help poly

I would write:

\(\displaystyle f(x)=(x-2)(x-4)(x-6)(x-8)+15=x^4+ax^3+bx^2+cx+399=0\)

I would then look at the prime factorization of 399.

\(\displaystyle 399=3\cdot7\cdot19\)

We then find:

\(\displaystyle f(3)=0,\,f(7)=0\)

Hence:

\(\displaystyle f(x)=(x-3)(x-7)(x^2+dx+19)=(x^2-10x+21)(x^2+dx+19)=\)

\(\displaystyle x^4+ax^3+bx^2+(21d-190)x+399=x^4+ax^3+bx^2-((-2-4)(-6)(-8)+(-6-8)(-2)(-4))x+399\)

\(\displaystyle x^4+ax^3+bx^2+(21d-190)x+399=x^4+ax^3+bx^2-300x+399\)

Equating linear coefficients, we find:

\(\displaystyle 21d-190=-300\)

\(\displaystyle d=-10\)

and so:

\(\displaystyle f(x)=(x-3)(x-7)(x^2-10x+19)=0\)

Now you may more easily find the roots.
 
ibanez1608 said:
(x-2)(x-4)(x-6)(x-8)=-15

should i open?
how do i start?
Click to expand...

ibanez1608,

I'm going to show a method for a benefit to certain other users. I expect this is over your head at this point.
This may come in handy for you some day if it clicks.

I think this is more complicated and many more steps than MarkFL's, but it cuts down on the factoring
by having a smaller number to factor/break apart.

Because the factors on the left-hand side are consecutive factors which differ by the same amount,
there can be a shift made with some other variable, say t, and rational numbers, a and b,
where b = a + 2 so that there will be symmetry and eliminate both first-degree and third degree terms.


Let's get it to look like the form:


(t + b)(t + a)(t - a)(t - b) = -15

toward a goal of

\(\displaystyle (t^2 - b^2)(t^2 - a^2) = -15\)



To convert the four binomial factors in the original equation, we could set


\(\displaystyle t + \ b = x - 2\) \(\displaystyle \ \ \ \ \ \)and
\(\displaystyle t - \ b = x - 8\)
--------------



Solve for b:

(I'll do it by elimination.)



\(\displaystyle \ \ \ \ \ \ \ t + b \ = \ \ \ \ \ \ \ x - 2\)
\(\displaystyle -1(t - b) = -1(x - 8)\)
----------------------



\(\displaystyle \ \ \)t + b = \(\displaystyle \ \) x - 2
-t + b = -x + 8
---------------

2b = 6

b = 3


So that means we let x - 2 = t + 3 *

Then x - 4 = t + 1

and x - 6 = t - 1

and x - 8 = t - 3


*From this, x = t + 5


Substitute all of this into the original equation:


(t + 3)(t + 1)(t - 1)(t - 3) = -15 ===>

\(\displaystyle (t^2 - 9)(t^2 - 1) = -15 \ \) ===>

\(\displaystyle t^4 - t^2 - 9t^2 + 9 = -15 \ \) ===>

\(\displaystyle t^4 - 10t^2 + 24 = 0 \ \) ===>

\(\displaystyle (t^2 - 4)(t^2 - 6) = 0 \ \) ===>

\(\displaystyle t^2 - 4 = 0 \ \ \ or \ \ \ t^2 - 6 = 0 \ \) ===>

\(\displaystyle t^2 = 4 \ \ \ or \ \ \ t^2 = 6 \ \) ===>

\(\displaystyle t = \pm \sqrt{4} \ \ \ or \ \ \ t = \pm \sqrt{6} \ \) ===>

\(\displaystyle t = \pm 2 \ \ \ or \ \ \ t = \pm \sqrt{6} \ \)



Now, substitute back using x = t + 5 to get the four solutions.
 
ibanez1608 said:
(x-2)(x-4)(x-6)(x-8)=-15

should i open?
how do i start?
Click to expand...
I believe that it is possible to simplify Mark's solution.

When dealing with polynomials, it is frequently helpful to narrow the range of possible solutions.

So first I would observe that if x is real, it must be greater than two. Why? And secondly I would observe that if x is real, it must be less than 8. Why?

OK we now know that if x is real 2 < x < 8.

Now that I have some general idea what I am looking for, I'd do what Mark did.

\(\displaystyle (x-2)(x-4)(x-6)(x-8)=-15 \implies f(x) = (x-2)(x-4)(x-6)(x-8) + 15 = 0 .\)

Now by inspection f(x) is a quartic polynomial with integer coefficients, which means that the rational root theorem may apply. To apply the rational root theorem I need to know only the coefficient of the highest power and the constant term. The constant term is
(-2)(-4)(-6)(-8) + 15 = 384 + 15 = 399. The coefficient of the highest power is obviously 1 so I am dealing with the integer root theorem, which says that if there is a rational root, it will be 1, a prime factor of 399, or one of their additive inverses. But the additive inverses and 1 can be ignored because we already know that the answer (if it exists) must be between 2 and 8. And 399 has only 3 prime factors, namely 3, 7, and 19. But 19 can be ignored as a solution because 19 > 8. See how handy it can be to know where to look.

Now go back to the original equation and try 3 and 7.

(3 - 2)(3 - 4)(3 - 6)(3 - 8) = 1 * (-1) * (- 3) * (- 5) = - 15. So 3 is a solution.

(7 - 2)(7 - 4)(7 - 6)(7 - 8) = 5 * 3 * 1 * (-1) = - 15. So 7 is a solution.

Now I'd expand:

\(\displaystyle f(x) = (x-2)(x-4)(x-6)(x-8) + 15 = (x^2 - 6x + 8)(x^2 - 14x + 48) + 15 =\)

\(\displaystyle x^4 + (-14 - 6)x^3 + (48 + 8)x^2 + (-6 * 48 - -14 * 8)x + 8 * 48 + 15 =\)

\(\displaystyle x^4 - 20x^3 + 56x^2 - 400x + 399 = 0.\) Notice that the coefficient of \(\displaystyle x^3 = -20.\)

Now Mark uses the fundamental theorem of algebra and so would I.

\(\displaystyle f(x) = (x - 3)(x - 7)(px^2 + qx + r) = (x^2 - 10x + 21)(px^2 + qx + r) = 0.\)

But we already know that the coefficient of \(\displaystyle x^4 = 1\ and\ 21r = 399 \implies p = 1\ and\ r = \dfrac{399}{21} = 19.\)

\(\displaystyle So f(x) = (x^2 - 10x + 21)(x^2 + qx + 19) = x^4 + (q - 10)x^3 + z.\)

So \(\displaystyle q - 10 = - 20 \implies q = -10.\)

Now using the zero product property I know that the real solutions are 3, 7 and the real solutions (if there are any) such that

\(\displaystyle 2 < x < 8\ and\ x^2 - 10x + 19 = 0.\)

\(\displaystyle x^2 - 10x + 19 = 0 \implies x = \dfrac{10 \pm \sqrt{100 - 4 * 1 * 19}}{2} = \dfrac{10 \pm \sqrt{6 * 4}}{2} = 5 \pm \sqrt{6}.\)

\(\displaystyle 0 < \sqrt{6} < 3 \implies - 3 < - \sqrt{6} < 0 < \sqrt{6} < 3 \implies 2 < 5 - \sqrt{6} < < 5 < 5 + \sqrt{6} < 8.\)

Now CHECK

\(\displaystyle (5 - \sqrt{6} - 2)(5 - \sqrt{6} - 4)(5 - \sqrt{6} - 6)(5 - \sqrt{6} - 8) = (3 - \sqrt{6})(1 - \sqrt{6})(-1 - \sqrt{6})(-3 - \sqrt{6}) =\)

\(\displaystyle (3 - \sqrt{6})(- 3 - \sqrt{6})(1 - \sqrt{6})(-1 - \sqrt{6}) = (- 9 + 6)(-1 + 6) = - 3 * 5 = - 15.\) That checks.

\(\displaystyle (5 + \sqrt{6} - 2)(5 + \sqrt{6} - 4)(5 + \sqrt{6} - 6)(5 + \sqrt{6} - 8) = (3 + \sqrt{6})(1 + \sqrt{6})(-1 + \sqrt{6})(-3 + \sqrt{6}) =\)

\(\displaystyle (3 + \sqrt{6})(- 3 + \sqrt{6})(1 + \sqrt{6})(-1 + \sqrt{6}) = (- 9 + 6)(-1 + 6) = - 3 * 5 = - 15.\) That checks.

This is not an easy problem. If you are given a similar problem on a test, you will need to know how to solve it efficiently. I find my method of solving it to be a bit simpler than Mark's, but it really is using his ideas plus the idea of putting some bounds on the possible answers before starting. The bounds let you reduce the number of possibilities that need to be tested.
 
Last edited:
Denis said:
Plus x <> 2 or 4 or 6 or 8 .....
Click to expand...

Nope!!

2 < x < 8 means that \(\displaystyle x \ne 2\) and \(\displaystyle x \ne 8\)

only new information needs to be added is \(\displaystyle x \ne 4\) and \(\displaystyle x \ne 6\)

To the corner....
 
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