Find the remainder when f of x is divided by x - k and ƒ of x equals 2x3 plus 3x2 plus 4x plus 18 and k equals -2?
Find the remainder when f(x) is divided by (x - k)
ƒ(x) = 2x3 + 3x2 + 4x + 18; k = -2
(x - k)
= (x - (-2))
= (x + 2)
x + 2 = 0
x = -2
By Remainder Theorem
ƒ(x) = 2x3 + 3x2 + 4x + 18
ƒ(-2) = 2(-2)3 + 3(-2)2 +4(-2) + 18
= 2(-8) + 3(4) + 4(-2) +18
= -16 + 12 -8 +18
Thus, the remainder is 6
No number can satisfy these conditions: To have a remainder of 1 when divided by 6, the number must be odd (as all multiples of 6 are even and an even number plus 1 is odd) To have a remainder of 2 when divided by 8, the number must be even (as all multiples of 8 are even and an even number plus 2 is even) No number is both odd and even. → No…
What number can be divided by 4 with remainder 1 divided by 5 with remainder 2 divided by 6 with a remainder 3?
Using the remainder theorum determine the remainder when xcubed PLUS 3xsquared - x - 2 is divided by x PLUS 3 TIMES x PLUS 5 theres brackets around x PLUS 3 and seperatly around x PLUS 5?
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