cx + d) + 1

ax^2 + bx + 6 = 2cx^2 + (c + 2d)x + (d + 1)

Comparing Coefficients on both sides,

2c = a
c = a / 2 ——-> 1

c + 2d = b ——-> 2

d + 1 = 6
d = 5 ——-> 3

Substitute c = a / 2 & d = 5 in equation 2,

(a/2) + 10 = b

a + 20 = 2b

a - 2b = -20 ——-> 4


Part II
———
Let Quotient be “cx + d”

2bx^2 + 6x + a = (3x - 1) (cx + d) + 2

2bx^2 + 6x +a = 3cx^2 + (3d - c)x + (2 - d)

Comparing Coefficients on both sides,

3c = 2b

c = (2/3)*b ——-> 5

3d - c = 6

3d - (2/3)*b = 6

3d = 6 + (2/3)*b

d = 2 + (2/9)*b ——-> 6

2 - d = a

2 - 2 - (2/9)*b = a

a = -(2/9)*b

9a + 2b = 0 ——-> 7

Solving equations 4 & 7,

a - 2b = -20

9a + 2b = 0

Adding both the equations, we get

10a = -20

a = -2

b = 9

Therefore answer is a = -2 & b = 9