Find the value of k so that x-3 is a factor of 6x^3-kx^3-5x+63

Question

Find value K so that x-3 will be factor of the given problem above.

Answer 1

Assuming a typo in the equation...

Should it be 6x^3-kx^2-5x+6 ?

divide (x-3) into this polynomial and you get 6x^2+(18-k)x+[3(18-k)-5] with remainder of

3[3(18-k)-5]+6 which must equal 0

3[3(18-k)-5]+6=0

3[54-3k-5]+6=0

3[49-3k]+6=0

3[49-3k]=-6

49-3k=-2

49+2=3k

51=3k

k=51/3=17

or more simply, use Mark M.'s method: f(3) must equal 0 so replace each x with 3 to get the value of k

Answer 2

Do you mean 6x³ - kx² - 5x + 6?

 

If so, then, by the Factor Theorem, x-3 is a factor of              f(x) = 6x³ - kx² - 5x+ 6 if f(3) = 0.

 

f(3) = 162 - 9k - 15 + 6 = 0

 

               -9k = -153

 

                k = 17

 

Answer 3

(x - 3)(Ax^2 + Bx - 21) = (6-k)x^3 - 5x + 63

 

Ax^3 + Bx^2 - 21x - 3Ax^2 - 3Bx + 63 = (6-k)x^3 - 5x + 63

 

Ax^3 + (B - 3A)x^2 -(21+3B)x = (6-k)x^3 - 5x

 

21+3B = 5

     B = -16/3

B - 3A = 0

     A = -16/9

 

6-k = A = -16/9

 

k = 16/9 + 6 = 16/9 + 54/9 = 70/9

 

ANS: k = 70/9