For the remainder of the division of x3−2x2+3kx+18 by x−6 to be equal to zero, k must be equal to
0
1
5
272
−272
A
272
B
0
C
1
D
5
E
−272
Open in App
Solution
Verified by Toppr
Remainder theorem: remainder when polynomial P(x) is divided by (x−a) is P(a)
Therefore remainder when P(x)=x3−2x2+3kx+18 is divided by x−6 is
=P(6)=63−2(62)+2k(6)+18=0 (remainder is given zero here )
⇒36−12+2k+3=0
⇒k=−272
Was this answer helpful?
0
Similar Questions
Q1
For the remainder of the division of x3−2x2+3kx+18 by x−6 to be equal to zero, k must be equal to
View Solution
Q2
the polynomial P of x is equal to 4 x3-2 X2 + px + 5 and q of x is equal to x3+ 6 X2 + P leaves remainder A and B respectively when divided by X + 2 find the value of p if a + b is equal to 0
View Solution
Q3
Use synthetic division and the remainder theorem to determine f(1/2) given that f(x)=−3x5+x3−2x2+x−3
View Solution
Q4
If dividend = x4+x3−2x2+x+1, divisor =x−1 and remainder =2, then the quotient will be .
View Solution
Q5
Use synthetic division method for performing the following division. Write the result in the form Dividend = Divisor x Quotient + Remainder.