Yea i have another polynomial problem..I tried it 3 times sigh....
Solve the equation $\displaystyle x^{4}+2x^{3}-21x^{2}-22x+40=0$ given that the roots are in an arithmetic progression. Let the roots be a-3d, a-d, a+d, a+3d
Thanks a lot
Yea i have another polynomial problem..I tried it 3 times sigh....
Solve the equation $\displaystyle x^{4}+2x^{3}-21x^{2}-22x+40=0$ given that the roots are in an arithmetic progression. Let the roots be a-3d, a-d, a+d, a+3d
Thanks a lot
Use Cardano Vieta relations.
Fernando Revilla
You know that $\displaystyle \displaystyle a-3d, a-d, a+d, a+3d$ are all roots of the polynomial $\displaystyle \displaystyle P(x) = x^4 + 2x^3 - 21x^2 - 22x + 40$.
So $\displaystyle \displaystyle P(a-3d) = 0, P(a - d) = 0, P(a + d) = 0, P(a + 3d) = 0$.
Substitute the values and you will end up with a system of equations to solve for $\displaystyle \displaystyle a$ and $\displaystyle \displaystyle d$.
Those are the relations I was talking about:
$\displaystyle x_1+x_2+x_3+x_4=4a=\dfrac{-2}{1}=-2$
so, $\displaystyle a=-1/2$
Could you continue?
Fernando Revilla
Hello, aonin!
$\displaystyle \text{Solve the equation: }\:x^4+2x^3-21x^2-22x+40=0$
$\displaystyle \text{ given that the roots are in an arithmetic progression.}$
$\displaystyle \text{Let the roots be: }\: a-3d,\;a-d,\;a+d,\;a+3d$ . . . . but why?
The terms would be: .$\displaystyle a-2d,\;a-d,\;a+d,\;a+2d$
Why force the A.P. to go up by twice some number?
And do they really expect us to plug those four expressions into the polynomial?
Using the Factor Theorem: .$\displaystyle (x+5)(x+2)(x-1)(x-4) \:=\:0$
Therefore, the roots are: .$\displaystyle x \;=\;\text{-}5,\:\text{-}2,\:1,\:4$
Certainly the equation can be easily solved using Ruffini's rule. However the sense of the problem is surely to use the fact that the roots are in arithmetic progression. An this, in order to see if the student knows the relations between roots an coefficients. In this case the relations:
$\displaystyle \begin{Bmatrix} x_1+x_2+x_3+x_4=-2\\x_1x_2x_3x_4=40\end{matrix}$
lead to $\displaystyle a=-1/2$ and to the biquadratic equation:
$\displaystyle \left(\dfrac{1}{4}-9d^2\right)\left(\dfrac{1}{4}-d^2\right)=40$
Of course it is not necessary to write the difference of the progression as $\displaystyle 2d$ , but it is correct.
Fernando Revilla