what must be added to x^4+2x^3-2x^2+x-1 so that the result is exactly divisible by x^2+2x-3
tssuyambulingam:
pls i m crying oh............oh.............oh........................................
answer me pls;;;;;;;;;;;;;;;;;;;;;;;;;;;
Answers
Answered by
6
Since the divisor polynomial of degree 2 (quadratic), we add a polynomial of degree 1. M x + K.
P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)
Q(x) = x² + 2x - 3
- 3 has factors 3 and -1 whose sum is 2, coefficient of x.
Q(x) = (x + 3) ( x - 1)
If P(x) is to be exactly divisible by Q(x) then P(x) must be exactly divisible without a reminder by x + 3 as well as x - 1. Then P(-3) and P(1) must both be zero.
P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)
P(1) = 1 + 2 - 2 + 1 - 1 + M + K = 0
M + K = -1 ---- (1)
P(-3) = 81 - 54 - 18 - 3 - 1 - 3 M + K = 0
2 M - K = 5 --- (2)
Solve the two equations: adding them: 3 M = 4
M = 4/3
K = - 7/3
Hence, we can add: 4/2 x - 7/3 to the given polynomial to have the given polynomial exactly divisible by the divisor polynomial.
P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)
Q(x) = x² + 2x - 3
- 3 has factors 3 and -1 whose sum is 2, coefficient of x.
Q(x) = (x + 3) ( x - 1)
If P(x) is to be exactly divisible by Q(x) then P(x) must be exactly divisible without a reminder by x + 3 as well as x - 1. Then P(-3) and P(1) must both be zero.
P(x) = x⁴ + 2 x³ - 2 x² + x - 1 + (M x + K)
P(1) = 1 + 2 - 2 + 1 - 1 + M + K = 0
M + K = -1 ---- (1)
P(-3) = 81 - 54 - 18 - 3 - 1 - 3 M + K = 0
2 M - K = 5 --- (2)
Solve the two equations: adding them: 3 M = 4
M = 4/3
K = - 7/3
Hence, we can add: 4/2 x - 7/3 to the given polynomial to have the given polynomial exactly divisible by the divisor polynomial.
Similar questions