Find all the rational zeros of the polynomial?
827041
Obviously 1 isn't worth trying because 90 is so much bigger than the other terms, but you should check that\nP(2) = 0, and so\nP(x) = (x - 2)(x^3 + x^2 - 21x - 45)\n[You know how to do long division to get this?]\nSub -3 in the second factor and find\nP(-3) = 0\nand so x + 3 is a factor, and then\nP(x) = (x - 2)(x + 3)(x^2 -2x  -15)\nand you should be able to finish it from there.
945878
I'm assuming you are looking for values of b that will give rational roots. That is, the discriminant will be a perffect square or 0. If this is not what you wanted, I'm sorry.\n\n\n3x^2 + bx -3 =0\nb^2 + 36= perfect square if b=0 or b=8\n\n5x^2 +bx +1=0\nb^2-10 =perfect square for no value of b\n\n-3x^2 +bx -3=0\nb^2 -36 = perfect  square fo r b=6 and b=10
760715
If it is divisible by x - 3, then (x - 3) is a factor of that polynomial, and if x - 3 = 0, x = 3.  \n\nIf we substitute x = 3 to the polynomial, then it must yield 0 (since 3 is a root).\nx³ + 3x² + kx - 24\n\nThus,\n(3)³ + 3(3)² + k(3) - 24 = 0\n\nThen,\n27 + 27 + 3k - 24 = 0\n\nTranspose,\n3k = -30\n\nThus,\nk = -10\n\nThe polynomial then becomes\nx³ + 3x² - 10x - 24 = 0\n\nWe then divide it by x - 3, to factor it out.  The easiest method goes like this\nx³ - 3x² + 6x² -18x + 8x - 24 = 0\n\nNow, we group\n(x³ - 3x²) + (6x² - 18x) + (8x - 24) = 0\n\nWe factor out from each group\nx²(x - 3) + 6x(x - 3) + 8(x - 3) = 0\n\nWe then factor out x - 3\n(x - 3)(x² + 6x + 8) = 0\n\nWe can factor the quadratic trinomial\n(x - 3)(x + 2)(x + 4) = 0\n\nTherefore, your polynomial is also divisible by (x + 2) and (x + 4).\n\n^_^
789094
divide by x^2\n6(x^2+(1/x^2))+ (x)+(1/x)+10 = 0\nput x+(1/x)= t\n\nthen solve for t and then solve for x\nyou'll find that there are no roots.!
729558
I worked this problem. I did get both sides to zero. I don't even know how to begin explaining it.\n\nFor the sake of typing this with some readability, let (k+1) = b and (k+2) = c.\n\nFirst you subtract (2^c)/(k+3) from both sides to get 0 on one side. Then to get a common denominator, you multiply top and bottom of the first term by (k+3) and top and bottom of the third term by (k+2). You get:\n\n[ (k+3)(2^b) + (k+1)(2^b) - (k+2)(2^c) ] / [ (k+2)(k+3) ] = 0\n\nThen you distribute out the terms:\n\n[ k(2^b) + 3(2^b) + k(2^b) + (2^b) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you factor out 2^b from the first four terms:\n\n[ (2^b)(k + 3 + k + 1) - k(2^c) - 2(2^c) ] / [ (k+2)(k+3) ]\n\nThen you combine like terms on the left four. On the right two, you reduce the exponent by one and multiply the term by two. Since c is one more than b, this turns (2^c) into (2^b). You can do this since it's an exponent on two:\n\n[ (2k + 4)(2^b) - 2k(2^b) - 4(2^b) ] / [ (k+2)(k+3) ]\n\nYou can factor 2^b out of the right two terms to get:\n\n[ (2k + 4)(2^b) - (2k + 4)(2^b) ] / [ (k+2)(k+3) ] = 0\n\nThe two terms in the numerator cancel out:\n\n0 / [ (k+2)(k+3) ] = 0\n\nAnd:\n\n0 = 0\n\nI must admit that I liked this problem. I don't do math for fun, but every once in a while I like a challenge.\n\nI hope this helped.
875938
5\n\ntake the highest power in the polynomial, that exponent is your leading exponent, and is used as your "n"\n\nso 5 zeros.
