3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
-7.5?
x^5+2x^4+4x^2+2x-3
3x2(x2 - 3)
(2x)ysquared
3x4 plus 5x3 plus x2 - 5 divided by x 2 =[(3x4) + (5x3) + (x2 - 5)]/x2 =(12 + 15 + x2 -5)/x2 =(27 - 5 + x2)/x2 =(22 + x2)/x2
-7.5?
(3x4 + 2x3 - x2 - x - 6)/(x2 + 1)= 3x2 + 2x - 4 + (-3x - 2)/(x2 + 1)= 3x2 + 2x - 4 - (3x + 2)/(x2 + 1)where the quotient is 3x2 + 2x - 4 and the remainder is -(3x + 2).
x2+7x+12
it is 3. You are doing APEX right?
x2 plus 3x plus 3 is equals to 0 can be solved by getting the roots.
x^5+2x^4+4x^2+2x-3
x6 + 3x4 - x2 - 3 = 0(x6 + 3x4) - (x2 + 3) = 0x4(x2 + 3) - (x2 + 3) = 0(x2 + 3)(x4 - 1) = 0(x2 + 3)[(x2)2 - 12] = 0(x2 + 3)(x2 + 1)(x2 - 1) = 0(x2 + 3)(x2 + 1)(x + 1)(x - 1) = 0x2 + 3 = 0 or x2 + 1 = 0 or x + 1 = 0 or x - 1 = 0x2 + 3 = 0x2 = -3x = ±√-3 = ±i√3 ≈ ±1.7ix2 + 1 = 0x2 = -1x = ±√-1 = ±i√1 ≈ ±ix + 1 = 0x = -1x - 1 = 0x = 1The solutions are x = ±1, ±i, ±1.7i.
3x2(x2 - 3)
(2x)ysquared
-130
(x+5)^2 and therefore x = -5