Question

Seja P(x) =x⁴-(2a-1)x³+(b-3)x²-3x um polinômio de raízes 2 e - 1. Determine a e b

Answer 1

Se 2 e -1 são raízes do polinômio, quando substituirmos esses valores no lugar do x, então P(x) = 0.

$\mathsf{P(2)\Rightarrow2^4-(2a-1)2^3+(b-3)2^2-3\cdot2=0}$

$\mathsf{16-(2a-1)\cdot8+(b-3)\cdot4-6=0}\\\\\mathsf{16-16a+8+4b-12-6=0}\\\\\mathsf{-16a+4b=-6~~~(-1)}\\\\\boxed{\mathsf{16a-4b=6}}$

$\mathsf{P(-1)\Rightarrow(-1)^4-(2a-1)\cdot(-1)^3+(b-3)cdot(-1)^2-3\cdot(-1)=0}$

$\mathsf{1-(2a-1)\cdot(-1)+(b-3)\cdot1-3cdot(-1)=0}$

$\mathsf{1+2a-1+b-3+3=0}$

$\boxed{\mathsf{2a+b=0}}$

Montando um sistema:

$\begin{cases}16a-4b=6\\2a+b=0\to b=-2a\end{cases}$

Substituindo b:

$16a-4\cdot(-2a)=6\\\\16a+8a=6\\\\a=\dfrac{6}{24}\\\\\boxed{a=\dfrac{1}{4}}$

$2a+b=0\\\\2\cdot\dfrac{1}{4}+b=0\\\\\dfrac{1}{2}+b=0\\\\\boxed{b=-\dfrac{1}{2}}$