How many real roots does a polynomial of third degree have?
484948
A 3rd degree polynomial can have a maximum of 3 real roots, but it must have at least 1 real root I believe.
885151
The zeros tell us that x-1, x-2 and x-4 are factors, so the polynomial is  \nf(x) = k*(x-1)(x-2)(x-4) where k is some constant.\n\nf(0) = k*(-1)(-2)(-4) so you just choose the value of k that makes this -16.\n\nIf 3 + i is a root then so is 3 - i, and multiplying\n(x - (3+i))(x - (3-i)) gives\n\nx^2 - 6x + 10, so this is your quadratic polynomial.\n\nObviously the linear factor is x - 4 since  4 is a root.  Leading coeff 1 means you don't have to bother to multiply these factors by some other constant coefficient.
941571
1) f(x) = A(x-1)(x-2)(x-4) for some A. Substitute in 0 to get -16 = A*-1*-2*-4, so A = 2. So 2(x-1)(x-2)(x-4).\n2) Whenever you have a polynomial with real coefficients and a+bi is a root, you also have a-bi is a root. So you know the 3 roots are 4, 3+i, 3-i. Thus f(x) = (x-4)(x-(3+i))(x-(3-i)). Multiplying those last two together, you get x^2 - x(-3-i-3+i) + (3+i)(3-i) = x^2 - 6x + (9-i^2) = x^2-6x+10. So f(x) = (x-4)(x^2-6x+10).\n3) I don't think you wrote the question down properly. Expanding the left hand side gives 3x^3 + (-a-27)x^2 + (9a+42)x - 14a. You need a=5 for the first part to work, but that doesn't give what you wrote.
142543
Polynomial= multiple terms, i.e. 3x^2+7x-3\nSo I suppose addition of them would be like\n(4x^2+7x-5)+(9x^2+9x+7+y)\n=4x^2+7x-5+9x^2+9x+7+y\n=13x^2+16x+2+y\nor\n(4x^2+7x-5)-(9x^2+9x+7+y)=\n(4x^2+7x-5)-1(9x^2+9x+7+y)=\n(4x^2+7x-5)+ (-1*(9x^2+9x+7+y))=\n(4x^2+7x-5)+(-9x^2-9x-7-y)=\n4x^2+7x-5-9x^2-9x-7-y=\n-5x^2-2x-12-y.\n\nMaybe now you can do your own.\nNote that these methods only work with adding/subtracting.
799253
Polynomial is a general term for any expression, including monomial, binomial, trinomial, etc.  Monomials have just one term with no + or -, such as x or 3x or 5x^6.  Binomials are two monomials added or subtracted, such as 3x + 5 or 5x^2 - 6x.  The degree of a polynomial is the highest exponent of any of its variables, assuming there is only one variable.  So for x^3 + 8x^2 + 3x - 4, the degree is 3.
355086
2x^3 + x - 2 has only one real root and it is positive.\n(I used Decartes Rule of Signs to figure out the number of possible roots.)\nUsing my graphing calculator I get x = .83512235
593998
When you have a polynomial, the coefficients are integers.  For example:\n\nax^2 + bx + c = 0\n\nThis is the general formula for a second degree polynomial (x is to the 2, hence second degree).  In this case, a, b, and c are the integers.\n\nAlso, if you add other terms, such as x^3, they would also be multiplied by integers.
923617
First, factor out 3x from the equation, so 0= 3x(x^2-2x-3).\n\nThen, factor the polynomial in the parenthesis, so 0= 3x(x-3)(x+1)\n\nAfter that, use the zero product property to set each part of the the equation to zero, so 0=3x, 0=x-3, and 0=x+1.\n\nSolve these three to get x=0, x=3, and x=-1, respectively.
188735
x^256-1 is a polynomial of degree 256. It has 2 pure real roots (+1, -1)  and 254 complex roots. They are all of the form (cos(n2pi/256) + isin(n2pi/256)) n = 0, 1, ...255.\nIt has 256 factors of the form (x-a) where a is one of the 256 roots.
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
715426
x^4 + 5x^3 - 30 / x^2 - 3\n\nThe important thing is that, especially with division, polynomials need place value so 0 coefficients apply.\n\nRewrite as follows\n\nx^4 + 5x^3  + 0x^2 + 0x - 30 / x^2 + 0x - 3\n\n.................... x² + 5x + 3\n................. ____________________\nx² + 0x - 3 )x^4 + 5x³ + 0x² + 0x - 30\n.................. x^4 + 0x³ - 3x²\n.................  ------------------ .....↓\n........................... 5x³ + 3x² + 0x\n........................... 5x³ + 0x² - 15x\n........................... ------------------ ....↓\n.................................... 3x² + 15x - 30\n..................... ............... 3x² + 0x - 9\n.................................... ------------------\n....................... ................... 15x - 21\n\nSo x^4 + 5x³  - 30 = (x²  - 3)(x² + 5x + 3) + 15x - 21\n(Or quotient is x² + 5x + 3 and the remainder is 15x - 21)
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.
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!!
539132
both roots are complex numbers
800072
It has one real solution.  Irrational roots are always in groups of two, so there could have been one or three real solutions.\n\nThe real solution is -0.531447857.  Unfortunately, there is not any easy way to find this one other than guess and check.\n\nYou will also note that plotting the line x^3-4x^2+7x+5=y is always increasing.  (The first derivative is always positive).  This tells you that you are guaranteed to only have one real solution.\n\nAnd 2.26... doesn't work
854837
Show it's a zero just means you sub x = -3 in the polynomial and get the result zero.  If that happens, you can divide the polynomial by \nx -(-3),  i.e. x + 3, and then find the zeros of the remaining expression.  Want to do it yourself?  Good.  Then don't scroll down to see what I've done.\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n-3      2     5     -6    -9\n\n              -6       3      9\n\n         2    -1     -3      0\n\nP(x) = (x + 3)(2x^2 - x - 3)\n       =  (x + 3)(x + 1)(2x - 3)\n\nZeros are -3, -1, 3/2
653663
Do you mean real AND imaginary? Or just all imaginary?  Well, either way, since we \nhave a polynomial with four factors, we will factor by grouping to simplify.\n\nf(x)=(x^3)-(9x^2)+26x-24\n\nf(x)=[(x^3)-(9x^2)+26x]-24\n\nf(x)=x[(x^2)-9x-26)]-24\n\nf(x)=(x-24)[(x^2)-9x+26]\n\nWe have one real solution. Our linear factor is (x-24), so our Real solution is x=24.\n\nNow we need to use the quadratic formula:\n\n{-b(+or-)[The Square Root of:(b-4ac)]}ALL over 2a.\n\n[(x^2)-9x+26] a=1, b=-9, c=26\n\n-(-9) ± √(-9-(4*1*26))ALL divided by 2\n\n9 ± √(-9-104)ALL divided by 2\n\n9 ± √(-113)ALL divided by 2\n\nSo our two imaginary solutions are:\n\n [9+√(-113)]/2] and [9-√(-113)]/2]\n\nRecall that i=√(-1)\n\nSo, our first solution [9+√(-113)]/2]=9+i√(113)or 9+√(113)i\n\nand our second solution [9-√(-113)]/2]=9-i√(113)or 9-√(113)i\n\nTherefore, All Solutions, Real and Imaginary, for \n\nf(x)=(x^3)-(9x^2)+26x-24 are:\n\nx=24, x=9+√(113)i, and x=9-√(113)i\n\n:)