SOLUTION: Im totally stumped. Find the value of k that makes the linear expression a factor of a cubic expression. kx^3-2x^2+x-6;x+3 could you show the steps or explain how? Thanks

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Im totally stumped. Find the value of k that makes the linear expression a factor of a cubic expression. kx^3-2x^2+x-6;x+3 could you show the steps or explain how? Thanks      Log On


   

Question 28598: Im totally stumped.
Find the value of k that makes the linear expression a factor of a cubic expression.
kx^3-2x^2+x-6;x+3
could you show the steps or explain how?
Thanks
Found 2 solutions by longjonsilver, venugopalramana:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
for x+3 to be a factor, then putting x=-3 into the original, will give an answer of zero.

so kx%5E3-2x%5E2%2Bx-6 becomes
+k%28-3%29%5E3-2%28-3%29%5E2%2B%28-3%29-6+=+0+
k(-27)-2(9)+(-3)-6 = 0
-27k - 18 - 3 - 6 = 0
-27k - 27 = 0
-27k = 27
--> k = 27/-27
--> k = -1

jon.

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Find the value of k that makes the linear expression a factor of a cubic expression.
kx^3-2x^2+x-6;x+3
could you show the steps or explain how?
HAVE YOU BEEN TAUGHT REMAINDER THEOREM..THEN IT IS EASY..OK LET ME EXPLAIN THE THEOREM IN SIMPLE NUMBERS AND THEN IN NORMAL STATEMENT
SUPPOSE WE DIVIDE 34 WITH 5 WE GET 6 AS QUOTIENT AND 4 AS REMAINDER..WE CAN WRITE THE RELATION BETWEEN THEM AS
34=5*6+4...THIS TRUE FOR ALL DIVISIONS..NOW IN ALGEBRA LET US USE
F(X) FOR 34 THE DIVIDEND
(X-A) FOR 5 THE DIVISOR
Q(X) FOR QUOTIENT 6
AND R FOR REMAINDER 4...SO WE GET OUR EQN.AS
F(X)=(X-A)*Q(X)+R.........................I
AS YOU KNOW IF THE REMAINDER IS ZERO WE SAY THE DIVISOR IS A FACTOR OF DIVIDEND.SO TO KNOW WHETHER A DIVISOR IS FACTOR OR NOT WE HAVE TO FIND THE REMAINDER AND CHECK WHETHER IT IS ZERO OR NOT.NOW IF WE PUT IN EQN.X=A..WE GET
F(A)=(A-A)*Q(A)+R
F(A)=R...SO WE CAN FIND R WITHOUT DIVISION BY FINDING F(A)....THIS IS REMAINDER THEOREM...NOW LET US USE IT...
WE HAVE
F(X)=kx^3-2x^2+x-6............AND x+3 AS THE DIVISOR.SET X+3=0..SO...X=-3
HENCE REMAINDER IS F(-3) AND IT SHOULD BE XERO FOR X+3 TO BE A FACTOR...
SO F(-3)=K*(-3)^3-2(-3)^2+(-3)-6=0
=-27K-18-3-6=0
-27K=27
K=-1