A polynomial of degree 3 that has three real zeros, only one of which is rational?
A rational number can be expressed as the ratio of two integers,like 5/1 or 10/3, versus an irrational number, like e or π or √2, that goes on and on forever with non repeating numbers.
A sixth degree polynomial function has r distinct real zeros. What are the possible values of r?
Hi, The maximum number of different solutions a 6th degree polynomial can have is 6. However, since a polynomial like x² + 9 = 0 has no real roots, a polynomial can have from 0 to 6 distinct real zeros. <==ANSWER I hope that helps!! :-)
How many real roots does a polynomial of third degree have?
A 3rd degree polynomial can have a maximum of 3 real roots, but it must have at least 1 real root I believe.
What are the nature of roots of f(x) if it is a polynomial with rational coefficients and of degree 3 and also touches the x-axis?
As it is a cubic polynomial it must have a real root. Since the curve touches the x axis one real root in guaranteed. Now if a function touches the x axis then that particular point must be a repeated root of the function. So now we have two real roots (but not distinct) .Since the other root can not be imaginary( imaginary roots always occur in pairs) ,the other root must also be real. So the polynomial has three real roots, two of which are same.Say for example this graph. It touches the x axis at origin. So the origin is a repeated root of the function. Also it must have another real root which is at x=-2.The equation to the curve is thus [math](x-0)(x-0)(x+2)=x^2(x+2)=x^3+2x^2[/math]Hope that was useful.Okay so whether the roots are rational or not starts from here.The polynomial must be of the form [math](x-a)^2*(x-b)[/math] as we have discussed above. Expanding this we get [math]x^3-x^2*(b+2a)+x*(x^2+2ab)-a^2*b=0[/math]Its given that the coefficients are rational. So b+2a must be rational. So either [math]b+2a=0[/math] or both are rational.If both are rational the roots must be rational.In the other case b=-2aThe constant terms is [math]-a^2*b[/math] which is also rational.Substituting b we get the coefficient as [math]-2a^3.[/math] For it to be rational a has to be rational. This implies that b also has to be rational because b=-2a. So a and b are both rational, which means that the roots are rational.So in either case, we will have rational roots.
Is it correct that every polynomial equation of odd degree must have at least one real root?
Yes, it is absolutely correct.First of all, you should know, complex roots appear in conjugate pairs in case of a polynomial equation with real coefficients. Now, let, f(x) is a polynomial with degree 'n'(n is odd). So, f(x) will have n roots. Since, n is odd therefore (n-1) is even. Now, as complex roots appear in conjugate pairs only i.e. they appear even no. of times only so if we consider f(x) to have minimum real roots, f(x) can have only 1 real root & the rest roots will be complex no.s[since, (n-1) is even]. e.g. If we take, n=3, the minimum no. of real roots that f(x) can have is 1 & in this case the other two roots would be complex no.s(conjugate to each other). But, if we say that f(x) has 2 real roots & 1 complex root, it is not possible. Because, 1 complex root can't be paired with its conjugate. Or all 3 roots will be real.Similarly, if n=5, f(x) will have 1 real, 4 complex roots or 3 real, 2 complex roots or all 5 roots will be real and so on. And a polynomial with degree 1(here, n=1) has only one 1 real root.So, an odd degree polynomial equation can have at least one real root. Also, it can have more than one real root but in such a way that the remaining roots(complex) can be paired. Or all of the roots would be real numbers.Aliter :Suppose [math]f(x)[/math] is an odd degree polynomial in [math]x,[/math] such that[math]f(x)=a_0x^n+a_1x^{n-1}+a_2x^{n-2}+…+a_{n-1}x+a_n;[/math] where, [math]a_0(\ne 0),a_1,a_2,…,a_{n-1},a_n \in \mathbb{R} \, \& \, n[/math] is an odd non-negative integer.Clearly,[math]\displaystyle\lim_{x \to +\infty} f(x)=+\infty, f(0)=a_n \, \& \, \displaystyle\lim_{x \to -\infty} f(x)=-\infty.[/math]If [math]a_n[/math] is positive, then [math]f(x)=0[/math] has a root lying in the open interval [math](-\infty,0),[/math] and if [math]a_n[/math] is negative, then [math]f(x)=0[/math] has a root lying in the open interval [math](0,\infty).[/math]Thus, we can conclude that,Every polynomial equation of an odd degree has at least one real root whose sign is opposite to that of its last term.
Is it possible for a 3nd degree polynomial with rational coefficients to have no real zeros?
No. Complex zeros always come in complex conjugate pairs. There's always an even number of complex roots. Since 3 is odd there must at least one real zero.
How do you write a polynomial function of the least degree that has rational coefficients, a leading coefficient of 1, and given zeros 3 and 4 + I?
if 4 + i is a zero then the conjugate of 4 - i is also a zero.This makes this a 4th degree polynomial with factors (x - 1)(x - 3)(x - 4 - i)(x - 4 + i)(x^2 - 4x + 3) (x^2 + 16) = x^4 - 4x^3 + 19x^2 - 64x + 48
How do you write a polynomial function of f of the least degree that has a leading coefficient of 1 and the given zeros of -2, 3, and 6?
I can prove to you that if you have a polynomial with three roots, it is either f(x)=0 or has a degree of at least three.Since there is the added constraint that the leading coefficient be 1, that excludes f(x)=0.Since you have the roots, you can easily construct a polynomial of degree three with those roots: f(x)=(x+2)(x-3)(x-6). You should be able to clearly see that f(x)=0 at those roots.With the constraint that the leading coefficient be 1, there are no other cubic polynomials that meet those constraints, so that is the only way to write it. That polynomial times any other nonconstant polynomial with a leading coefficient of 1 will also meet those requirements, but its degree will be higher than three.
If p(x) is the polynomial of degree 4 with leading coefficient as three such that P(1) = 2,p(2) = 8, P(3) = 18 P(4) = 32, then how do you find the value of P(5)?
Let P(x) = 3x^4 + ax^3 + bx^2 + cx + d. Putting x= 1,2,3 and 4 and equating with the given values of P(1), P(2), P(3) & P(4) respectively you will get 4 equations asa+b+c+d = - 1 ..... (1)8a +4b+2c+d = - 40 ...... (2)27a+9b+3c+d = - 225 ...... (3) and64a+16b+4c+d = -736 ..... (4)Solving these equations you may get (please check)a=-17, b=-10, c= 188 and d= -162 Put these values in P(5) i.e. 1875+125a+25b+5c+d to get the answer.
If f(x) is a 4th degree polynomial, and has 3 complex numbers as roots, what is the other root?
You have a quartic equation,[math]x^4+ax^3+bx^2+cx + d=0[/math],and you know three of its roots, [math]x _ 1[/math], [math]x _ 2[/math], and [math]x _ 3[/math], and want to find its fourth root, [math]x _ 4[/math].Rewrite the quartic equation as[math](x-x _ 1)(x-x _ 2)(x-x _ 3)(x -x _ 4)=0[/math].Equate the two versions of the quartic equation, equate the coefficients of [math]x[/math] raised to the four different powers, and find [math]x _ 4[/math].