51216
x = [-(-4) +/- root((-4)^2 - 4x1x(-5))]/2x1\n  = (4 +/- 6)/2\n  = 5 or -1
660785
i don't fucking know.\n\nEDIT: waitaminnit i'm gonna give it a try!\n\nx^3-4x^2-7x+10 = 0\n\nfor x = -1\n-1 -1 +7+10 != 0\nso x != -1\n\nfor x = 1\n1-1-7+10 != 0\nso x != 1 either!! ...goshdarnit.\n\nx^3-4x^2-7x+10 = 0\nx(x^2-4x) - 7x = -10\nx^2-4x - 7 = -10/x (we divided both sides by x)\n..ok...i think i've lost the plot sorry.
1164377
There are zeros at 2, -3/2 and -6, so\n\nf(x) = (x-2)*(x+3/2)*(x+6)\n\nf(x) = x^3 + 11/2*x^2 - 6*x -18
877357
#1\n\nt=0 or -3\n\n#2\n\np^2 - (1/4)^2 which is p^2 - 1/16
38209
actually polynomial functions make me happy...they are actually pretty easy.  You evaluate each term (thats the stuff separated by the addition (+) or subtraction (-) sign) individually.  So if you had sin(x) -x^3 + 3x\n\nthen you have the following\n\ncos(x) -3x^2 +3\n\nhope this helps!!
1122465
Just plug in 1 for x.\n√(1-1) = √0 = 0
712942
x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0\nx^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 - 10x^3 = 0\nSo (x - 1)^6 - 10x^3 = 0\nSo ((x - 1)²)³ - (³√10x)³ = 0\n\nSo ((x - 1)² - ³√10x)((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0\n\nSo ((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0 N0 real roots (Min value near x = 0.2 just more than 0.87)\n\nOR (x - 1)² - ³√10x = 0 \nie x² - (2 + ³√10)x + 1 = 0\nie x = ½[(2 + ³√10) ± √(4 + 4³√10 + ³√100 - 4)\n= ½[(2 + ³√10) ± √(4³√10 + ³√100)]\n\nSOLUTION\nx = ½[(2 + ³√10) ± √(4³√10 + ³√100)]
913484
are you talking about a polynomial :\na + bx +cx^2 + dx^3 +...\n\nbecause those a,b,c,d... do not appear any where on the graph... you have to look at the intercepts and the extreama, and the inflection points to get any good idea about the shape of the graph
673027
In order to find a polynomial equation with real coefficients that has roots 3-i, you need to understand that if 3-i is a root, then, 3+i is also a root because 3+i is the conjugate of 3-i. Then, you should multiply (3-i) with (3+i) using foil, or, distributive properties. You also need to understand that the imaginary number i\n\n\n(3-i)(3+i) 9+3i-3i-i² = 9-(-1)=10
621340
The first one is (x-2)(x+9)(x-7).\nHowever, for the second part, everyone else has forgotten that imaginary zeroes come in CONJUGATE PAIRS.\nSo there are actually 5 zeroes to part 2, 4i, -5, -2i, 2i, -4i.\n\nSo the solution would be (x-4i)(x+4i)(x+5)(x-2i)(x+2i)\n(x^2+16)(x+5)(x^2+4)\nx^5 + 5x^4 + 20x^3 +100x^2 + 64x + 320\nViola...
1032854
given the zeroes of y:\nif x = -4, 0 or 2, then y = 0.  Therefore it passes through\n(-4, 0)\n(0,0)\n(2,0)\n\nIn other words -4, 0 and 2 are x-intercepts of the polynomial.\n\nThe polynomial (3rd degree) can be expressed as:\ny = x³ + ax² + bx + c\nIt can also be expressed as\ny = (x - x1)(x - x2)(x - x3) \nwhere n is a constant.\n\nWhen you substitute x1 to the expression\ny = (x1 - x1)(x1 - x2)(x1 - x3)\ny = (0)(x1 - x2)(x1 - x3)\ny = 0\n\nTherefore, x1 is a zero\n\nThe same goes with x2 and x3. Therefore you can put the zeroes into our expression\n\ny = (x - (-4))(x - 0)(x - 2)\ny = (x + 4)(x)(x - 2)\ny = x(x + 4)(x - 2)\nor\ny = x(x² + 2x - 8)\ny = x³ + 2x² - 8x\n\nThat is your 3rd degree polynomial! Hope this helps^_^