write a first degree factor of p(x)
Answers
Answer:
According to the factor theorem, if (x−a) is a factor of f(x), then f(a)=0
Hence, (x−1) will be a factor of p(x)=6x
3
+3x
2
when p(1)=0
p(1)=6(1)
3
+3(1)
2
=6+3=9
=0
∴ p(1)
=0
Hence, (x−1) is not a factor of p(x)=6x
3
+3x
2
(b) Let f(x) be the required polynomial.
p(x)=6x
3
+3x
2
Let the first degree polynomial to be added be ax+b.
f(x)=p(x)+ax+b
If x
2
−1 is a factor of the f(x), then f(−1)=0 and f(1)=0 [x
2
−1=(x+1)(x−1)]
⇒f(−1)=p(−1)+a(−1)+b=0
⇒6(−1)
3
+3(−1)
2
−a+b=0
⇒−6+3−a+b=0
⇒a−b=−3....(i)
Also, f(1)=p(1)+a(1)+b=0
⇒6(1)
3
+3(1)
2
+a+b=0
⇒a+b=−9....(ii)
Adding (i) and (ii), we get:
2a=−12
∴ a=−6
Substituting a=−6 in (i), we get:
b=−3
∴ ax+b=−6x−3
Hence, −6x−3 is the first degree polynomial that must be added to p(x) to get a polynomial that has (x
2
−1) as a factor.
Step-by-step explanation:
1degree. - liner
2- quadratic
3-cubic
4-bi quadratic
5-quintic
pleace make brainlist
answer