Question

x^2+px+q(q≠0) is a factor of x^3-ax^2+bx-c..find q(a-p)=?​

Answer 1

P(x)=ax2+bx+c

P(x)=ax2+bx+cHence, for real roots,

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+c

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, 

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0Hence, P(x).Q(x) has atleast 2 real roots .

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0Hence, P(x).Q(x) has atleast 2 real roots .If ac>0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0Hence, P(x).Q(x) has atleast 2 real roots .If ac>0Then

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0Hence, P(x).Q(x) has atleast 2 real roots .If ac>0Thend2+4ac>0

P(x)=ax2+bx+cHence, for real roots,b2−4ac≥0Q(x)=−ax2+dx+cHence, for real roots,d2+4ac≥0Now, ac=0Now if ac<0Then,b2−4ac>0Hence, P(x).Q(x) has atleast 2 real roots .If ac>0Thend2+4ac>0Hence, P(x).Q(x) has atleast 2 real roots .