Show that the given value of c is a zero of P(x) and find all other zeros?
831674
1. replace x with -3, calculate and see if you get 0\n2. divide by (x+3) and solve the remaining quadratic equation
865971
p(x)=-3x^2 (x-1) (x+1/3) (x+2)^3\n\n1) what are the roots and multiplicities? \nroots - multiplicity\n0 .............2\n1..............1\n-1/3 .........1\n-2 ............3\nbasically the multiplicity of a root A is the exponent of the factor (x-A)\n.
747036
0.12 \nP(A or B) = P(A) + P(B) - P(A and B) = P(A) + P(B) - P(A)P(B|A) = 0.1 + 0.1 - 0.1(0.8)
54176
The instruction is to write a polynomial function with the zeros 0, 1, and 1+i. Please keep in mind that when a zero consists radicals or imaginary numbers, the conjugate of that zero is also a zero. For exmaple, here 1+i is a zero. Then 1-i must also be a zero.\n\nThis problem is like working backwards. Instead of factoring the polynomial, you are asked to do it backwards, starting with the zeros. Since 0, 1, and 1+i and 1-i are zeros, you would expect the last step of the factoring to look something like this:\n\nx(x-1)[x-(1+i)][x-(1-i)]\n\nNow simplify this expression a bit. It becomes:\n\nx^2-x[(x-1-i)(x-1+i)]\n\nSimplify this expression even further and you get:\n\nx^4-3x^3+4x^2-2x
914435
x > 5\n\n- - - - - -s-
898700
x = 4 ==> (x - 4) = 0\nx = -2 ==> (x + 2) = 0\n(x - 4)(x + 2) = 0\nx^2 - 2x - 8 = 0
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
304995
1{p+5}/u-6\n\nThe 1 can be distributed into the {p+5}\n\n{p+5}/u-6
114568
I don't know if this will be specific enough for you...but\nif x is within range of P (a matrix) ...when multiplying P by a constant, it gets distributed....hence multiplying x term and its corresponding solution.  Therefore, if c>=x, then cP-xx>=0  [If c=x, it will be =0, if c>x, it will be positive.\n\nMatrix P: {1x + 0y = a      cP [ cx    + c(0)y = ca\n           0x + 1y = b]          c(0)x + cy    = cb(note that I have already worked matrix P down to solution form of\n  [1x + 0y = a\n   0x + 1y = b]\nIf x>=0, and c>=0, then cx>=0 and xx>=0.  Hence, if c>=x, then cx>=xx, so cP-xx>=0\n\nIf you didn't mean x as a variable, but instead another matrix...disregard this solution
1066492
x = -4\n10points thank u!
852832
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)\nf(x): a = 1, b = 0, c = -4\ng(x): a = 1, b = -1, c = -12
511329
Start by multiplying both sides by D this gives you\n(D*P) = (2*s*t)\n\nthe divide both sides by 2t which gives you\n(D*P)/(2*t) = s\n\nor\n\ns = (D*P)/(2*t)
255958
Substitute 2 for x and solve for c.\n2^2-3(2)+c=0\nRemember do powers, then multiplication, then subtraction.
622612
Joel, you need some help.  This is one of the easiest questions I have ever heard!  You should know this, man!  You're gonna fail your test if you keep cheating on here and don't actually figure out how it works.  Good luck, brotha'\n\nL = -2 or L(x) = -2
594239
ok so you have p^2 + p, you can factor out p from both parts of the expression, it will end up like this: p(p + 1).  Hope that works.
654517
x^2-2x+1\n(x-1)(x-1)\nso zero at x=1\n\nx^2-x=6\nx^2-x-6=0\n(x-3)(x+2)=0\nzero at x= 3, and x=-2\n\nx^2-3x-4\n(x-4)(x+1)\nzero at x=4  or x=-1\n\nx^2+8x+15\n(x+3)(x+5)\nzero at x=-5 or x=-3\n\nx^2-6x+5\n(x-5)(x-1)\nzero ar x=5, or x=1\n\nx^2+9x+20\n(x+4)(x+5)\nso zero at x=-4 or x=-5
950494
The rational zero theorem says that all possible rational zeros of a polynomial with integer coefficients, can be obtained by dividing the factors of the constant (in this case, it is -3) by the factors of the integer that is multiplied by the highest power of x (in this case, it it 2).\nAll factors of -3 are -1, +1, -3, +3.\nAll factors of 2 are -1, +1, -2, +2.\n\nThus, all possibilities are given by :\n-1/-1, -1/+1, -1/-2, -1/+2,\n+1/-1, +1/+1, +1/-2, +1/+2,\n-3/-1, -3/+1, -3/-2, -3/+2,\n+3/-1, +3/+1,+3/-2, +3/+2\n\nwhich leads to ±1, ±3, ±1/2, ±3/2 as the 8 possibilities.