Question

find the possible values of c if x+2 is a factor of x^3+4cx^2+(c+1)^2 x-6
Factorisation of Polynomials.
Plz solve this problem it is very important​

Answer 1

Answer:

Let p(x) = x3 + ax2 + bx +6

 (x-2) is a factor of the polynomial x3 + ax2 + b x +6

p(2) = 0                                                                                       

p(2) = 23 + a.22 + b.2 +6 =8+4a+2b+6 =14+ 4a+ 2b = 0

7 +2 a +b = 0

b = - 7 -2a -(i)                                                                              

x3 + ax2 + bx +6 when divided by (x-3) leaves remainder 3.

p(3) = 3                                                                                      

p(3) = 33 + a.32 + b.3 +6= 27+9a +3b +6 =33+9a+3b = 3

11+3a +b =1 => 3a+b =-10 => b= -10-3a -(ii)                                    

Equating the value of b from (ii) and (i) , we have

 (- 7 -2a) = (-10 - 3a)

a = -3                                                                                          

Substituting a = -3 in (i), we get

b = - 7 -2(-3) = -7 + 6 = -1

Thus the values of a and b are -3 and -1 respectively

Answer 2

Answer:

6

Step-by-step explanation:

1. (x^3 + 4cx^2 + xc^2+2cx+x-6)/(x+2)
2. You must take coefficients of each polynomials
3. 1     4c      c^2 + 2c + 1       -6

-2     1     4c-2     c^2 + 2c + 1 - 8c + 4     -2c^2 - 4c - 2 + 16c - 8 -6

then

-2c^2 - 4c - 2 +16c -8 = 0

-2c^2+12c -16 = 0

D = 144 - 128= 16

c1 = (-12 +4)/-4 = 2

c2 = (-12-4)/-4 = 4

c1 + c2 = 2+4 = 6