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

Answer:

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

=> $x^{4}+4x^{3}-x^{2} -16x-12 = 0$

=> ........

Step-by-step explanation: