Question

If 2x³ + ax + bx - 6 has x-1 as a factor and leaves a remainder 2 when divided by x-2, find the value of 'a' and 'b'?

Answer 1

Answer:

Let P(x) = 2x^3 + ax^2 + bx - 6

If x-1 is a factor of P(x) then the sum of the coefficients of P(x) = 0

So 2 + a + b - 6 = 0 => a+b = 4 ……..(1)

When P(x) is divided by x-2 then the remainder is 2.

Then by remainder theorem P(2) = 2

2×2^3 + a×2^2+2b-6 = 2

=> 4a + 2b = -8 => 2a + b = -4 ……(2)

(2) - (1) gives a = -8 and using this in (1) we get b = 12

Step-by-step explanation:

hope this helps you

Answer 2

Answer:

At the balancing condition, Anticlock wise moment must be equal to clock wise moment.

Clockwise moment, about the balancing point is

= moment by 80 gf

=80gf×(80−X)cm

=6400gfcm−80Xgfcm.....(1)

Anticlock wise moment about the balancing point is

= moment by 10 gf + moment by 50 gf

=10gf×(X−10)+50gf×(X−50)

=10Xgfc,−100gfcm+50Xgfcm−2500gfcm.....(2)

At the balancing condition, Clock wise at equilibrium = Anti clock wise moments.

6400−80X=100X−100+50X−2500

6400+2600=140X⇒140X=8000

⇒X=1408000=64.28

Scale is balanced at 64.28 cm from the beginning.