Say x^4-5x^2-10x-12 = f(x)(x+2)+c.
If we substitute x = -2 something magical happens:
(-2)^4-5(-2)^2-10(-2)-12 = f(-2)(-2+2)+c
16-20+20-12 = f(-2)*0+c
4 = c
So the remainder must be 4.
I assume you meant x^4 + 5x^2 +10x + 12. The remainder is 28
The result is a polynomial q(x) whose order is one fewer than the order of p(x) and a remainder term of the form b/(x + a).
The answer would be 100, with 34 remainder.
x3-x2+5x-1 with remainder 7, which the final answer would be written as:x3-x2+5x-1+[7/(4x+3)]
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 number exists that has a remainder of 1 when divided by 6, and 2 when divided by 8.
-7
bghgbh
I assume you meant x^4 + 5x^2 +10x + 12. The remainder is 28
The result is a polynomial q(x) whose order is one fewer than the order of p(x) and a remainder term of the form b/(x + a).
-3
8900
True.
The answer would be 100, with 34 remainder.
x3-x2+5x-1 with remainder 7, which the final answer would be written as:x3-x2+5x-1+[7/(4x+3)]
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 number exists that has a remainder of 1 when divided by 6, and 2 when divided by 8.
29From superscot 85: Not so, 29/5 = 5 remainder 4. The correct answer is 57 (plus all multiples of 60, eg 117, 177 etc)
2x^3 - 3x^2 + 4x - 3