Question

Cho 2 đa thức :f (x)=3x^4+2x^2-2x^4+x^2-5x

g (x)=x^4-x^2-2x +6+3x^2

Tìm đa thức h(x) sao cho h(x )+g(x)=f(x)

Tính h (-1/3) h (3/2)

Tìm nghiệm đa thức h(x)

Answer 1

h(x) + g(x) = f(x)

=> h(x)= f(x) - g(x) = \(3x^4+2x^2-2x^4+x^2-5x-\left(x^4-x^2-2x+6+3x^2\right)=x^2-3x-6\)\(h\left(-\dfrac{1}{3}\right)=\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)-6=\dfrac{-44}{9}\)

\(h\left(\dfrac{3}{2}\right)=\left(\dfrac{3}{2}\right)^2-3\cdot\dfrac{3}{2}-6=-\dfrac{33}{4}\)

\(x^2-3x-6=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{6}\\x=\dfrac{3-\sqrt{33}}{6}\end{matrix}\right.\)