1.

If LCM of the two polynomials (x^2 + 3x)(x^2 + 3x + 2) and (x^2 + 6x + 8)(x^2 + kx + 6) isx(x + 1)(x + 2)^2 (x + 3)(x + 4), then find k.​"please answer fast"

Answer»

\mathfrak{\huge \underline{Question:}}

If~<klux>LCM</klux>~of~the~two~polynomials~\\\\ (x^2 + 3x)(x^2 + 3x + 2) ~and~ (x^2 + 6x + 8)(x^2 + kx + 6)\\\\is~x(x + 1)(x + 2)^2 (x + 3)(x + 4), then find ~\bf k.

\mathfrak{\huge \underline{Answer:}}

\huge \boxed{\fcolorbox{black} {pink}{K=5}}

\mathfrak{\huge \underline{Explanation:}}

FACTORIZE both the polynomial:

(x^2 + 3x)(x^2 + 3x + 2) = \bf x(x+3)(x+1)(x+2) \\\\(x^2 + 6x + 8)(x^2 + kx + 6)=\bf(x+2)(x+4)(x^2+kx+6) \\\\ LCM~of~both~the~polynomial = x(x+3)(x+1)(x+2)(x+4)(x^2+kx+6).

But, LCM GIVEN is

x(x + 1)(x + 2)^2 (x + 3)(x+4)

So,

x(x+3)(x+1)(x+2)(x+4)(x^2+kx+6)=x(x + 1)(x + 2)^2 (x + 3)(x + 4)\\\\ \cancel{x(x+3)(x+1)(x+2)(x+4)}(x^2+kx+6) =\cancel{x(x + 1)(x + 2)(x + 3)(x + 4)}(x+2) \\\\ x+2 is~ a ~factor ~of ~x^2+kx+6

So, x+2= 0, x= - 2(by remainder THEOREM)

Putting the value of x=-2~ in~ x^2+kx+6, we GET

4-2k+6=0

10-2k=0

2k=10

k=5


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