Question
The polynomial ax3+bx2+x–6 has (x+2) as a factor and leaves a remainder 4 when divided by (x–2). Find a and b.
Answer: Option A
:
A
Let
p(x)=ax3+bx2+x−6
By using factor theorem, (x+2) can be a factor of p(x)only when p(−2)=0
p(−2)=a(−2)3+b(−2)2+(−2)−6=0
⇒−8a+4b−8=0
∴−2a+b=2...(i)
Also when p(x) is divided by (x−2) the remainder is 4.
∴p(2)=4
⇒a(2)3+b(2)2+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2...(ii)
Adding equations (i)and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i)we get
−2a+2=2
⇒−2a=0⇒a=0
Hence,
a=0andb=2
Was this answer helpful ?
:
A
Let
p(x)=ax3+bx2+x−6
By using factor theorem, (x+2) can be a factor of p(x)only when p(−2)=0
p(−2)=a(−2)3+b(−2)2+(−2)−6=0
⇒−8a+4b−8=0
∴−2a+b=2...(i)
Also when p(x) is divided by (x−2) the remainder is 4.
∴p(2)=4
⇒a(2)3+b(2)2+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2...(ii)
Adding equations (i)and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i)we get
−2a+2=2
⇒−2a=0⇒a=0
Hence,
a=0andb=2
Was this answer helpful ?
More Questions on This Topic :
Question 2. Degree of a constant polynomial is _____.....
Question 4. If x+x−1=11, evaluate x2+x−2.....
Latest Videos
Latest Test Papers
Submit Solution