There exists a polynomial $f(x)$ and a constant $k$ such that

(x^2 - 2x - 5) f(x) = 2x^4 + 19x^3 + kx^2 - 15x - 1.

What is $k?$

booboo44 Feb 13, 2024

#1**+2 **

Looking at the terms of both sides, we see that a quadratic * f(x) is a quartic, so f(x) must be a quadratic.

We additionally know that the coefficient of the quadratic term of f(x) is 2, because the only way 2x^{4} can be formed, is the multiplying of the quadratic terms, and since on quadratic term is x^{2}, the other term must be 2x^{4}/x^{2} = 2x^{2}.

Similarly, applying this same trick, we know that the constant term of f(x) must be 1/5, because only constant terms * constant terms can create constant terms, and on constant term is -5, so the other must be 1/5, so that -5*1/5 = -1. Another way to visualize this is by plugging in 0 as x.

Now we need the linear term.

We know f(x) so far as 2x^{2}+ax+1/5.

To find the linear term, we consider that (x^{2}-2x-5)*f(x) is the quartic.

(x^{2}-2x-5)*(2x^{2}+ax+1/5)=2x^{4}+19x^{3} ...

From here we can directly get a. We see that the cubic term 19x^{3} on the right, should also be equal to the cubic term of the expansion on the left.

We notice that cubic terms in the expansion can only be formed from quadratic terms * linear terms.

The cubic terms on the expansion on the left is -2x*(2x^{2}) = -4x^3 and x^{2}*ax, so the cubic term in the expansion is (-4+a)x^{3},

We equate the two cubic terms getting, -4+a = 19. a = 23.

But we see that (x^{2}-2x-5)(2x^{2}+23x+1/5) is not the correct expansion, and polynomials can't have negative exponents, I believe that this question is not possible.

hairyberry Feb 13, 2024