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
Since the polynomial has three zeroes it has a degree of 3. So we have:(x - -5)[x - (2 + i)][x - (2 - i)]= (x + 5)[(x - 2) - i][(x - 2) + i]= (x + 5)[(x - 2)2 - i2]= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]= (x + 5)(x2 - 4x + 4 + 1)= (x + 5)(x2 - 4x + 5)= x(x2 - 4x + 5) + 5(x2 - 4x + 5)= x3 - 4x2 + 5x + 5x2 - 20x + 25= x3 + x2 - 15x + 25Thus,P(x) = x3 + x2 - 15x + 25
x2 + 3x - 40 = x2 + 8x - 5x - 40 = x(x + 8) - 5(x + 8) = (x - 5)(x + 8)
30
x2 + 1x - 30 = (x + 6) (x - 5)
The factors of x2 are x and x.
f(x) = x2-4x-5 = x2-5x+x-5 = x(x-5)+1(x-5) = (x+1)(x-5) Factors of f(x) are (x+1) and (x-5).
x2 - 3X - 10 = (X - 5)(X + 2)So, (X2 - 3X -10) / (X-5) = X + 2
x2 + 6x + 5 = (x + 1)(x + 5)
10x + 5 unless you know what x & y are * * * * * Perhaps this answer will be of help: x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). If you are willing to go to complex roots, then, x3 + x2 + 5x + 5 = (x + 1)(x2 + 5). = (x + 1)(x + i√5)(x - i√5); in which case, x = -1 or ±i√5.
5(x2 - 25) 5(x + 5)(x - 5)
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
Since the polynomial has three zeroes it has a degree of 3. So we have:(x - -5)[x - (2 + i)][x - (2 - i)]= (x + 5)[(x - 2) - i][(x - 2) + i]= (x + 5)[(x - 2)2 - i2]= (x + 5)[x2 - 2(x)(2) + 22 - (-1)]= (x + 5)(x2 - 4x + 4 + 1)= (x + 5)(x2 - 4x + 5)= x(x2 - 4x + 5) + 5(x2 - 4x + 5)= x3 - 4x2 + 5x + 5x2 - 20x + 25= x3 + x2 - 15x + 25Thus,P(x) = x3 + x2 - 15x + 25
32
(x2 - x - 20) / (x2 + 8x + 16) = [(x - 5)(x + 4)] / (x + 4)2 = (x - 5) / (x + 4)
if x2 = 5 the value of x has to be 5 divided by 2 which is 2.5 x = 2.5
x2 - 6x + 5 = (x - 1)(x - 5)