Question

If f(x) = x2 + 6x + a, g(x) = x2 + 4x + b, h(x) = x2
+ 14x + c and the LCM of f(x), g(x) and h(x) is (x +
8)(x - 2)(x + 6), then find a + b + c. (a, b and c are
constants).
(a) 20
(b) 16
(c) 32
(d) 10

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Answer 1

Answer:

Step-by-step explanation:

Answer 2

Therefore the value of (a + b + c) = 20. option (a) 8s correct.

If f(x) = x² + 6x + a, g(x) = x² + 4x + b , h(x) = x² + 14x + c and the LCM of f(x), g(x) and h(x) is (x +

(x +8)(x - 2)(x + 6).

We have to find the value of a + b + c.

• It is very logical question. We know LCM is the least common multiple of given numbers or functions.
• Here we have to arrange f(x) , g(x) and h(x) in such a way that they contain (x + 8) , (x - 2) and (x + 6) in their factors.

f(x) = x² + 6x + a ,

let's take, a = -16

⇒ x² + 6x - 16 = x² + 8x - 2x - 16 = (x + 8)(x - 2)

g(x) = x² + 4x + b ,

let's take , b = -12

x² + 4x - 12 = x² + 6x - 2x - 12 = (x + 6)(x - 2)

h(x) = x² + 14x + c ,

let's take , c = 48

x² + 14x + 48 = x² + 8x + 6x + 49 = (x + 8)(x + 6)

Now We can see, All the equations are arranged in such a way that they contain only (x + 8), (x - 2) and (x + 6) , don't they ?

So the LCM of them is (x + 8)(x - 2)(x + 6).

hence a = -16, b = -12 , c = 48

∴ a + b + c = -16 - 12 + 48 = 20

Therefore the value of (a + b + c) = 20.

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