Answer to Question #261183 in Algebra for Master J

Question

Given 𝑓(𝑥) = 𝑎𝑥³ + 2𝑥² + 𝑏𝑥 − 20 where 𝑥 − 2 and 𝑥 − 5 are two of its factors determine the real values of 𝑎 and 𝑏. 

Answer 1

f(x) = ax³+2x²+bx-20

Let's find value of f(x) at x = 2, x = 5

f(2) = a*(2)³+2*(2)²+b*2-20

f(2) = 4a + b - 6

f(5) = a*(5)³+2*(5)²+b*5-20

f(5) = 25a + b + 6

Let's add f(2) + f(5) and solve for b

4a + b - 6 + 25a + b + 6 = 29a + 2b

29a +2b = 0

b = -29a / 2

Let's subtract f(2) - f(5) to find a

4a + b - 6 - (25a + b + 6) = -21a - 12

-21a - 12 = 0

a = 12 / (-21)

a = -(4/7)

find b

b = (-29 * (-4/7)) / 2

b = 58/7

Answer: a = - 4/7, b = 58/7