Question

If 2x³+5x²+5x+k is exactly divisible by x²+x+1 . find the value of k.​

Answer 1

Answer:

f(x) = 2x³ +6x² -4x+9 = 0

Comparing the above equation with ax³ +bx² +cx+d = 0, we get,

a = 2, b = 6, c = -4, d = 9

Now,

pqr = -d/a

3r = -9/2

r = -9/6 = -3/2

r = -1.5

Answer 2

Answer

factorofx

factorofx 2

factorofx 2 −5x+4

factorofx 2 −5x+4=(x−1)(x−4)

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0ork=12+3=15

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0ork=12+3=15andP(4)=0

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0ork=12+3=15andP(4)=0256−320+16+4k−12=0

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0ork=12+3=15andP(4)=0256−320+16+4k−12=04k=+60

factorofx 2 −5x+4=(x−1)(x−4)hencex=1and4arezeroofgivenpolynomial∴P(1)=01−5+1+k−12=0ork=12+3=15andP(4)=0256−320+16+4k−12=04k=+60∴k=15

: