Question

P(x) = x^4+ax^3-x^2+bx-12 has factors of (x-2) and (x+1). Solve the equation P(x)=0​

Answer 1

because P(x) has factors of (x-2) and (x+1) => x = 2 and x = -1 are the solutions of P(x)

so we have:

`\left \{ {{2^(4)+2^(3)a-2^(2)+2b-12 =0} \atop {1-a-1-b-12=0}} \right.\\=>\left \{ {{8a+2b=0} \atop {a+b=-12}} \right.\\<=>\left \{ {{4a+b=0} \atop {a+b=-12}} \right.\\<=>\left \{ {{a=4} \atop {b=-16}} \right.`

=>
`P(x)=x^(4) + 4x^(3)-x^(2) -16x-12`

With P(x) = 0

=>
`x^(4)+4x^(3)-x^(2) -16x-12 = 0`

=> ........

Explanation: