Question

x³-3x²+2x-6-xy-3y solve this sum using factorization method.

Answer 1

Answer:

P(x)=x3−3x2+2x−6,P(3)=33−3(3)2+2(3)−6=0  

So  x−3  is a factor and  x=3  is a zero of above polynomial.

So,  P(x)=(x−3)(x2+ax+b)=x3+(a−3)x2+(b−3a)x−3b  

Comparing coefficient of  x2,  

a−3=−3,(a=0)  

Comparing coefficient of  x0,  

−3b=−6,(b=2)  

P(x)=(x−3)(x2+2)=0  

x=3,±i2–√  

So given polynomial has only one real zero  x=3.

Step-by-step explanation: