Questions on Algebra: Functions, Domain, NOT graphing answered by real tutors!

You can put this solution on YOUR website!
The general polynomial with these roots is:


let a=1, then






:)

You can put this solution on YOUR website!
Form a polynomial f(x) with real coefficients having the given degree and zeros.
Degree 5; zeros: 9, -i, -8+i
-----
f(x) = (x-9)(x+i)(x-i)(x+8 - i)(x+8 +i)
----
= (x-9)(x^2+1)((x+8)^2+1)
----
= (x-9)(x^2+1)(x^2 + 16x + 65)
------------
etc.
Cheers,
Stan h.
================

You can put this solution on YOUR website!
Solve the equation in the real number system.
X^4-3x^3+12x^2-48x-64=0
***
use rational roots theorem:
...0..|....1.....-3......12....-48.....-64
...1..|....1.....-2......10....-38....-102
...2..|....1.... -1......10....-38....-140
...3..|....1.......0......12....-12...-100
...4..|....1......1.......16.....16.......0 (4 is a root)
====================
...0...|....1......1.......16.....16
..-1..|....1.....0........16......0 (-1 is a root)
f(x)=(x-4)(x+1)(x^2+16x)
roots are:-1, 4,and 2 non-real roots

You can put this solution on YOUR website!
Find functions f and g such that the given function h can be expressed in the form h=fog
h=(x-3)^2 +x-3
----
g(x) = (x-3)^2+x
f(x) = x-3
========
Then h(x) = f[g(x)] = f[(x-3)^2+x] = (x-3)^2+x-3
==============
Cheers,
Stan H.
==============

You can put this solution on YOUR website!
F(x)=3x^6-27x^4+x^2-7
the divisor is x+3 so the zero of the divisor=-3
now F(-3)=3(-3)^6-27(-3)^4+(-3)^2-7=3*729-27*81+9-7=2
so the Remainder is 2

You can put this solution on YOUR website!
Just some help but no further:

The needed set of zeros is
{ 9, -i, +i, -8+i, -8-i }

The factored form is
You have some multiplications and simplifications to do. Avoid extra difficulties through careful pre-simplification and associating:


Continue the multiplications and simplifications.

You can put this solution on YOUR website!
Degree of 2, possible two zeros, and you must expect the conjugate of your given zero is also a zero. Your TWO zeros are 2+2i AND 2-2i.

Equation is


THE MULTIPLICATION GIVES DIFFERENCE OF TWO SQUARES








EDITED: Corrected mistake.

You can put this solution on YOUR website!

F(x)=x^3-8x^2+29x-52

The possible rational zeros are ± the factors of 52,
which are ±1,±2,±4,±13,±26, and ±52

You can keep trying those and the only one that gives
a zero remainder and is therefore a rational zero is 4.

4|1 -8  29 -52
 |   4 -16  52
  1 -4  13   0

So we have factored F(x) this way:

F(x) = (x-4)(x²-4x+13)

To find the other zeros we set the factor
x² - 4x + 13 = 0 and use the quadratic formula:


x = 

x =  

x = 

x = 

x = 

x = 

x = 

x = 2 ± 3i

So the other zeros are 2 + 3i and 2 - 3i

So the factored form of F(x) is

F(x) = (x - 4)[x - (2 + 3i)][x - (2 - 3i)]

or 

F(x) = (x - 4)(x - 2 - 3i)(x - 2 + 3i)

Edwin

You can put this solution on YOUR website!


Refer to this:

You can put this solution on YOUR website!
Use the rational zero theorem to list all possible rational zeros for the given function
f(x)=x^3-4x^2-19x-14
-----
I graphed it and found a zero at x = -1.
Use synthetic division to find the others.
------
-1)....1....-4....-19....-14
.......1....-5....14....|..0
----
Factor the Quotient which is x^2-5x+14
----
= (x-7)(x+2)
------
Zeros are x = -1, -2, 7
==========================
Cheers,
Stan H.
===================