The remainder when x5 + kx2 is divided by (x- 1)(x- 2)(x- 3) contains no terms in x2. find k without performing division.
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Answered by
66
(x – 1) (x – 2) (x – 3) = x3 – 6x2 + 11x – 6
Let the quotient and remainder when
x5 + kx2 is divided by (x – 1) (x – 2) (x – 3) be
(ax2 + bx + c) and dx + e
Now
x5 + kx2 = (x3 – 6x2 + 11x – 6) (ax2 + bx + c) + dx + e
Comparing the coefficients of x5, x4, x3 and x2 we get
a = 1
b – 6 a = 0 ⇒ b = 6
c – 6 b + 11 a = 0 ⇒ c = 25
– 6 c + 11 b – 6 a = k
⇒k = –90
Hence, the value of k is –90
hope Helpful mark as brilliancy...
Let the quotient and remainder when
x5 + kx2 is divided by (x – 1) (x – 2) (x – 3) be
(ax2 + bx + c) and dx + e
Now
x5 + kx2 = (x3 – 6x2 + 11x – 6) (ax2 + bx + c) + dx + e
Comparing the coefficients of x5, x4, x3 and x2 we get
a = 1
b – 6 a = 0 ⇒ b = 6
c – 6 b + 11 a = 0 ⇒ c = 25
– 6 c + 11 b – 6 a = k
⇒k = –90
Hence, the value of k is –90
hope Helpful mark as brilliancy...
Anonymous:
Plz explain answer In detail
Answered by
34
Answer:
k=-90
Step-by-step explanation:
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