2007-02-27 22:07:35 · 4 answers · asked by sikirutaye 1 in Science & Mathematics ➔ Mathematics
Can someone help solve this eqn X3-2x2+kx-10 has a remainder 3 when divided by x+2?
2007-02-27 22:12:21 · update #1
Can someone help solve this eqn X3-2x2+kx-10 has a remainder 3 when divided by x+2
Let f(x) = X3-2x2+kx-10
By the remainder theorem f (-2) = 3 (since f(x) has a remainder 3 when divided by x+2)
Therefore
(-2)^3 - 2(-2)^2+k(-2) - 10 = 3
ie -8 - (-8) - 2k - 10 = 3
So 2k = -13
k = -6.5
2007-02-27 22:15:47 · answer #1 · answered by Wal C 6 · 0⤊ 0⤋
By synthetic division I get
(x^3 -2x^2 + kx - 1-)/(x + 2) = x^2 - 4x + (k+8) with a remainder term of 2k-26. If that's equal to 3, then k is 29/2.
If you're looking for the roots of the equation, plug in the value for k and solve it using Cardano's Formula (which was actually first discovered by Ferro, an earlier Italian mathematician)
HTH ☺
Doug
2007-02-27 22:22:58 · answer #2 · answered by doug_donaghue 7 · 0⤊ 1⤋
so you are told to solve for the value of k given such condition.
Use the remainder theorem
replacing x=-2 and equating the equating to 3
(-2)^3 - 2(-2)^2 + k(-2) - 10 = 3
-8 -8 -2k -10 =3
-2k =26+3
k= 29/3
2007-02-27 22:17:23 · answer #3 · answered by duntoktomee 2 · 0⤊ 1⤋
Replace every equation in the above eqn by x+2 and then put it equal to 3 now solve for x.
2007-02-27 22:13:44 · answer #4 · answered by Anonymous · 0⤊ 1⤋