# Math.......graphing polynomials 2?

Math.......graphing polynomials?

y = (x-1)^2(x^3+1)

a)the x intercept

b)the y intercept

c)the multiplicity of each zero

d)the degree of the poynomial

Relevance

a) There are TWO x-intercepts. They occur when y = 0; that is when x = 1 or x = - 1.

b) The y-intercept is analogously, just one: y = (-1)^2(0 + 1) = 1.

c) The multiplicity of y = 0 is TWO roots x = 1, and THREE roots of x = - 1. At x = 0, there is ONE root y = 1.

d) The degree of the polynomlal is degree 5 in x, and 1 in y.

Live long and prosper.

• 3 years ago

i'm working in this problem and that i'm hoping to have an answer quickly. as much as date version: i visit end smci's artwork and and tutor here: There are no longer any ideas of the posed problem with m = n = 2 or m=a million, n = 2. enable's handle his equations separately. If m =a million, n = 2, smci has to unravel d = 2(c-a million) ± ?((c-3)^2-6) in integers. enable u = c-3. For d to be an integer u^2-6 could be a sq., say v^2. yet u^2-v^2 = 6 is impossible mod 4, so there are no longer any values of d that artwork. the 1st case, m = n = 2, is extra exciting. we could resolve d² -(3c+2)d +(2c-4) = 0 (*) which calls for 9c^2 + 4c + 20 = t^2, say, and that's a sq. for c = 2, which yields d = 8. Equating like powers of x (and correcting misprints) as smci does, we get a = mn = mg+cf + de b = dg me + nc = 0 mf + ce + nd = 0 df + cg = 0 Now me + nc = 0 yields e = -2. next mf + ce + nd = 0 yields 2f -4 +sixteen = 0 f = -6 ultimately, df + cg =0 yields -40 8 + 2g = 0 g = 24. regrettably, mg + cf + de = 20, no longer 4. If we plug each thing returned in and cancel a 4, we get x^5 + 5x^2 + 40 8 = (x^2+x+4)(x^3-x^2-3x+12), which seems to be new, yet would not fulfill the circumstances of the problem. next, me + nc = 0 implies e = -c So mf -c^2 +nd = 0 and m = n = 2 implies c is even. Now we tutor here effect: If c is even, the only integer ideas of 9c^2 + 4c + 20 = t^2 are c= 2, t = ±8. to work out this notice that c even implies d even, so we are able to positioned c = 2u, t = 2v and get 9u^2 + 2u + 5 = v^2. If we finished the sq. of the left-hand area and multiply by 9, we get (9u+a million)^2 + 40 4 = 9v^2. = (3v)^2. enable x = 9u+a million, y = 3v and we get y^2 - x^2 = 40 4. (y+x)(y-x) = 40 4. hence y+x and y-x could the two be divisors of 40 4. If we try all achieveable pairs, the only ones yielding integer ideas are (22,2) and (-2,-22). So y + x = 22 y - x = 2 yields y = 12, x = 10, u = a million, v = 4, c = 2. and y + x = -2 y - x = -22 yields y = -12 x = 10 and returned, c = 2. hence there are no longer any ideas of the problem with m = n = 2.