Question

इन फार्मास्यूटिकल प्रिपरेशन जिंक सल्फेट इज यूज्ड एस एम सी क्यू​

Answer 1

Explanation:

(a) If (x+1) is a factor of p(x)=6x3+3x2, then p(−1)=0

P(−1)=6(−1)3+3(−1)2=−6+3=−3=0

Hence, (x+1) is not a factor of p(x)=6x3+3x2

(b) Let f(x) be the required polynomial.

p(x)=6x3+3x2

Let the first degree polynomial to be added to ax+b.

f(x)=p(x)+ax+b

If x2−1 is a factor of the f(x), then f(−1)=0 and f(1)=0

f(−1)=p(−1)+a(−1)+b

⇒0=6(−1)3+3(−1)2−a+b

⇒0=−6+3−a+b

⇒a−b=−3....(i)

Also, f(1)=p(1)+a(1)+b

⇒0=6(1)3+3(1)2+a