Question

If (x – 1) is a factor of Ax 3 + Bx 2 – 36x + 22 and 2 B = 64 A , find A and B

Answer 1

Solution:-
x - 1 = 0
x = 1
Let p(x) = Ax³ + Bx² - 36x + 22
p(1) = A(1)³ + B(1)² - (36 × 1) + 22
⇒ A + B - 36 + 22 =0
⇒ A + B - 14 = 0
⇒ A + B = 14 ...........equation (1)
2^B = 64^A
⇒ 2^B = (2⁶)^A
⇒ B = 6A
⇒ A = B/6
Putting the value of A = B/6 in the equation, we get
B/6 + B = 14
⇒ B + 6B = 84
⇒ 7B = 84
⇒ B = 12
Putting the value of B = 12 in the equation A + B = 14, we get
A + 12 = 14
⇒ A =14 - 12
⇒ A = 2
Hence A = 2 and B = 12
Hence, x - 1 is a factor of 2x³ + 12x² - 36x + 22 and if we divide it by x - 1, we will get 2x² + 14x - 22 as quotient and 0 as remainder.
Let p(x) = 2x³ + 12x² - 36x + 22
p(1) = 2(1)³ + 12(1)² - (36 × 1) + 22
⇒ 2 + 12 - 36 + 22
⇒ 36 - 36 = 0

Answer 2

Answer:

A=2 & B= 12.These are the vales of a and b