If the product of two roots is 6 then the solution of the equation $x 4 + x 3 − 16 x 2 − 4 x + 48 = 0$ is _______.

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(x - 2)(x - 3)(x + 2)(x + 4) = 0

(x - 1)(x + 2)(x + 1)(x + 4) = 0

(x + 2)(x + 4)(x + 2)(x + 5) = 0

(x + 2)(x + 5)(x + 2)(x - 4) = 0

answer is A.

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Detailed Solution

Let’s say that

p (x) = x 4 + x 3 − 16 x 2 − 4 x + 48 = 0

.
Since, if the root is given to be 6, there are many possibilities for the roots, like 1 and 6, -1 and -6, 2 and 3, √12 and √3, 1.2 and 0.5, and even some complex numbers are also possible which can give the product as 6.
We will use the trial method to pick a number and check whether it is a root of the given polynomial or not by making use of the factor theorem.
Obviously, picking very large or very small values will not help because then

x 4

will become either very large or very small as compared to the remaining part of the polynomial.
Let’s pick the numbers 2 and 3 (whose product is 6), and check whether they satisfy the factor theorem or not.

p (2) = 24 + 23 − 16 × 22 − 4 × 2 + 48 = 16 + 8 − 64 − 8 + 48 = 0 p (3) = 34 + 33 − 16 × 32 − 4 × 3 + 48 = 81 + 27 − 144 − 12 + 48 = 156 − 156 = 0

Therefore, both 2 and 3 are the roots of the polynomial.
The factor theorem also tells us that both (x - 2) and (x - 3) must be the factors of the given polynomial.
Since the degree of p(x) is 4, we must have another polynomial, say g(x), of degree 2 as its factor.