contestada

the factors of x²+kx+8 are(x+a)(x+b), where a and b are integers. show that there are two possible values of k

Respuesta :

I hope this helps you





x.x+x.b+x.a+a.b


x^2+(b+a).x+a.b


a.b=8

a=1 b=8


a+b=k


k=1+8=9


a=2 b=4


k=2+4=6


a= -1 b= -8


k= -1-8=-9


a= -2 b= -4


k= -2-4=-6





Otras preguntas