Question

The product of all real integral solutions of (x2-3x+3)*2-16x+48
=1 is​

Answer 1

There is an error in the question the given function in its correct form shall be x3−3x2−16x+48=0

The product of the real integral solution is (x−4)(x+3)(x+4).

GIVEN: x3−3x2−16x+48=0
TO FIND: The product of the real integral solution
SOLUTION:
As we are given in the question,

x3−3x2−16x+48=0

Since the given function is cubic, that is its degree is 3.

Therefore, the same function will have three solutions.

For the first one, we may use the trial and error method,

We put x = -3

f(x) =  x3−3x2−16x+48

f(-3) = (-3)3 - 3(-3)2 - 16(-3) + 48

f(-3) = -27 + 27 -48 + 48

f(-3) = 0

Therefore,

x = -3 is one of the solutions.

Therefore, one integral bracket is x+3.

Now, performing the long division followed by the middle-term split,

We get the solution as,

(x−4)(x+3)(x+4).

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