Find Factor(s) of P(X)?
447887
You probably mean\np(x) = 2x^4 +5x^3 – 3x^2 -2x -6\nThe answer is e) Both b,d\nbecause when you multiply the factors you should get the original expression and (x+3)(2x^3 –x^2 -2)= 2x^4 +5x^3 – 3x^2 -2x -6\nThis was quickly seen because you need x^4 which is only obtainable by x multiplied by x^3 in your problem :)\nHope this helps
254834
if you use quadratic formula you get the factors of \n\nx= ( ( 3 +/- square rt.(5) ) / -6 )
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.
697928
First you group the 'like terms'; in your example, the first two and the last two. Then you see the common factor and pull it out\n\ny^2(x^2-1) - 15 (x^2-1)\nNow you see the common term in both and pull that out\n(x^2-1) (y^2-15)\n\nthus you have your two factors of the polynomial
339712
If 3 is a zero then P(3) = 0.\nP(3) = 27 + 36 - 21 - 10 = 32 \nThus 3 is not a zero.\nSince 3 is not a zero, then x-3 cannot be a factor.\nPossible rational factors are the positive and negative factors of 10\n1,2,5,10\nP(5) = 180\nP(10) = 1320\nThere are no rational factors greater than 3.\nAlso, there is not a zero between 3 and 4.\nC.\nThe zeros are -5, -1, and 2\nP(x) = (x+5)(x+1)(x-2)
26326
A procedure for finding the quadratic factors for the complex conjugate roots  of a polynomial P(x) with real coefficients.\n\n[x-(a+ib)][x-(a-ib)]=\nx^2+2ax+(a^2+b^2)=\nx^2+Bx+C.
902478
wow its been a long time.. i use a calculator to do that now.. but here goes one thing i remember...\n\nlets say we have a polynomial like this\n\nx^2 -3x-10  and we need to factor it..\n\nwe take a look at what factors of -10 we could use that would add up to -3  in this case there are two way to get -10 (-2*5) or (2*-5)  if we add the two sets together one gives us -3 and the other gives us 3 so we can throw out the -2*5 terms.\n\nSo now we lay out the factors.... we need two x's to get x^2 so we need two x's in the first terms (remember FOIL?)\n\nlay the terms out so they give us the correct values\n\n(x+2)(x-5)  now we can use FOIL (firs, outer, inner, last) to check ourselves...\nX*X = x^2\nX*-5 = -5X\n2*x =2x\n2*-5=-10\n\nso we get [x^2] +[-5x] +[2x] +[-10]\nor x^2 -5x+2x-10\ncombine like terms now\n-5x+2x =-3x\n\nwhich gives us\nx^2-3x-10\n\nhope that helps
762587
use remainder theorem\nhttp://en.wikipedia.org/wiki/Remainder_theorem\n"When p(x) is divided by (x-a), the remainder is p(a)"\nUR QUESTION:\np(x) = x^3+3x+k\n(x-a) = (x-1)\np(a) = 6\nhence, p(1)=6\nso, (1)^3 + 3(1) + k = 6\nhence, 1+3+k = 6\nso, k = 2\n\nhence, k = 2
92762
A = P + Prt\n\n==> P(1 + rt)\n\n==> P = A/(1 + rt)\n\n<done>
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)
219588
3x^3 + 2x^2 + x - 6\n= 3x^3 + 5x^2 - 3x^2 + 6x - 5x - 6\n= x(3x^2 + 5x + 6) - 1(3x^2 + 5x + 6)\n= (3x^2 + 5x + 6)(x - 1) \n\nThe factors are : (3x^2 + 5x + 6) and (x - 1) \n\nAlternatively, using polynomial division : \n(3x^3 + 2x^2 + x - 6) divided by (x-1) yeilds \n(3x^2 + 5x + 6)
617264
Since each term has at least x in it, divide out the x by distributive property:\n\nx(x^2 - 4xy + 4y^2)\n\nThen factor the rest by FOIL:\n\nx(x-2y)(x-2y)
1002101
factor 9x^2y^2-9y^2\n9y^2(x^2-1)\n9y^2(x+1)(x-1)
503348
No, the previous answer is wrong.\n\nGroup it like this:\n\n2p + pq -6q -3q^2\n\nthen factor like this:\np(2+q) -3q(2+q)\n\nfactor out the 2+q to get:\n(2+q)(p-3q)\n\nFOIL it out and you will see it is correct.
884668
f(x)=-4x^3+5x^2+8\nf(-3)=-4(-3)^3+5(-3)^2+8\n=108+45+8 is not equal to zero\nso x+3 is not a factor
503153
(x+7)(x+3)\n(x-11)(x+2)\nforget this one.  its AC factoring though\nthis ones AC too
1029303
if you are asking for the factors,it is\n(x-15y)*(x+15y)