Ngoài 1 nghiệm là 2, đa thức f(x)=2x^4 – 9x^3 -18x^2 + 71x -30 còn nghiệm nào nữa? 06/12/2021 Bởi Samantha Ngoài 1 nghiệm là 2, đa thức f(x)=2x^4 – 9x^3 -18x^2 + 71x -30 còn nghiệm nào nữa?
`f(x)=2x^4-9x^3-18x^2+71x-30=0` `⇒(2x^4-4x^3)-(5x^3-10x^2)-(28x^2-56x)+(15x-30)=0` `⇒2x^3(x-2)-5x^2(x-2)-28x(x-2)+15(x-2)=0` `⇒(x-2)(2x^3-5x^2-28x+15)=0` `⇒(x-2)[(2x^3+6x^2)-(11x^2+33x)+(5x+15)]=0` `⇒(x-2)[2x^2(x+3)-11x(x+3)+5(x+3)]=0` `⇒(x-2)(x+3)(2x^2-11x+5)=0` `⇒(x-2)(x+3)[(2x^2-10x)-(x-5)]=0` `⇒(x-2)(x+3)[2x(x-5)-(x-5)]=0` `⇒(x-2)(x+3)(x-5)(2x-1)=0` \(⇒\left[ \begin{array}{l}x-2=0\\x+3=0\\x-5=0\\2x-1=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}x=2\\x=-3\\x=5\\x=\dfrac{1}{2}\end{array} \right.\) Vậy nghiệm của `f(x)` là `2; -3; 5; 1/2` Bình luận
$\begin{array}{l}f(x) = 2x^4 – 9x^3 -18x^2 + 71x -30\\ \to f(x) = 2x^4 -x^3 – 8x^3 + 4x^2 – 22x^2 + 11x + 60x – 30\\ \to f(x) = x^3(2x – 1) – 4x^2(2x – 1) – 11x(2x – 1) + 30(2x-1)\\ \to f(x) = (2x-1)(x^3 – 4x^2 – 11x + 30)\\ \to f(x) = (2x- 1)[x^3 – 5x^2 + x^2 – 5x -6x + 30]\\ \to f(x) = (2x-1)[x^2(x-5) + x(x-5) – 6(x-5)]\\ \to f(x) = (2x-1)(x-5)(x^2 + x – 6)\\ \to f(x) = (2x-1)(x-5)(x-2)(x+3)\end{array}$ Bình luận
`f(x)=2x^4-9x^3-18x^2+71x-30=0`
`⇒(2x^4-4x^3)-(5x^3-10x^2)-(28x^2-56x)+(15x-30)=0`
`⇒2x^3(x-2)-5x^2(x-2)-28x(x-2)+15(x-2)=0`
`⇒(x-2)(2x^3-5x^2-28x+15)=0`
`⇒(x-2)[(2x^3+6x^2)-(11x^2+33x)+(5x+15)]=0`
`⇒(x-2)[2x^2(x+3)-11x(x+3)+5(x+3)]=0`
`⇒(x-2)(x+3)(2x^2-11x+5)=0`
`⇒(x-2)(x+3)[(2x^2-10x)-(x-5)]=0`
`⇒(x-2)(x+3)[2x(x-5)-(x-5)]=0`
`⇒(x-2)(x+3)(x-5)(2x-1)=0`
\(⇒\left[ \begin{array}{l}x-2=0\\x+3=0\\x-5=0\\2x-1=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}x=2\\x=-3\\x=5\\x=\dfrac{1}{2}\end{array} \right.\)
Vậy nghiệm của `f(x)` là `2; -3; 5; 1/2`
$\begin{array}{l}f(x) = 2x^4 – 9x^3 -18x^2 + 71x -30\\ \to f(x) = 2x^4 -x^3 – 8x^3 + 4x^2 – 22x^2 + 11x + 60x – 30\\ \to f(x) = x^3(2x – 1) – 4x^2(2x – 1) – 11x(2x – 1) + 30(2x-1)\\ \to f(x) = (2x-1)(x^3 – 4x^2 – 11x + 30)\\ \to f(x) = (2x- 1)[x^3 – 5x^2 + x^2 – 5x -6x + 30]\\ \to f(x) = (2x-1)[x^2(x-5) + x(x-5) – 6(x-5)]\\ \to f(x) = (2x-1)(x-5)(x^2 + x – 6)\\ \to f(x) = (2x-1)(x-5)(x-2)(x+3)\end{array}$