Đặt phép chia, rồi tính ra. kq= -9
Đúng thì tick cho mình 1 cái nha.
\(q\left(x\right)=x^2-2x-3=\left(x+1\right)\left(x-3\right)\)
\(f\left(x\right)=x^4+px^2+g\)
Để \(f\left(x\right)⋮q\left(x\right)\) thì \(f\left(-1\right)=0;f\left(3\right)=0\)(Theo định lí Bezout)
\(f\left(-1\right)=0\Leftrightarrow1+p+g=0\left(1\right)\)
\(f\left(3\right)=0\Leftrightarrow81+9p+g=0\left(2\right)\)
\(\left(2\right)-\left(1\right)=0\)
\(\Leftrightarrow\left(81+9p+g\right)-\left(1+p+g\right)=0\)
\(\Leftrightarrow80+8p=0\)
\(\Leftrightarrow8\left(10+p\right)=0\)
\(\Leftrightarrow10+p=0\)
\(\Leftrightarrow p=-10\)
